Haritha

Equilibrium

You have learnt about several aspects of a chemical reaction so far. You know, for example, that certain proportions of reactants react to form particular proportions of products and that these proportions are specified in a chemical equation. In the previous chapter, you learnt about the energy changes accompanying a chemical reaction. But there is one aspect of a chemical reaction that we have not considered at all.

Suppose we take reactants in exactly the required proportions (in the ratio of their respective number of moles in the balanced equation) and mix them together. Is it necessary that the reaction will proceed to completion? In other words, is it necessary that all of the reactants will be converted into products?

• We assume this in stoichiometric calculations (chemical arithmetic), but this is not always true. Quite often, a chemical reaction stops after a while and the resultant mixture contains reactants and products. It is said that the reaction has reached a state of chemical equilibrium.
• This state remains unchanged and the composition of the mixture remains the same, unless something forces a change. This something could be a change in temperature, pressure, or concentration of the reactants.
• It could even be a change in the concentration of the products, which could be achieved by removing the products (already formed) from the reaction site. These considerations, i.e., how to change the state of chemical equilibrium, are very important for industrial chemists. You can well imagine that the productivity of an industrial process may depend on these factors.
• When a reaction system reaches a chemical equilibrium, all the properties of the system, i.e., temperature, pressure, concentration of reactants and concentration of products, remain constant over time. What actually happens in such a system is that two opposing processes take place at the same time. Just as the reactants react to form the products, the products combine to produce the reactants.

And there comes a time when the forward process and the reverse process occur in such a way that the reaction seems to come to a standstill. At equilibrium, the rates at which the two opposing processes occur become equal.

• A double arrow (\(\rightleftharpoons\)) is used to denote an equilibrium, whereas a single arrow (-») is used to denote a reaction which proceeds to completion. The extent to which the chemical reactions proceed may be different. Some reactions proceed nearly to completion and only a negligible concentration of the reactants is left.
• There are reactions in which the concentrations of reactants and products are comparable at equilibrium. Lastly, there are a few reactions in which most of the reactants remain unchanged at equilibrium, i.e., small amounts of products are formed.
• Our discussion so far has been centred around chemical reactions, so you may be tempted to believe that a state of equilibrium can be reached only in chemical processes. But this is not true.
• A state of equilibrium may be reached in a physical process as well. When the forward process and the reverse process occur simultaneously at exactly the same rate, a state of equilibrium is reached.

Equilibrium In Physical Processes

You have already come across physical processes which reach a state of equilibrium. A liquid in equilibrium with its own vapour inside a closed vessel is a system in which the processes of evaporation and condensation proceed at the same rate. The examples of physical equilibrium are solid \(\rightleftharpoons\)liquid, liquid \(\rightleftharpoons\) gas, solid \(\rightleftharpoons\) solution and gas solution.

Solid-liquid equilibrium: If you drop a few ice cubes into a glass of water, all the ice changes into water after some time. This physical process or change proceeds in a particular direction until it is complete, i.e., until all the ice melts into water. Now suppose you pour a mixture of ice cubes and water at 273 K and normal atmospheric pressure into a perfectly insulated flask.

• The system, comprising the mixture of ice and water in the flask, will not exchange heat with the surroundings and a state of equilibrium will be reached, in which neither the temperature, the pressure, nor the composition of the mixture will change with time.
• It is not as though nothing will happen inside the flask, but there will be no way of telling that something is happening since there will be no outward indication.
• If you could become really small and enter the world of the molecules of water and ice, you would be able to see that what really happens is that molecules of ice go into the liquid state and molecules of water collide and stick to ice, constantly.

The two processes go on simultaneously and at the same rate. Consequently, neither the mass of ice nor the mass of water changes.

To put this in terms of change in free energy, when an ice-water system is in equilibrium at 273 K and at 1 atm pressure, ΔG = 0. For the process, ice \(\rightleftharpoons\) water

ΔG < 0 when T > 273 K, and

ΔG > 0 when T < 273 K.

• The system comprising ice and water at 1 atm pressure can be at equilibrium only at 273 K. The solid and liquid phases of any (pure) substance at 1 atm pressure can be in a state of equilibrium only at a particular temperature.
• This temperature at which the solid and liquid phases of a pure substance are at equilibrium at a pressure of 1 atm is called the normal melting point or the normal freezing point of the substance.

The same definition but at a pressure of 1 bar refers to standard freezing point of a substance. The equilibrium between the solid and liquid phases of a substance at its melting point is a dynamic equilibrium, in which the activity in one direction is balanced by the activity in the reverse direction. Let us recall the characteristics of a system in dynamic equilibrium.

1. The forward and reverse changes (processes) occur simultaneously and at the same rate, so that there is no change of mass of either side of the equilibrium.
2.  ΔG = 0

Liquid-vapour equilibrium: You have already come across this kind of equilibrium while studying vapour pressure. It can be demonstrated easily by performing a simple experiment. Take a small quantity of water at room temperature in a closed vessel connected to a manometer.

• Take care to see that the vessel has been evacuated first. To begin with, the level of mercury in both limbs of the manometer will be the ⇒ same. Then the level of mercury in the right limb of the manometer will rise slowly and the level of mercury in the left limb will fall.
• The level of water in the vessel falls gradually. There comes a time when the levels of mercury in tire two limbs become constant and so does the level of water in the vessel.

• Initially, there is no water vapour in the vessel and the level of mercury in both limbs is the same. As the water starts evaporating and more and more molecules escape into the gaseous phase, the pressure exerted by the water vapour increases, leading to a drop in the mercury level in the left limb and a rise in the right limb.
• But with the accumulation of vapour molecules in the vessel, the reverse process (condensation) starts. Some vapour molecules revert to the liquid phase. Condensation occurs at a lower rate than evaporation, to start with. But after a while, the two rates become equal and a state of equilibrium is established.

The pressure exerted by the vapour molecules becomes constant, as shown by the steady level of mercury in the manometer. The level of water in the vessel becomes constant because there is no net evaporation. The number of molecules entering the gas phase at any time is the same as the number of molecules entering the liquid phase.

∴ \(\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

The pressure exerted by the water vapour at equilibrium is called the equilibrium vapour pressure or simply the vapour pressure of water at the given temperature. In general, the pressure exerted by the vapour of a liquid in equilibrium with the liquid at a particular temperature is called the vapour pressure of the liquid at that temperature.

• Vapour pressure changes with temperature but is independent of the volume of the liquid. It does not depend on the volume of the container either.
• Different liquids have different vapour pressures at the same temperature and the liquid which has a higher vapour pressure is more volatile, or boils at a lower temperature. For instance, if the vapour pressure of liquid A is 3.2 kPa and that of another liquid B is 10.3 kPa at the same temperature, B is more volatile than A, or B has a lower boiling point than A.
• This should not be difficult to understand if you bear in mind that the vapour pressure of a liquid rises with temperature and that its boiling point is that temperature at which its vapour pressure becomes equal to the atmospheric pressure.
• Consider the two liquids A and B. The vapour pressure of B is higher than that of A at a particular temperature. As the temperature rises, the vapour pressures of both the liquids will rise.

Obviously, the vapour pressure of B will become equal to the atmospheric pressure at a temperature lower than that at which the vapour pressure of A will become equal to the atmospheric pressure. Consequently, the boiling point of B will be lower than that of A.

Solid-vapour equilibrium: In case of sublimable solids, the particles may escape occasionally into the vapour phase and establish vapour pressure.

• The particles which are at the surface of the solid and which have, at any point of time, higher average kinetic energy than the others, may escape into the vapour phase. If the solid is kept in a closed cylinder, particles are not able to escape and add to the vapour phase.
• Eventually, a point is reached when the particles from the vapour phase start returning to the solid phase. As in the case of liquid-vapour equilibrium initially, this reverse process is slow but after some time, an equilibrium is reached whereby rate of return of the particles to the solid phase becomes equal to the rate of escape.
• A dynamic equilibrium is thus established. The vapour pressure now corresponds to the equilibrium vapour pressure, which is characteristic of the solid.
• The tendency of the particles of the solid to escape depends on the intermolecular forces. Tire larger the intermolecular interaction, the less will be the vapour pressure of that solid.

Temperature also has a considerable effect on the vapour pressure of solids. The higher the temperature, the more energetic are the particles and the higher is the vapour pressure.

Iodine is a volatile solid. When solid iodine is placed in a closed vessel and the vessel is heated, the region above the solid becomes violet due to iodine vapours. The equilibrium is represented as \(\mathrm{I}_2 \text { (solid) } \rightleftharpoons \mathrm{I}_2 \text { (vapour). }\)

Another example is dry ice (solid CO₂); it sublimes directly from solid to gas at atmospheric pressure.

• So far we have been considering closed systems. Systems we encounter in our everyday lives are (generally) not closed. The atmosphere, into which water vapour from lakes and rivers and other water bodies escapes, is an open system.
• Though the rate of evaporation of a liquid at a particular temperature in an open system is the same as that in a closed system, equilibrium can never be reached in an open system.
• This is because the rate of condensation can never be equal to the rate of evaporation in an open system, in which the vapour molecules get dispersed in a large volume.
• The water molecules which escape into the atmosphere, for instance, do not remain confined to a small space, they, get dispersed by the wind.
• How much water vapour is there in the air (what we refer to as humidity) depends on several factors, such as the presence of water bodies in the area, the temperature and the velocity of the wind.
• At places close to the sea, especially when the wind velocity is not high, the accumulation of water vapour in the air leads to acute discomfort.

Solid-solution equilibrium: It is true that sugar dissolves in water. But you cannot dissolve an indefinite amount of sugar in a particular amount of water. You can, of course, dissolve more sugar in a particular amount of water if you heat the solution.

• This is what they do in sweet shops when they prepare sugar syrups. But when the syrup cools, crystals of sugar separate from the solution because when the solution (syrup) cools, the solubility of the solute (sugar) in the solvent (water) decreases.
• You may remember from your science courses in junior classes that a solution in which more solute cannot be dissolved is called a saturated solution.
• The amount of a solute that has to be dissolved in a certain amount of a solvent to prepare a saturated solution at a particular temperature is called the solubility of the solute in that solvent at that temperature.
• Solubility is usually expressed in terms of the number of grams of a solute that dissolves in 100 mL of the solvent.
• The solubility of a solid in a liquid depends on temperature, but we will come to that later. Pressure has hardly any effect on the solubility of a solid in a liquid.

What happens when a solution becomes saturated is that a dynamic equilibrium is established between the molecules of the solute in the solid state and the molecules of the solute in the solution.

Solute (in solution) \(\rightleftharpoons solute\) (solid)

• The process by which molecules of the solid solute go into solution is called dissolution. The reverse process by which molecules of the solute in the solution revert to the solid state is called precipitation.
• When a solute is added to a solvent, there are no molecules of the solute in the solvent to begin with. Slowly, as the solute dissolves in the solvent, the reverse process begins.

At first, the process of dissolution proceeds at a faster rate than the process of precipitation. Than there comes a point when the two processes proceed at the same rate. This happens when equilibrium is reached.

Rate of dissolution = rate of precipitation

It is not as though activity comes to a standstill in the solution, but the number of molecules of the solute going into the solution (in a certain time) becomes the same as the number of molecules returning to the solid state.

• This can be demonstrated with the help of an experiment. Prepare a saturated sugar solution in a beaker in such a way that some undissolved sugar remains at the bottom of the beaker.
• If you add some sugar-containing radioactive carbon to the solution, in a while the solution and the undissolved sugar contains sugar which has radioactive carbon.
• This shows that though the solution is saturated, the process of dissolution continues and that the rate of dissolution must be equal to the rate of precipitation, since the amount of sugar left undissolved remains the same.
• In other words, the experiment demonstrates the dynamic nature of the equilibrium reached.

Gas-solution equilibrium: You must have heard people referring to drinks like Pepsi or Mirinda as carbonated or aerated water.

• Well, strictly speaking, such drinks also contain sugar and other additives and it would be more correct to refer to soda as aerated water.
• Nonetheless, it is true that drinks like Pepsi, too, have carbon dioxide dissolved under pressure. It is this carbon dioxide which escapes with a hissing sound when a bottle of coke or soda is opened.
• The fizz that you enjoy in such drinks is also because of the dissolved carbon dioxide.
• Aerated drinks are manufactured by dissolving carbon dioxide in them at a high pressure. Unlike solids, the solubility of gases (in liquids) increases with pressure.

The effect of pressure on the solubility of a gas in a liquid is given by Henry’s law (William Henry was a British physician and chemist), which says that the mass of a gas dissolved in a given mass of a solvent at a particular temperature is proportional to the pressure of the gas above the solvent.

• The solubility of a gas in a liquid decreases as temperature increases. When an aerated drink is bottled there is a dynamic equilibrium between the gas molecules above the solution (tire drink) and the gas molecules in the solution.
• This equilibrium holds as long as the pressure is maintained. When the bottle is opened, the pressure above the solution falls and some of the dissolved gas escapes to reach a new equilibrium.
• If an aerated drink is left open to the air for some time, it turns ‘flat’ because most of the dissolved gas escapes as the pressure above the drink (solution) becomes equal to the atmospheric pressure.

Henry’s law can be expressed mathematically as follows.

∴ m∝ p or m = kp,

where m is the mass of the gas dissolved in the solution, p is the pressure of the gas above the solution and k is a constant of proportionality known as the Henry constant.

• Let us look at this another way. The mass of the gas dissolved in the solution will be related to the concentration of the gas in the solution.
• The pressure exerted by the gas above the solution will depend on the concentration of the gas above the solution. This gives us another way of expressing Henry’s law.
• Concentration of gas in solution x concentration of gas above solution

or \(\frac{\text { concentration of gas in solution }}{\text { concentration of gas in gaseous phase }}=\text { constant. }\)

Example: Suppose 0.30 g of iodine is stirred in 100 mL of water at 288 K until equilibrium is reached. Hoiv much iodine would dissolve in the water, if the solubility of iodine in water at 288 K is 0.0011 mol L^-1 [Another way of expressing this is I₂ (aq) at equilibrium = 0.0011 mol L^-1 at 288 K]. Suppose another 100 mL of water is added to the solution after equilibrium is reached with 0.30 g of iodine and 100 mL of water. How much iodine would be left undissolved and what would be the concentration of iodine in the solution?
Solution:

The solubility of iodine (concentration at equilibrium) is given as 0.0011 mol L^-1 at 288 K.

This means at equilibrium mass of iodine dissolved in 1 L of water = 0.0011 x 254 ≅ 0.279 g s 0.28 g [molar mass of I₂ = 254 g]

∴ mass of iodine dissolved in 100 mL of water = 0.028 g.

If another 100 mL of water is added after equilibrium has been established, another 0.028 g of iodine would dissolve.

∴ mass of undissolved iodine – 0.30 – (0.028 x 2) = 0.30 – 0.056 = 0.244 g.

The concentration of iodine in the solution would simply be the solubility of iodine at equilibrium i. i.e., 0.28 gL^-1.

Characteristics of equilibrium in physical processes: We have discussed the attainment of an equilibrium state in relation to four different physical processes. They all have certain characteristics in common.

1. An equilibrium can be reached only in a dosed system, i.e., a system which neither gains matter from nor loses matter to the surroundings. For instance, in the case of a liquid—gas system, equilibrium is not reached if the vapour is allowed to escape from the vessel.
2. The equilibrium reached is dynamic in nature, i.e., there is a balance between two opposing processes occurring at the same rate.
3. At equilibrium, the measurable properties of the system become constant (under a given set of conditions). For instance, in the case of evaporation, the vapour pressure of the liquid is constant at a given temperature.
   • In solid-liquid equilibrium, the melting point (the temperature at which the two phases coexist provided no heat is exchanged) is constant for a particular pressure.
   • For the dissolution of a solid in a liquid, the solubility is constant at a given temperature. For the dissolution of a gas in a liquid, the concentration of the gas in the solution is constant for a particular pressure of the gas above the solution at a given temperature.
4. At equilibrium, the concentrations of substances in the liquid and gaseous phase bear a constant ratio at a particular temperature. For the dissolution of C02 in water, for example, \(\frac{\mathrm{CO}_2(\mathrm{aq})}{\mathrm{CO}_2(\mathrm{~g})}\) is constant at a given temperature. This constant is called the equilibrium constant and its magnitude indicates the extent to which the process proceeds before equilibrium is reached. For example, the greater the value of \(\frac{\mathrm{CO}_2(\mathrm{aq})}{\mathrm{CO}_2(\mathrm{~g})}\) greater is the extent to which CO₂ disiolves in water.

Equilibrium In Chemical Systems

We started this chapter with a discussion on whether a chemical reaction must always proceed to completion. There are some chemical reactions which proceed to completion, i.e., at equilibrium the concentration of the, reactants is negligible compared to the concentration of the products. (In reality, no reaction proceeds to completion in the true sense, but we need not concern ourselves with this now.) An example of this type of reaction is

⇒ \(\mathrm{BaCl}_2(\mathrm{aq})+\mathrm{Na}_2 \mathrm{SO}_4(\mathrm{aq}) \longrightarrow \mathrm{BaSO}_4(\mathrm{~s})+2 \mathrm{NaCl}(\mathrm{aq})\)

We could also call this reaction an irreversible reaction, since the reaction proceeds in one direction and the reverse process, i.e., the products reacting to form the reactants under the same conditions, does not take place. A few irreversible reactions are given below.

⇒ \(2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_2\)

⇒ \(\mathrm{AgNO}_3(\mathrm{aq})+\mathrm{NaCl}(\mathrm{aq}) \longrightarrow \mathrm{NaNO}_3(\mathrm{aq})+\mathrm{AgCl}(\mathrm{s})\)

⇒ \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_2(\mathrm{~s}) \longrightarrow 2 \mathrm{MgO}(\mathrm{s})\)

Let us now consider a reaction which is reversible, i.e., a reaction in which the reactants react to form products under certain conditions and the products combine to form the reactants under the same conditions.

For instance, iron reacts with steam to form hydrogen and iron oxide, and heated iron oxide reacts with hydrogen to produce iron and steam.

⇒ \(3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2\)

If a reversible reaction is carried out in such a way that one of the products escapes, the reaction becomes irreversible, or proceeds in only one direction.

• If, for instance, in the reaction between metallic iron and steam, the hydrogen formed is allowed to escape, the reverse reaction cannot take place and the process becomes virtually irreversible, or proceeds to completion.
• If, on the other hand, the reaction is carried out in a closed vessel, from which hydrogen cannot escape, both forward and backward reactions take place simultaneously and there comes a time when an equilibrium is established.
• At equilibrium, the reaction appears to have stopped, but what happens is that the forward and reverse reactions occur at the same rate.

For instance, when hydrogen and iodine are heated in a closed vessel at 717 K, they react to form hydrogen iodide. However, since the hydrogen iodide is not allowed to escape, the reverse reaction occurs. Hydrogen iodide decomposes to form hydrogen and iodine.

⇒ \(\underset{\text { colourless }}{\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})} \rightleftharpoons \underset{\text { viclet }}{2 \mathrm{HI}(\mathrm{g})}\)

Initially, the reaction mixture is deep violet because of the presence of iodine.

• The intensity of the colour decreases, as the reaction proceeds, and after a while, it becomes steady, indicating that an equilibrium has been reached.
• At equilibrium, it appears that the reaction has stopped, despite the presence of the reactants.
• But in reality, the rate at which hydrogen and iodine combine to form hydrogen iodide becomes equal to the rate at which hydrogen iodide decomposes.
• When we talk about the time it takes for the colour of the
  reaction mixture that we have just been discussing to become
  steady, we are talking about a finite time interval.
• However, to the naked eye, the change seems to be instantaneous. It is possible, though, to determine this time interval with modem instruments.
• It is also possible to establish the dynamic nature of the equilibrium using radioactive reagents.

Characteristics of chemical equilibrium

1. Consider a general reversible reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\)

• Initially, the concentrations of A and B are maximum and the concentrations of C and D are zero. As the reaction proceeds, the concentrations of A and B decrease and the concentrations of C and D increase.
• With this, the rate of the forward reaction starts decreasing, while the rate of the backward reaction starts increasing, until the two rates become equal, and equilibrium is reached.
• At this point the concentration of A and B, and C and D become constant.
• That the concentrations of the reactants and the products become constant at equilibrium can be demonstrated by connecting a manometer to a closed vessel containing CaCO₃.

If the vessel is heated, CaCO₃ decomposes to yield CaO and CO₂.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

To begin with, the pressure due to CO₂ keeps increasing. Then the level of mercury in the manometer becomes steady, indicating that the pressure has become constant. This shows that the concentration of CO₂ produced becomes constant (as those of CaCO₃ and CaO) at equilibrium.

2. In the reaction that we have just considered, the pressure exerted by CO₂ becomes constant at equilibrium. In the reaction between hydrogen and iodine, the colour of the reaction mixture becomes constant.

In general, the observable properties of the system become constant at equilibrium. Actually, this follows from the fact that the concentrations of the reactants and products become constant at equilibrium.

3. A chemical equilibrium can be attained only in a closed system. This should be obvious. If the system is open, one or more of the products may escape, making the reverse reaction impossible.

4. The rates of the forward and backwards reactions become equal at equilibrium. We have discussed this point at length already.

If this were not so, the reaction between hydrogen and iodine, for instance, would not appear to come to a standstill, despite the presence of the reactants.

The dynamic nature of chemical equilibrium can be demonstrated in the synthesis of ammonia by the Haber process.

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)

Known amounts of hydrogen and nitrogen are taken at high temperature and pressure, which react to form ammonia.

• The concentrations of all the three substances respectively are determined at regular intervals and a graph of the molar concentrations against time is plotted.
• The graph shows that initially the concentration of the reactants, i.c„ hydrogen and nitrogen decreases whereas that of the product ammonia increases.
• That equilibrium is reached is depicted in the graph when, after a certain time, the composition of the reaction mixture is constant.
• In order to prove the dynamic nature of equilibrium, the experiment is performed again in the same set of conditions except for the use of deuterium in place of hydrogen.
• The reaction mixture in this case, at equilibrium, has the same composition as that in the prior one except for the presence of deuterium in place of hydrogen (i.e., D₂, N₂ and ND₃).
• Once the equilibrium is attained in both cases, the two reaction mixtures are mixed together and left. After some time the concentration of ammonia in this new mixture is found to be the same as before.
• However, mass spectrometer analysis revealed that all forms containing hydrogen as well as deuterium are present—NH₃, NH₂D, NHD₂, ND₃, H₂, HD, D₂—in the reaction mixture.
• This is only possible when the bonds are continuously breaking and being made at equilibrium or the forward and reverse reactions are taking place.

Had the reaction stopped then there would have been no mixing of the isotopes. This proves that the equilibrium is dynamic in nature.

5. The equilibrium can be approached from either direction. For instance, in the case of \(\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2 \rightleftharpoons 3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O}\), it would make no difference whether we started with iron oxide and hydrogen, or metallic iron and steam.

• The fact that equilibrium can be approached from either direction can be proved by performing a simple experiment. Dinitrogen tetroxide (N₂O₄) is a colourless gas and nitrogen dioxide (NO₂) is reddish brown in colour.
• Dinitrogen tetroxide decomposes almost completely to nitrogen dioxide at 373 K. N₂O₄ is stable at 273 K and exists almost entirely as pure N₂O₄, which is colourless.

⇒ \(\underset{\text { collourless }}{\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g})} \rightleftharpoons \underset{\text { reddish brown }}{2 \mathrm{NO}_2(\mathrm{~g})}\)

If you take two identical flasks A and B, fill them with nitrogen dioxide and keep them at room temperature, both will actually contain a pale brown mixture of N₂O₄ and NO₂.

• Now place flask A in a bath maintained at 273 K and flask B in a vessel of boiling water. The gas in flask A will turn colourless, indicating that it is mostly N₂O₄, while the colour of the gas in flask B will deepen to reddish brown, showing that it is mostly NO₂.
• If you then transfer both the flasks to a bath maintained at room temperature (298 K), the gas in flask A will turn pale brown and the reddish brown colour of flask B will fade, until both appear the same colour.
• The fact that the gaseous mixtures in both flasks attain the same colour indicates that, at equilibrium, both flasks contain a mixture of N₂O₄ and NO₂ in the same proportion, which means that equilibrium can be approached from either direction.

The fact is also demonstrated by Figure, which graphically represents the reaction \(\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}\) we have considered earlier in the chapter.

Whether we start the reaction from H₂ and I₂ or from HI, the composition of the reaction mixture at equilibrium is constant. The figure represents the forward and the backward reaction in the same graph.

• When we start the reaction with an equimolar mixture of H₂ and I₂, as you can see in Figure, the concentration of these two decreases (curve a) while that of HI increases (curve b). Both of them finally reach a stable concentration at equilibrium.
• When we move from right to left in the graph, we are actually starting with HI and its concentration decreases with time (curve c) while that of H₂ and I₂ increases (curve d), both of them stabilising at the equilibrium point.

6. You know that a catalyst increases the rate of a reaction. In the case of a reversible reaction, a catalyst increases the rates of the forward and reverse reactions to the same extent.

Hence it does not alter the state of equilibrium. That is to say, the concentrations of the reactants and products at equilibrium are the same with or without the catalyst. All that happens is that the state of equilibrium is attained faster.

7. To repeat what you already know, at equilibrium, ΔG = 0.

Law Of Mass Action

We have been discussing the fact that at equilibrium, the rates of the forward and reverse reactions become equal. But can one determine these rates? In 1863 two Norwegian chemists, C. M. Guldberg and P. Waage, came up with a law on the basis of experimental observations.

• This law, known as the law of mass action, relates the rate of chemical reaction with the concentrations of the reactants. It says that the rate at which a chemical reaction occurs at a given temperature is proportional to the product of the active masses of the reactants.
• You must remember that Guldberg and Waage came up with this law in connection with ideal gases and that the law is strictly correct only for ideal gases.

The active mass of a reactant is nothing but its molar concentration, or the number of moles dissolved per litre of the solution. For example, suppose a solution of HCl has x g of HCl dissolved in y L of solution. Then the concentration of the solution = \(\frac{x}{y} \mathrm{gL}^{-1}\)

= \(\frac{x}{36.5 \times y} \mathrm{~mol} \mathrm{~L}^{-1}\)(molar mass of HCl = 36.5)

Thus the active mass of the HCl solution, represented as [HCl], is x/36.5 y M, where M stands for molar concentration or mol L^-1.

Let us consider a reaction between two reactants A and B.

A + B → products

According to the law of mass action, the rate of reaction ∝ [A] [B] or rate = k[A][B],

where [A] and [B] are the concentrations of the reactants A and B respectively, and A is a proportionality constant, called the velocity constant or rate constant.

Now consider a reaction 3C + 4D → products

you could think of this reaction as C + C + C + D + D + D + D → products then rate = k[C][C][C][D][D][D][D] = k[C]³[D]⁴.

In general, for any reaction aA + bB → products

∴ rate of reaction = k[A]^a [B]^b.

Thus, the law of mass action can also be expressed as follows.

The rate of a reaction is proportional to the product of the concentrations of the reactants, each raised to the power equal to its coefficient (number of moles of the species) in the balanced chemical equation representing the reaction.

Law Of Chemical Equilibrium

This law follows from the law of mass action. Consider the following general reversible reaction.

aA + bB \(\rightleftharpoons\)cC + dD

If we apply the law of mass action to this reaction at equilibrium, rate of forward reaction = \(k[\mathrm{~A}]_{\mathrm{eq}}^a[\mathrm{~B}]_{\mathrm{eq}}^b\),

where k is the rate constant for the forward reaction and [A]_eq and [B]_eq are the molar concentrations of the reactants A and B at equilibrium.

Rate of reverse reaction = \(k^{\prime}[C]_{\mathrm{eq}}^c[\mathrm{D}]_{\mathrm{eq}}^d\),

where k’ is the rate constant for the reverse reaction and [C]_eq and [D]_eq are the molar concentrations of the products C and D at equilibrium.

At equilibrium, rate of forward reaction = rate of reverse reaction or \(k[\mathrm{~A}]_{\mathrm{eq}}^a[\mathrm{~B}]_{\mathrm{eq}}^b=k^{\prime}[\mathrm{C}]_{\mathrm{eq}}^c[\mathrm{D}]_{\mathrm{eq}}^d\)

or, \(\frac{k}{k^{\prime}}=\frac{[\mathrm{C}]_{\mathrm{eq}}^c[\mathrm{D}]_{\mathrm{eq}}^d}{[\mathrm{~A}]_{\mathrm{eq}}^a[\mathrm{~B}]_{\mathrm{eq}}^b}=K_c\)

Generally the subscript ‘eq’ is used to denote concentration at equilibrium. The subscript ‘c’ indicates that K_c is expressed in concentrations of mol L^-1.

K_c must be a constant at a given temperature since k and Ak’ are constants for a given temperature. Guldberg and Waage called it the equilibrium constant.

Thus, at equilibrium, the product of the molar concentrations of the reaction products, each raised to the power equal to its stoichiometric coefficient in the balanced chemical equation, divided by the product of the molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient, is constant at a given temperature. This is referred to as the law of chemical equilibrium.

The equilibrium constant for the reaction \(4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text { is } K_{\mathrm{c}}=\frac{\left[\mathrm{NO}^4\left[\mathrm{H}_2 \mathrm{O}\right]^6\right.}{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}\)

The symbols used to denote phases in a chemical reaction are ignored while writing the expression for the equilibrium constant.

Reaction quotient: For any reversible reaction at any stage other than equilibrium, the ratio of the molar concentrations of the products to that of the reactants, where each concentration term is raised to the power equal to the stoichiometric coefficient of the substance concerned, is called the reaction quotient, Q.

It is symbolised as Q_c. while molar concentrations are considered and Q_p while partial pressures are considered for any reaction.

For a general reaction \(a\mathrm{~A}+b \mathrm{~B} \rightleftharpoons c \mathrm{C}+d \mathrm{D}\)

which is not at equilibrium  \(Q_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{~A}]^a[\mathrm{~B}]^b}\).

1. If Q_c > K_c, the value of Q_c will tend to decrease to reach the value of K_c (towards equilibrium), and the reaction will proceed in the reverse direction.
2. If Q_c < K_c, it will tend to increase, and the reaction will proceed in the forward direction.
3. If Q_c = K_c, the reaction is at equilibrium.

So, if you know the relative concentrations of reactants and products at a particular stage of a reaction, you can predict the direction in which the reaction will proceed.

Also, if you know the initial concentrations of the reactants and the value of K_c, you can predict the composition of the reaction system at equilibrium.

Now consider the following reaction. \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\)

Here \(Q_c=\frac{\left[\mathrm{CO}_2\right]\left[\mathrm{H}_2\right]}{[\mathrm{CO}]\left[\mathrm{H}_2 \mathrm{O}\right]}\) and Kc = 0.64 at 800°C. The initial conditions for a reaction can vary.

Here we have discussed three possibilities for the given reaction. The values taken are from different experiments. There is no need for you to memorise these values, the discussion will help you learn how to predict the direction of a reaction.

1. Only the reactants are present, say. [CO] = 0.0243 M and [H₂O] = 0.0243 M.

Since the concentration of products is zero, the reaction quotient Q_c = 0, i.e., Q_c < K_c.

The reaction thus proceeds in the forward direction till Qc reaches a value equal to K_c.

At equilibrium, the concentrations are [CO]_eq = 0.0135 M,[H₂O]_eq = 0.0135 M,[CO₂]_eq = 0.0108 M,[H₂]_eq = 0.0108 M.

∴ \(K_c=\frac{0.0108 \times 0.0108}{0.0135 \times 0.0135}=0.640\)

2. Only the products arc present, say. [CO₂] = 0.0468 M,[H₂] = 0.0468 M.

Q_c >K_c and hence the reaction proceeds in the reverse direction, forming CO and H₂O.

At equilibrium the concentrations are [CO]_eq = 0.0260 M,[H₂O]_eq = 0.0260 M,[CO₂]_eq = 0.0208 M,[H₂]_eq = 0.0208 M.

∴ \(K_2=\frac{(0.0208)(0.0208)}{(0.0260)(0.0260)}\)=0.640

3. The reactants and products are present in the following concentrations.

[CO]= 0.0094 M.[HO] =0.055 M,[CO₂] = 0.0005 M,[H₂]= 0.0046 M.

then, \(Q_c=\frac{(0.0006)(0.0046)}{(0.0094)(0.0065)}=0.044\)

As Q_c. <K_c, the reaction will proceeds in the forward direction and as expected, the concentrations at equilibrium are found to be [CO]_eq = 0.0074 M; [H₂O]_eq = 0.0035 M; [CO₂]_eq =0.0025 M and [H₂]_eq = 0.0066 M resulting in a K_c value of 0.64.

Types of equilibrium: A chemical equilibrium is called homogeneous if all the reactants and products are present in the same phase. Some examples of homogeneous equilibrium follow.

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})\rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)

⇒ \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

In a heterogeneous, chemical equilibrium, on the other hand, the reactants and products are not all in the same phases A few examples follow.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})\)

⇒ \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+4 \mathrm{H}_2(\mathrm{~g})\)

∴ \(\mathrm{CrO}_4^{2-}(\mathrm{aq})+\mathrm{Pb}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{PbCrO}_4(\mathrm{~s})\)

Gas-phase reactions: When all fee reactants and products of a reaction are gaseous, the reaction is called a gas-phase reaction and it is a case of homogeneous equilibrium.

In the case of such reactions, the equilibrium constant can be expressed either in terms of molar concentrations or partial pressures.

When expressed in terms of molar concentrations it is denoted by K_c and if it is expressed in terms of partial pressures, K_p is used. If A, B, X and Y are gases in the reaction the equilibrium constant can be expressed as follows.

⇒ \(a \mathrm{~A}+b \mathrm{~B} \rightleftharpoons x \mathrm{X}+y \mathrm{Y}\)

∴ \(K_p=\frac{\left(p_{\mathrm{X}}\right)^x\left(p_{\mathrm{Y}}\right)^y}{\left(p_{\mathrm{A}}\right)^a\left(p_{\mathrm{B}}\right)^{\mathrm{b}}}\)

In this expression, p_A, p_B, p_x and p_y are the partial pressures of A, B, X and Y, expressed in atmospheres or pascals.

At constant temperature, the pressure of a gas is proportional to its concentration.

In terms of concentrations, the equilibrium constant can be expressed as

∴ \(K_c=\frac{[X]^z[Y]^y}{[A]^a[B]^b} \quad \text { or } \quad \frac{\left(C_X\right)^x\left(C_Y\right)^y}{\left(C_A\right)^a\left(C_B\right)^b} \text {, }\)

where C_A, C_B, C_x and C_y are the molar concentrations of A, B, X and Y respectively.

If A, B, X and Y are ideal gases, pV = nRT

or \(p=\frac{n}{V} R T\)

or p =n/VRT = CRT [n/V = number of moles per litre or mol/dm³]

Then \(p_{\mathrm{A}}\)=\(\mathrm{C}_{\mathrm{A}} R T\) and \(p_{\mathrm{B}}\)=\(\mathrm{C}_{\mathrm{B}} R T\).

⇒ \(p_{\mathrm{X}}\)=\(C_{\mathrm{X}} R T\) and \(p_{\mathrm{Y}}\)=\(\mathrm{C}_{\mathrm{Y}} R T\).

∴ \(K_p=\frac{\left(C_X R T\right)^x\left(C_Y R T\right)^y}{\left(C_A R T\right)^a\left(C_B R T\right)^b}=\frac{\left(C_X\right)^x\left(C_Y\right)^y}{\left(C_{\mathrm{A}}\right)^a\left(C_B\right)^b} R T^{(x+y)-(a+b)}\).

But \(K_c=\frac{\left(C_X\right)^x\left(C_Y\right)^y}{\left(C_{\mathrm{A}}\right)^a\left(C_B\right)^b}\).

∴ \(K_p=K_c R T^{\Delta n}\),

where Δn = (x + y) – (a + b) = no. of moles of products – no. of moles of reactants in the balanced chemical equation.

Consider the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) at equilibrium. If the equilibrium constant for the reaction is K_c then K_p would be K_p = K_cRT_Δn,

where Δn = no. of moles of HI- no. of moles of H₂ and I₂ = 2- 2 or 0.

Therefore, K_p =K_c for the reaction at equilibrium.

However, this is not always the case. Consider another example, \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\) at equilibrium.

Here the number of moles of reactants is more than that of the product or Δn = 2 – 4 or -2.

Therefore, the equilibrium constant in terms of partial pressure Kp for the above reaction would be K_p = K_c(RT)^-2.

If the concentration is in mol L^-1 and p is in bar then R = 0.0831 L bar K^-1 mol^-1.

Nature of equilibrium constant

1. The equilibrium constant, as we have defined it so far, is the ratio of the product of the molar concentrations of the reaction products to that of the reactants, with each concentration term raised to the power equal to the stoichiometric coefficient of the substance in the balanced chemical equation. You may argue that if this is so, K should have units.

• Though the equilibrium constant was initially derived in terms of concentrations, strictly speaking, concentrations or partial pressures can be used only for ideal gases. However, nowadays, equilibrium constants are expressed in dimensionless quantities.
• This is possible only when the quantities (concentration and partial pressures) are measured with respect to a corresponding standard state.
• The standard state for a pure gas is taken to be 1 bar and for solute the standard state (C₀) is 1 molar solution. Thus a pressure of 4 bar in terms of standard state is 4, a number without a unit.
• Therefore the numerical value of the equilibrium constant depends on the standard state chosen and both K_p and K_c are dimensionless quantities.

2. The equilibrium constant has a definite value for a particular reaction at a given temperature. This value does not depend on the initial concentrations of the reactants.

If the reaction is reversed, the tire value of K becomes the inverse of the value of the forward process. For instance, suppose the value of the equilibrium constant for the combination of hydrogen and iodine at 720 K is K_c.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) ; \quad K_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \text {. }\)

Then the value of the equilibrium constant, say K’_c, for the decomposition of hydrogen iodide at the same temperature would be 1/K_c.

⇒ \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) ; \quad K_c^{\prime}=\frac{1}{K_c}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}\).

3. It is all very well to say that the value of the equilibrium constant remains the same for a particular reaction at a given temperature. But a reaction may be expressed in more ways than one. For instance, the reaction between hydrogen and iodine can be expressed in the following ways.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)….(1)

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})\)…(2)

⇒ \(n \mathrm{H}_2(\mathrm{~g})+n \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 n \mathrm{HI}(\mathrm{g})\) …..(3)

If the equilibrium constant for (1) is \(K_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

then the equilibrium constant for (2) is \(K_c^{\prime}=\frac{[\mathrm{HI}]}{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}=\sqrt{K_c} \text { or } K_c^{1 / 2}\)

and the equilibrium for  (3) is \(K_c^n=\frac{[\mathrm{HI}]^{2 n}}{\left[\mathrm{H}_2\right]^n\left[\mathrm{I}_2\right]^n}=K_c^n\).

Thus, the value of K_c depends on the coefficients of the reactants and products in the equation being considered.

4. If an equation is written in two steps and the values of the equilibrium constants for the two steps are K₁ and K₂ then the value of the equilibrium constant for the equation is the product of and K₂. For example, the reaction

⇒ \(\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 ; \quad K_c=\left[\mathrm{NO}_2\right]^2 /\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]^2\)

occurs in two steps.

⇒ \(\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; \quad K_1=\left[\mathrm{NO}^2 /\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]\right.\)

⇒ \(2 \mathrm{NO}+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 ; \quad K_2=\left[\mathrm{NO}_2\right]^2 /\left[\mathrm{NO}^2\left[\mathrm{O}_2\right]\right.\)

∴ \(K_1K_2=\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{NO}^2\left[\mathrm{O}_2\right]\right.}=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}=K_c\)

5. The value of equilibrium constant does not change if a catalyst is present.

6. The magnitude of K_c helps to predict the extent to which the reactants will be converted into the products, or the equilibrium composition of the reaction system.

• As products appear in the numerator of the equilibrium constant expression, and reactants are in the denominator, a large value of K_c or K_p indicates that the ratio of products to reactants is very large.
• In other words, the reaction proceeds nearly to completion. A very low value of K_c indicates that the reaction cannot proceed much in the forward direction.

In other words the value of K_c tells us whether the particular reaction can actually take place. Let us consider an example.

⇒ \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{FeSCN}^{2+}(\mathrm{aq})\)

At 298 K, the equilibrium constant for the forward reaction is 138, which shows that the forward reaction is favoured or can occur. For the reverse reaction the equilibrium constant is 1/138, which shows that this reaction does not proceed much to the left.

Similarly, for the reaction, \(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCl} \text { at } 300 \mathrm{~K}\) the equilibrium constant is 4.0 x 10³¹. This shows that the reaction goes virtually to completion.

The equilibrium constant for the reverse reaction \(2 \mathrm{HCl}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \text { is } \frac{1}{4.0 \times 10^{31}}=25 \times 10^{-30}\) indicating that the reverse reaction is less favoured. While a large value of K_p or K_c favours the formation of products, a very small value favours the reverse reaction or the formation of reactants.

Consider the reaction \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) . \mathrm{K}_c \text { is } 4.8 \times 10^{-31}\) at 298 K for this reaction. Here, at equilibrium, the ratio of concentration of products to reactants is small, indicating that the reverse reaction is favoured.

Examples of reactions having intermediate values of equilibrium constants are

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) K_c=57 \text { at } 700 \mathrm{~K}\)

⇒ \(\text { and } \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g})  \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) K_c=4.64 \times 10^{-3} \text { at } 298 \mathrm{~K}\).

In both cases, there are appreciable amounts of reactants and products. This may be verified for the first reaction by substituting the concentrations of H₂ and I₂ (say 0.01 M) in the equilibrium constant expression

⇒ \(K_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \cdot\)

⇒ \({[\mathrm{HI}]^2 } =K_c \cdot\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]\)

∴ \({[\mathrm{HI}] } =\sqrt{K_c \cdot\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\sqrt{57 \times 0.01 \times 0.01}=0.075 \mathrm{M}\).

It is worthwhile to mention here that the magnitude of an equilibrium constant of a reaction does not tell us anything about the rate of that reaction, i.e., how fast or how slowly the reaction will take place. It is quite possible that the equilibrium constant of a reaction may suggest completion but the rate is too low for all practical purposes.

We can make a generalisation about the extent to which a reaction proceeds based on the magnitude of K_c, as shown in Table.

Conventions regarding K

1. In the case of a homogeneous equilibrium involving products and reactants in the gaseous phase for example, \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

obtaining the expression for Kc or Kp is simple enough.

∴ \(K_{\mathrm{c}}=\frac{[\mathrm{HI}(\mathrm{g})]^2}{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right]} \text { and } K_p=\frac{p_{\mathrm{HI}}^2}{p_{\mathrm{H}_2} p_{\mathrm{I}_2}} \text {. }\)

2. If one of the products or reactants involved in a heterogeneous equilibrium is a solid or a liquid its concentration is taken to be unity, by convention.

The molar concentration of a pure solid or liquid is coastant. In other words for any amount of substance ‘X’, [X(s)] and [X(l)] are constant. But [X(g)] and [X(aq)] will vary with the volume. Consider the following equilibrium.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

⇒ \(K_{\mathrm{c}}=\frac{[\mathrm{CaO}(\mathrm{s})]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{CaCO}_3(\mathrm{~s})\right]}=\frac{1 \times\left[\mathrm{CO}_2(\mathrm{~g})\right]}{1}=\left[\mathrm{CO}_2(\mathrm{~g})\right]\)

or \(K_p=p_{\mathrm{CO}_2}\)

This fits with the fact that the pressure of CO₂ becomes constant when equilibrium is reached in the decomposition of CaCO₃ in a closed vessel.

It has been found experimentally that the pressure of CO₂ is 2.5 x 10⁴ Pa when the reaction for the decomposition of CaCO₃ reaches equilibrium at 1073 K. Therefore the equilibrium constant at 1073 K for the above reaction is

⇒ \(K_p=p_{\mathrm{CO}_2}=\frac{2.5 \times 10^4 \mathrm{~Pa}}{10^5 \mathrm{~Pa}}\) = 0.25

(since 1 bar =10⁵ Pa and while calculating the value of K_p, pressure should be expressed in bar as the standard state of a substance is its pure state at exactly 1 bar.)

3. Consider an equilibrium in an aqueous medium.

⇒ \(\mathrm{NH}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

In this case H₂O is the solvent and is present in a large quantity. Consequently, its concentration does not change much during the reaction and is taken to be constant (unity, by convention). In general, the concentration of a solvent is taken to be 1.

Thus, \(K_c=\frac{\left[\mathrm{NH}_4^{+}(\mathrm{aq})\right]\left[\mathrm{OH}^{-}(\mathrm{aq})\right]}{\left[\mathrm{NH}_3(\mathrm{aq})\right] \times 1}\).

4. In the light of what we have just discussed, let us reconsider the equilibrium between a liquid and its vapour in a closed system.

⇒ \(\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

⇒ \(K_c=\frac{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]}=\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]\)

or \(K_p=p_{\mathrm{H}_2 \mathrm{O}}\).

Now do you see why the vapour pressure of a liquid is constant at a particular temperature and why it does not depend on the amount of liquid present.

5. Remember that in a chemical equilibrium the concentration of a liquid is taken as 1 only if the liquid is present as a solvent, or in a large quantity. Consider the following reaction.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{l})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{l})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Here \(K=\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{l})\right]\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(\mathrm{l})\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})\right]}\)

The concentration of H₂O cannot be taken as 1 in this case.

Example: A reaction between hydrogen and iodine to produce hydrogen iodide at 675 K is carried out in a closed flask of volume 4 L. At equilibrium, the reaction system contains 0.6 mol of hydrogen, 0.6 mol of iodine and 2.4 mol ofhydrogen iodide. Calculate the equilibrium constant.
Solution:

For the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

∴ \(K_{\mathrm{c}}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

Given that,

[HI] = 2.4/4 = 0.6 mol L^-1.

[H₂] = 0.6/4 = 0.15 mol L^-1.

[I₂] = 0.6/4 = 0.15 mol L^-1.

Substituting these values in the expression for K_c, \(K_c=\frac{(0.6)^2}{0.15 \times 0.15}=16\)

The equilibrium constant, the reaction quotient and Gibbs energy: As already stated, the value of the equilibrium constant for a reaction does not depend on the rate of that reaction or vice versa.

• However, it is directly related to the thermodynamics of the reaction and, in particular, to the change in free energy ΔG. You already know from the previous chapter that if ΔG is negative for a reaction then the reaction is spontaneous and should proceed in the forward direction.
• If, on the other hand, ΔG is positive, the reaction is considered nonspontaneous. Then the reverse reaction will have a negative ΔG value and will proceed, converting the products of the forward reaction into reactants.
• Either way, a reaction should proceed in the spontaneous direction until it achieves equilibrium. At this point, there is no free energy left to drive the reaction ΔG becomes zero and the reaction attains equilibrium.

This thermodynamic view of equilibrium is expressed mathematically as \(\Delta G=\Delta G^{\ominus}+R T\) ln Q

When equilibrium is achieved ΔG = 0, Q = K and this equation becomes \(\Delta G =\Delta G^{\ominus}+R T \ln K=0\)

or \(\Delta G^{\ominus} =-R T \ln K\).

Rearranging the above equation, we get

ln \(K=-\frac{\Delta G^{\ominus}}{R T}\)

or \(2.303 \log K =-\frac{\Delta G^{\ominus}}{R T}\)

or \(\log K =-\frac{\Delta G^{\ominus}}{2303 R T}\).

Taking antilog of both sides K = \(e^{-\Delta G^{\ominus} / R T}\)

or \(K=\mathrm{antilog}\left(-\frac{\Delta G^{\ominus}}{2.303 R T}\right)\)

This equation provides an interpretation of the spontaneity of a reaction.

If \(\Delta G^{\ominus}<0 \text { then } \frac{-\Delta G^{\ominus}}{R T}\) is positive, and \(e^{-\Delta C^{\ominus} / R T}>1 \text {, i.e., } K>1\). Such a reaction is spontaneous and proceeds in the forward direction so that the products are present predominantly.

Similarly, a nonspontaneous reaction is one for which \(\Delta G^\ominus>0\) and K < 1 Such a reaction proceeds in the forward direction to such a small extent that very small amounts of products are formed.

Example 1. Consider the reaction \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})\). Amixture of SO₂, SO₃ and O₂ at equilibrium at 1000 K has [SO₂] = 3.8X10^-3 M, [SO₃] = 4.13X10^-3 M, and [O₂] = 4.3×10^-3 M. Calculate K_c and K_p of the reaction.
Solution:

⇒ \(K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}\)

Substituting the values in the expression, we get \(K_c=\frac{\left(4.13 \times 10^{-3}\right)^2}{\left(3.8 \times 10^{-3}\right)^2\left(4.3 \times 10^{-3}\right)}\)

(Note: The concentration must always be in ‘M’ or mol L^-1 for calculating Kc.]

⇒ K_c = 275.

To calculate K_p we must know Δn.

Δn = no. of moles of products- no. of moles of reactants. (considering only gaseous products and reactants)

Δn = 2- (2 + 1) = 2- 3 = -1

But K_p = K_c(RT)^Δn

= K_c(RT)^-1

= \(\frac{K_c}{R T}\)

∴ \(K_p=\frac{275}{(0.08312)(1000)}=3.3\). [While determining the value of K_p, pressure should be expressed in bar and the value of R must be in L bar K^-1 mol)^-1.]

Example 2. The concentration equilibrium constant, K_c for the reaction \(2 \mathrm{NO}_2 \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4\) at 25° C is 216. Find the pressure equilibrium constant, K_p, of the reaction at the same temperature.
Solution:

K_p = K_c(RT)^Δn

Δn =1- 2 =-1

∴ K_p =K_c(RT)^-1

= \(\frac{K_c}{R T}\)

= \(\frac{216}{(0.08312) \times(298.15)}\).

K_p = 8.72. [R = 0.0312 L bar K^-1 mol^-1]

Example 3. The value of K_c for the production of phosgene, the reaction being \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{~g})\), is 5 at 600 K. What are the equilibrium partial pressures ofthe gases if the initial pressures are as follows? \(p_{\mathrm{CO}}=0.265 \mathrm{bar}, p_{\mathrm{Cl}_2}=0.265 \mathrm{bar}, p_{\mathrm{COCl}_2}=0\).
Solution:

At first, K_p must be calculated from K_c.

K_p = K_c(RT)^Δn

Here, Δn =1- 2 = -1

∴ K_p = 5 x (0.08312 x 600)^-1 = 0.1

Let us now tabulate the partial pressures of all the gases. Suppose the partial pressure of COCl₂ at equilibrium is x bar.

Partial pressures:

⇒ \(\begin{array}{lccc}p_{\mathrm{CO}} p_{\mathrm{CO}_2} p_{\mathrm{COCl}_2} \\
\text { Initial } 0.265 \text { bar } 0.265 \text { bar }  0\end{array}\)

⇒ \(\begin{array}{lccc}
\text { Change } -x -x -x \\
\text { Equilibrium } 0.265-x 0.265-x x
\end{array}\)

∴ \(\dot{K}_p=0.1=\frac{p_{\mathrm{COC}_2}}{p_{\mathrm{CO} p_{\mathrm{Cl}_2}}}=\frac{x}{(0.265-x)^2}\)

0.1 = \(\frac{x}{\left(0.07-0.53 x+x^2\right)}\)

Rearranging, we get

0.1 x² – 0.053x + 0.007 = x

or 0.1 x² – 0.053x – x + 0.007 = 0

or 0.1 x² – 0.053 x + 0.007 = 0

or x = \(\frac{+1.053 \pm \sqrt{(1053)^2-4 \times 0.1 \times 0.007}}{2 \times 0.1}=\frac{1.053 \pm 1.0517}{0.2}=6.5 \times 10^{-3} \text { or } 10.52\)

x cannot be greater than 0.265 bar since even if all of CO and Cl₂ react, \(\mathrm{p}_{\mathrm{COCl}_2}\), cannot be more than 0.265 bar.

Choosing x to be 6.5 x 10^-3

at equilibrium, \(\mathrm{p}_{\mathrm{COCl}_2}\) =6.5 x 10^-3 bar

P_CO = 0-265 – 65 x 10^-3 = 0.2585 bar = pa

Example 4. The standard Gibbs free energy at 298 K for the reactions are given.

1. \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \Delta \mathrm{G}^{\ominus}=173.2 \mathrm{~kJ}\)
2. \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) [latex] \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \Delta \mathrm{G}^{\ominus}=-69.7 \mathrm{~kJ}\)

Calculate K_p at 298 K for the following reaction \(\mathrm{N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})\)

Solution:

First, we must find \(\Delta G^{\ominus}\), for the reaction of interest.

⇒ \(\begin{aligned}
\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) \quad \Delta \mathrm{G}^{\ominus}=173.2 \mathrm{~kJ} \\
2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \quad \Delta G^{\ominus}=-69.7 \mathrm{~kJ} \\
\overline{\mathrm{N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \Delta G^{\ominus}=-173.2-69.7} \\
=0.103 .5 \mathrm{~kJ} \text {. } \\
\end{aligned}\)

Thus, \(\Delta G^{\ominus}\) for the reaction in question is the sum of \(\Delta G^{\ominus}\) for the other two reactions.

∴ \(\Delta G^{\ominus}\) = 103.5 kJ.

∴ \(K_p=e^{-\Delta G^{\ominus} / R T}\)

But \(\frac{-\Delta G^{\ominus}}{R T}=\frac{-103.5 \times 1000}{8.314 \times 298}=-4178\)

∴ \(\quad K_p=e^{-41.78}=7.16 \times 10^{-19}\).

Example 5. \(\Delta_f G^{\ominus}\) for ammonia gas is -16.5 kJ mol1 at 298 K. Find the equilibrium constant for the reaction \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})\)
Solution:

2 mol of NH₃ is produced in the reaction but the \(\Delta_f G^{\ominus}\) value given is for only 1 mol. \(\Delta_f G^{\ominus}\) for the formation of 2 mol of NH₃ is equal to twice \(\Delta_f G^{\ominus}\)for 1 mol.

⇒ \(\Delta_f G^{\ominus}\) = 2 x -165 kJ = -33.0 kJ = -33,000 J.

Now, \(\frac{-\Delta_f G^{\ominus}}{R T}=\frac{33000}{8.314 \times 298}=13.32\)

∴ \(K_c=e^{-\Delta, G^4 / R T}=e^{+1332}=609,259.8=6.09 \times 10^5\).

Example 6. K_c for the hydrolysis of sucrose is 5.3 x 10¹² at 298 K. Find \(\Delta G^{\ominus}\) for the process sucrose + H₂O \(\rightleftharpoons\) glucose + fructose
Solution:

⇒ \(K_c=e^{-\Delta G^{\ominus} / R T}\)

ln \(K_c=-\Delta G^{\ominus} / R T\)

⇒ \(\Delta G^{\ominus}=-R T \ln K\)

= -8.314 x 298 x ln(53 x 10¹²) = -72,589.7 J.

∴ \(\Delta G^{\ominus}\)=-72.58 kJ.

Example 7. Consider the reaction \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\). Kp =23.6 at 500K for the reaction. Calculate the equilibrium partial pressures of the reactants and products if the initial pressure of PCl₅ is 0.56 bar and those of PCl₃ and Cl₂ are zero. How will the concentration of PCl₅ and PCl₃ change if more Cl₂ is added once equilibrium is reached?
Solution:

Let us tabulate the partial pressure—initial, change and at equilibrium.

⇒ \(p_{\mathrm{PC}} p_{\mathrm{PC}_3} p_{\mathrm{C}_2}\)

⇒ \(\begin{array}{lccc}
\text { Initial } 0.56  0  0 \\
\text { Change } -x +x +x \\
\text { Equilibrium } 0.56-x  x  x
\end{array}\)

Putting these values in the expression for K_p,

⇒ \(K_p=\frac{p_{\mathrm{PCl}_3} p_{\mathrm{Cl}_2}}{p_{\mathrm{PCl}_5}}\)

23.6 = \(\frac{x \cdot x}{0.56-x}\)

Rearranging, we get 23.6(056- x)- x² = 0

or x²– 23.6(056 -x) = 0

or x² 23.6 x 056 + 23.6x = 0

or x²+23.6x- 13.2 = 0.

∴ x = \(\frac{-23.6 \pm \sqrt{(23.6)^2+4 \times 13.2}}{2}\)

= \(\frac{-23.6 \pm 24.69}{2}=0.547 \text { or }-24.14\)

Taking the positive root, x = 0547, we get

⇒ \(p_{\mathrm{PCl}_5}\) = 0.56 -0547 = 0.013 bar

⇒ \(p_{\mathrm{PCl}_3}\) = 0.547 bar

⇒ \(p_{\mathrm{Cl}_2}\) = 0547 bar,

On adding Cl₂ at equilibrium, the reaction will shift towards the left \(p_{\mathrm{PCl}_5}\) and will increase and \(p_{\mathrm{PCl}_3}\) will decrease.

Factors Affecting Equilibrium

In this section, we shall discuss what happens to a system at equilibrium when certain conditions are changed. We shall consider various changes, one by one.

• What happens to the equilibrium when the temperature is changed, for example? Or what happens when the concentration of either the reactants or the products is changed?
• In 1888, the French chemist Henri Louis Le Chatelier proposed a general law on the behaviour of a system in equilibrium.
• This law, called Le Chatelier’s principle, states that if a system at equilibrium is subjected to a change which displaces if from the equilibrium, a net reaction will occur in a direction that counteracts the change.
• This law is applicable to all physical and chemical equilibria. Let us now discuss in detail the various factors that can affect an equilibrium.

Change in concentration: Let us consider the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) at equilibrium.

• If more of the reactant H₂ or I₂, or both, are added to the reaction system, the state of equilibrium will be disturbed and the tendency of the system would be to counteract this change so that equilibrium is established again.
• The only way this can happen is if more of the reactants react to form more of the products so that the ratio of the products of the concentrations the reaction products to that of the reactants, i.e., K, remains constant.
• Another way of saying this is, if the concentration of the reactants is increased, the equilibrium shifts in the forward direction, or the forward reaction is favoured.
• Applying Le Chatelier’s principle to a specific change (concentration), we may say, “When the concentration of any of the reactants or products in a reaction at equilibrium is changed, the composition of the equilibrium mixture changes so as to undo the effect of the concentration change.”

Let us study how equilibrium is re-established if we add H₂ at equilibrium. The concentration of H₂ increases and equilibrium is disturbed.

A new equilibrium will be set up in which the concentration of H₂ should be less than what it is after adding H₂. But the initial concentration of H₂ increases for the new equilibrium as shown in Figure.

• As you can see from the figure, till the time tx, the system is at equilibrium with the concentrations of H₂, I₂ and HI being given by the intercept of the three curves on the y-axis.
• At t₁, when some H₂ is added, its concentration increases as shown by the steep rise of the H₂ curve. The equilibrium is disturbed and the system responds to the change by forming more HI, thus decreasing the concentration of H₂ and I₂ than at time t₁. At t₂ a new equilibrium is established.

Let us consider the reaction quotient for this reaction \(Q_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

If hydrogen is added at equilibrium, the molar concentration of hydrogen increases and the equilibrium is disturbed. The denominator of the above equation increases and hence Q_c decreases or becomes less than K_c.

• Therefore, in order to make Q_c equal to K_c the numerator must increase or the denominator must decrease. This means more HI should be formed in order to counteract the disturbance in equilibrium.
• As a result more of H₂ and I₂ react to form more of HI to re-establish the equilibrium making the forward reaction more favourable.
• What if the concentration of hydrogen iodide is increased? Again there will be a tendency for the reaction system to counteract this change.
• This time, more hydrogen iodide will yield more hydrogen and iodine so that equilibrium is restored and K_c remains constant. That is to say, if the concentration of the products is increased, the equilibrium will shift in the reverse direction, or the reverse reaction will be favoured.

Changes in concentration (of either products or reactants) play an important role in the productivity of industrial processes. When ammonia is manufactured by the Haber process, for example, the equilibrium involved is the following.

⇒ \(3 \mathrm{H}_2(\mathrm{~g})+\mathrm{N}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)

The product, i.e., NH₃, is continuously removed from the site of the reaction. This disturbs the equilibrium and the system counteracts the change by producing more ammonia.

• Increasing the concentration of the reactants would have the same impact. But, obviously, the former is more economical.
• The large-scale production of CaO from CaCO₃ is another example: the constant removal of CO₂ from the lime kiln drives the reaction to completion.
• There are many applications of Le Chatelier’s principle in our everyday lives. For instance, on humid, cloudy days when there is hardly any breeze, we often dry clothes indoors, under the fan.
• When we do this, we are using Le Chatelier’s principle without knowing it. The water present in the clothes and the water vapour in the air in the vicinity of the clothes reach an equilibrium, which we disturb by turning on the fan.
• The artificial breeze (created by the fan) removes the water vapour from the immediate neighbourhood of the clothes and more water molecules from the clothes escape into the air to re-establish equilibrium.
• For the same reason, one feels more comfortable under the fan on a sultry day.
• A very vital life process depends on Le Chatelier’s principle. You know that haemoglobin (ITb) present in red blood corpuscles acts as a carrier of oxygen from the lungs to the tissues and as a carrier of carbon dioxide from the tissues to the lungs.

How does this happen? The blood that comes from the lungs and reaches the tissues has a high concentration of oxygen in comparison with the blood present in the tissues, where the partial pressure of oxygen is low.

This disturbs the equilibrium and in order to re-establish it, some of the oxyhaemoglobin (haemoglobin carrying oxygen) dissociates.

⇒ \(\mathrm{HbO}_2(\mathrm{~s}) \rightleftharpoons \mathrm{Hb}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g})\)

When the blood returns from the tissues to the lungs, more oxyhaemoglobin is formed.

• This happens because the concentration or partial pressure of oxygen in the lungs is high and to re-establish equilibrium, some of the haemoglobin in the blood returning from the tissues combines with oxygen to form oxyhaemoglobin.
• The removal of carbon dioxide from the tissues by haemoglobin happens in a similar manner.
• The partial pressure of carbon dioxide in the tissues is high, so some of it dissolves in the blood, which reaches the tissues from the lungs.

Carbon dioxide is released from the blood when it returns to the lungs, where the partial pressure of carbon dioxide is low.

⇒ \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{HCO}_3^{-}(\mathrm{aq})\)

The effect of concentration can also be demonstrated with the help of a simple experiment. You are familiar with the reaction

⇒ \(\underset{\text { yellow }}{\mathrm{Fe}^{3+}(\mathrm{aq})}+\underset{\text { colourless }}{\mathrm{SCN}^{-}(\mathrm{aq})} \rightleftharpoons \underset{\text { deep red }}{\mathrm{FeSCN}^{2+}(\mathrm{aq})}\)

⇒ \(K_c=\frac{\left[\mathrm{FeSCN}^{2+}(\mathrm{aq})\right]}{\left[\mathrm{Fe}^{3+}(\mathrm{aq})\right]\left[\mathrm{SCN}^{-}(\mathrm{aq})\right]}\)

If we take ferric nitrate solution and thiocyanate in a test tube in an appropriate proportion and shake, the colour changes from yellow to red due to the formation of FeSCN²⁺.

• This happens gradually and the colour does not change once equilibrium is attained. We can shift the equilibrium by adding or removing the reactant or product.
• If we add few more drops of thiocyanate solution in the test tube, Q_c becomes less than K_c. Therefore, the forward reaction is favoured and more FeSCN²⁺ is formed until Q_c = K_c and equilibrium is attained.
• At first, the intensity of the red colour decreases and then increases to deep red, at equilibrium.
• If on the other hand, we remove SCN^– by adding a reagent, say HgCl₂(aq) which forms a stable complex—[Hg(SCN)₄]^2-—the equilibrium shifts in the reverse direction so that the concentration of FeSCN²⁺( decreases because it dissociates to replenish SCN^–.
• A similar shift in the equilibrium will be observed if we add oxalic acid to the reaction mixture at equilibrium. Oxalic add reacts with Fe³⁺ to form a stable complex—[Fe(C₂O₄)₃]^3-.

Change in pressure: If one of the reactants or products in a reaction system is in the gaseous phase, pressure becomes an important factor governing shifts in equilibrium. In the case of a gas, increase in partial pressure amounts to increase in concentration.

• Let us consider a heterogeneous system first. \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CO}_2(\mathrm{aq})\)
• In this case increasing the partial pressure of CO₂(g) would mean an increase in the number of moles of CO₂ per unit volume (concentration).
• According to Le Chatelier’s principle the system would try to counteract the change to re-establish equilibrium. The only way this is possible is if more CO₂ dissolves in water. This is why the solubility’ of gases increases with increase in pressure.
• Now consider a homogeneous system in which all the reactants and products are in the gaseous phase. The reaction could lead to an overall increase in the number of moles, if the number of moles of the products is more than that of the reactants.

It could lead to an overall decrease in the number of moles, if the number of moles of products is less than that of the reactants. There is a third possibility—there may be no net change in the number of moles. Let us consider these cases one by one.

1. The reaction involved in the manufacture of ammonia by the Haber process is as follows. \(3 \mathrm{H}_2(\mathrm{~g})+\mathrm{N}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)

• This is obviously a case in which the forward reaction leads to a decrease in the number of moles.
• In such a case, if the total pressure is increased, say by decreasing the volume, the number of moles per unit volume increases and the system tries to nullify the change by decreasing the number of moles per unit volume (we already know that pV = constant and p ∝ moles of the gas).
• Since the forward reaction leads to a decrease in the number of moles, it is favoured and the equilibrium shifts in the forward direction.
• In short, increasing the pressure leads to an increase in the yield of the products. Obviously, if pressure is decreased, the reverse reaction will be favoured.

This can also be understood by considering reaction quotient. Let[H₂], [N₂] and [NH₃]be the molar concentrations at equilibrium in the reaction involved in the Haber process.

When the volume of the reaction mixture is halved the pressure as well as concentrations are doubled. Now, let us obtain the concentration quotient by replacing each value by its double.

⇒ \(Q_c=\frac{\left(2\left[\mathrm{NH}_3\right]\right)^2}{\left(2\left[\mathrm{H}_2\right]\right)^3\left(2\left[\mathrm{~N}_2\right]\right)}=\frac{1}{4} \frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{H}_2\right]\left[\mathrm{N}_2\right]}=\frac{K_c}{4}\)

As Q_c < K_c, the forward reaction is favoured.

2. Let us consider an equilibrium in which the forward reaction involves an increase in the number of moles.

⇒ \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)

Here an increase in pressure will favour the backward reaction, while a decrease in pressure will favour the forward reaction.

3. In certain reactions, for example, the ones which follow, the number of moles of the products is equal to that of the reactants.

⇒ \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})  \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Changes in pressure have no impact on such reactions.

When we study the effect of change of pressure on a heterogenous system, the solid or liquid reactants or products are ignored because the volume (and concentration) of a solid or a liquid is nearly independent of pressure.

For instance, in the following reactions carbon and iodine are solids.

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~s} \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

When the pressure on the reaction mixture is increased, the reverse reaction is favoured because the number of moles of the gases decreases in the reverse direction.

Change in temperature: Temperature affects the equilibrium constant K_c. The nature and extent of the change in K_c or K_p of a reaction due to the change in temperature depends on the enthalpy change of the reaction.

Suppose the forward reaction in a chemical equilibrium is exothermic then the reverse reaction will, naturally, be endothermic. Production of ammonia by the Haber process involves an exothermic reaction.

⇒ \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ; \Delta_r H^{\ominus}=-9238 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

• If such a reaction system attains equilibrium at a particular temperature and then heat is supplied to the system so that the temperature increases, the equilibrium gets disturbed.
• According to Le Chatelier’s principle, the system then tries to counteract the change in temperature and the only way it can do this is by absorbing heat.
• In other words, the equilibrium shifts to the left and the endothermic or reverse reaction is favoured.
• If, on the other hand, the temperature is decreased, the equilibrium shifts forward, favouring the forward or exothermic reaction, thus leading to high yield of ammonia.

In this context, refer to the experiment in Figure. NO₂ dimerises into N₂O₄ according to the following reaction.

⇒ \(2 \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) ; \Delta_r H^{\ominus}=-57.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {. }\)

Here the formation of N₂O₄ is favoured at a low temperature (ice bath) whereas that of NO₂ is favoured at high temperature (hot water).

This reaction is exothermic. On increasing the temperature, in order to relieve the stress of added heat, the system will tend to absorb this heat and the reverse reaction will be favoured.

• Since NO₂ is brown and N₂O₄ is colourless, the effect of increasing temperature on this reaction may easily be seen.
• Decreasing the temperature, say by immersing the vessel containing a mixture of NO₂ and N₂O₄ in an ice bath, favours the forward reaction.
• When the dissolution of a solid in water (or any other solvent) is accompanied by the absorption (endothermic) or release (exothermic) of heat, its solubility changes with temperature.
• When NH₄Cl dissolves in water, for example, heat is absorbed. The solubility of NH₄Cl increases with temperature because an increase in temperature favours the endothermic process.
• CaCl₂, on the other hand, dissolves in water with the evolution of heat. The solubility of CaCl₂, thus, decreases with increase in temperature.

The solubility of NaCl is almost unaffected by changes in temperature because there is very little heat change during the dissolution of NaCl.

Effect of a catalyst: A catalyst merely changes the time taken to reach the state of equilibrium, it does not change the equilibrium constant. It lowers the activation energy of the forward and reverse reactions by the same amount.

• The activation energy is the energy barrier that has to be overcome for the reaction to proceed. If a reaction mixture is not at equilibrium, a catalyst accelerates the rate at which equilibrium is reached but it does not affect the composition of the equilibrium mixture.
• Hence it does not appear in the balanced chemical equation or in the equilibrium constant expression.
• It increases the rates of the forward and reverse reactions to the same extent. Nonetheless, the fact that a catalyst decreases the time taken to reach equilibrium is very important for many industrial processes.
• For instance, iron-molybdenum is used as a catalyst in the production of ammonia from nitrogen and hydrogen because at low temperature, the rate of the reaction is very slow and at high temperature, the reverse reaction is favoured.
• The optimum conditions of temperature and pressure for the synthesis of NH₃ using a catalyst are 400-500°C and 130-300 atm.

In the manufacture of sulphuric acid by the contact process, the conversion of SO₂ to SO₃ is very important.

⇒ \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) ; K_{\mathrm{c}}=1.7 \times 10^{26}\)

The magnitude of the equilibrium constant indicates that the reaction should reach completion.

However, the oxidation of SO₂ is very slow but in the presence of the catalyst V₂O₅ (vanadium pentoxide), the rate of the reaction increases. But if a reaction has a very low K_c, a catalyst would be of no help.

Effect of inert gas: While studying the effect of pressure, we have considered changes that result from a change in volume (as volume and pressure are inversely proportional).

• But what if we keep the volume constant and increase the pressure by adding an inert gas. The inert gas is added just to change the pressure—it does not take part in the reaction.
• We could also think of adding an inert gas at constant pressure. If an inert gas is added to a reaction system containing reactants and products in the gaseous phase at equilibrium, the effect it will have on the equilibrium will depend on whether it is added at constant volume or at constant pressure.
• If the equilibrium is readied at constant volume, i.e., in a closed vessel, the addition of the inert gas will not change the molar concentrations (or partial pressures) of the reactants and products. Consequently, the state of equilibrium will not be affected.

If an inert gas is added to a system at equilibrium at constant pressure, the volume will increase.

• An increase in volume will lead to a decrease in the molar concentrations (or partial pressures) of all the reactants and products.
• If the reaction is such that the number of moles of the reactants is the same as that of the products, a decrease in the molar concentrations of all the reactants and products will make no difference to the equilibrium constant since it is the ratio of the product of concentrations of the products to that of the reactants.
• (We came across a similar situation while studying the effect of a change in pressure on an equilibrium.) If, however, the number of moles of the reactants is not the same as that of the products, an increase in the total volume will disturb the equilibrium.

Suppose the number of moles of tire reactants is less than that of the products, as in the following equilibrium.

⇒ \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)

In such a case, a decrease in the molar concentrations (due to increase in total volume) of all the products and reactants will lead the system to counteract the change and the equilibrium will shift in the direction of increasing the number of moles, i.e., the forward direction.

• We can look at it in another way. At equilibrium \(K_{\mathrm{c}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)
• If the concentrations of all the reactants and products decrease, the numerator will decrease more since there are two concentration terms in the numerator and Kc will decrease.
• For Kc to remain constant, the equilibrium will shift in the direction of an increase in the concentrations of products and decrease in the concentration of the reactants, i.e., in the forward direction.

Similarly, if a reaction is such that the number of moles of the reactants is more than that of the products, an increase in volume will push the equilibrium in the reverse direction.

⇒ \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})\)

For example, for the reaction an increase in volume will favour the reverse reaction. In general, the addition of an inert gas at constant pressure (to a system at equilibrium) makes the equilibrium shift in the direction that increases the number of moles.

Ionic Equilibrium

When certain compounds like NaCl, KCl or NaOH are dissolved in water, the resultant solution is a good conductor of electricity. The solution of NaCl contains Na⁺ and Cl^– ions surrounded by water.

⇒ \(\mathrm{NaCl}(\mathrm{s}) \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

⇒ \(\mathrm{KCl}(\mathrm{s}) \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{K}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

• The electrostatic attraction between the ion and water (ion-dipole interaction) results in the dissolution of such ionic compounds in water. The figure shows various interactions in the solution of KCI in water.
• When KCl is dissolved in water, the positive end of the polar water molecule (hydrogen) is attracted by negative chloride ions at the surface of the solid, similarly, the negative end of the water molecule (oxygen) is attracted by the positive potassium (K⁺) ions.
• The water molecules penetrate between K⁺ and Cl^– and surround them, thus reducing the strong interionic forces holding them together and allowing them to move as hydrated ions.
• This process is called ionisation, i.e., the process of breaking up of a compound into its ions in a solution.
• Though the exact definition of ionisation differs from that of dissociation, we have used both the terms to refer to the breaking up of an ionic substance into its constituent into ions for the sake of convenience.
• Michael Faraday in 1824 classified substances into electrolytes and nonelectrolytes. Nonelectrolytes do not conduct electricity.
• Electrical conductivity requires the existence of charged particles and in a solution of nonelectrolyte, the solute molecules retain their unbroken identity.

That is why though sucrose dissolves in water, it is not a conductor of electricity.

⇒ \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{~s}) \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})\)

Not all electrolytes conduct electricity with equal ease. Faraday categorised electrolytes as strong and weak. For instance, sodium chloride is a strong electrolyte whereas acetic acid is weak electrolyte.

Ionisation: All electrolytes do not ionise to the same extent. The extent to which an electrolyte ionises is called its degree of dissociation.

• More accurately, the fraction of the total number of molecules of the electrolyte in the solution which dissociates into ions is called the degree of dissociation of the electrolyte.
• A truly ionic compound dissociates completely in solution. For example, if 1 mol of NaCl is dissolved in 1 L of water, the solution will contain 1 mol of Na⁺ and 1 mol of Cl^– ions.
• Compounds which dissociate completely in solution are called strong electrolytes.
• Polar covalent compounds may also produce ionic solutions or solutions which conduct electricity.
• Depending on the degree of ionisation or dissociation, such solutes are strong electrolytes or weak electrolytes.

H₂SO₄, for example, dissociates completely in solution and is a strong electrolyte. Equations for the ionisation of strong electrolytes are written with a single arrow.

⇒ \(\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

⇒ \(\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

Weak electrolytes, like NH₃ and CH₃COOH (acetic acid), on the other hand, do not dissociate completely. Their solutions arc weakly or partly ionised.

In such cases there is an equilibrium between the un-ionised molecules of the electrolyte and the ions in solution. Consequently, the ionisation of such an electrolyte is represented with a double arrow.

⇒ \(\mathrm{NH}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})\)

• The degree of ionisation of weak electrolytes depends upon the nature of the electrolyte, temperature and dilution.
• We will not concern ourselves with the ionisation of strong electrolytes because such reactions proceed to completion and in this chapter we are examining reactions from the point of view of equilibrium.

Ionisation of weak electrolytes: An equilibrium between the (un-ionised) molecules of a weak electrolyte and its ions in solution is called an ionic equilibrium.

• If C is the concentration of an electrolyte in a solution, and a is the degree of dissociation, the number of moles of the electrolyte that dissociate is Cα.
• The degree of dissociation (α) of an electrolyte is the fraction of the total number of molecules of the electrolyte (in the solution) that ionise at equilibrium.
• For example, if the value of α = 0.5 or 1/2, it means that out of 1 mole of the electrolyte, 0.5 mol has ionised or there is 50% ionisation.
• So, if C is the concentration of the electrolyte, what is the concentration that is ionised? It is Cα.

Let us say that the concentration of some electrolyte XY in an aqueous solution is C and its degree of dissociation is a. At equilibrium, the ionisation of the electrolyte can be represented by

⇒ \(\mathrm{XY}(\mathrm{aq}) \rightleftharpoons \mathrm{X}^{+}(\mathrm{aq})+\mathrm{Y}^{-}(\mathrm{aq})\)

At equilibrium, the concentrations of X⁺ and Y^– must each be Cα. And the concentration of XY must be C -Cα = C(1 – α). (The concentration of X⁺ and Y^– are equal to the concentration of XY dissociated.)

Applying the law of chemical equilibrium

K = \(\frac{\left[\mathrm{X}^{+}(\mathrm{aq})\right]\left[\mathrm{Y}^{-}(\mathrm{aq})\right]}{[\mathrm{XY}(\mathrm{aq})]}\)

= \(\frac{(\mathrm{C} \alpha)(\mathrm{C} \alpha)}{C(1-\alpha)}=\frac{C \alpha^2}{1-\alpha}\).

For a weak electrolyte, a is so small compared to 1 that it can be neglected in approximate calculations. Then K = Cα²

or \(\quad \alpha^2=\frac{K}{C}\)

or \(\alpha=\sqrt{\frac{K}{C}}\)

⇒ \(\quad \alpha \propto \sqrt{\frac{1}{C}}\) (because K is a constant)

• Therefore, for a weak electrolyte, the degree of ionisation is inversely proportional to the square root of the molar concentration.
• This is known as Ostwald’s dilution law. As C approaches zero, or dilution approaches infinity, the degree of dissociation approaches unity.
• This means that the ionisation of a weak electrolyte will be more in a dilute solution than in a concentrated solution.

Example: Calculate the degree of dissociation of a 0.02-M solution of formic acid, given that K = 1.9 x 10^-4.
Solution:

The dissociation of formic acid in water can be represented as follows.

⇒ \(\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{HCOO}^{-}\)

Let α be the degree of dissociation. Then the concentrations of the ionic species at equilibrium will be as follows:

[H₃O]⁺ = 0.02α

[HCOO]^– = 0.02α

[HCOOH] = 0.02 (1 – α) = 0.02, assuming a to be negligible compared to 1.

Now K = \(\frac{(0.02 \alpha)(0.02) \alpha}{0.02}=1.9 \times 10^{-4}\) (given)

or \(\alpha^2=\frac{1.9 \times 10^{-4}}{0.02}\)

or \(\alpha=\sqrt{95} \times 10^{-2}=9.75 \times 10^{-2}\).

The degree of dissociation of formic acid = 0.0097.

Acid Base Equilibrium

Acids and bases are also electrolytes. You are quite familiar with some properties of acids and bases. You know, for example, that adds turn blue litmus red and that bases (also called alkalis) turn red litmus blue.

You know that adds react with active metals like zinc and magnesium to liberate hydrogen. You know that acids are sour and corrosive and that bases are bitter and slippery. And you also know that adds and bases neutralise each other.

This is essentially the concept chemists had of acids and bases in the olden days. You could call this concept, based on certain properties of adds and bases, the dassical concept.

Arrhenius theory: In 1884, S A Arrhenius, a Swedish chemist, proposed a new definition of acids and bases. It was based on his general theory of ionisation of electrolytes, which basically said that an electrolyte, on dissolution in water, dissociated into positively and negatively charged ions.

Arrhenius defined an acid as a substance which produces hydrogen ions (H⁺) when mixed with water. Bases, according to Arrhenius, are substances which produce hydroxyl ions (OH^–) when mixed with water. And the neutralisation of adds and bases is a reaction between H⁺ and OH^– ions.

⇒ \(\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Strong acids like HCl, HNO₃ and H₂SO₄ dissociate completely when dissolved in water.

⇒ \(\mathrm{HCl} \stackrel{\text { water }}{\longrightarrow} \mathrm{H}^{+}+\mathrm{Cl}^{-}\)

⇒ \(\mathrm{HNO}_3 \stackrel{\text { water }}{\longrightarrow} \mathrm{H}^{+}+\mathrm{NO}_3\)

Weak acids like acetic acid, carbonic acid and phosphoric acid dissociate only to a small extent when dissolved in water.

⇒ \(\mathrm{CH}_3 \mathrm{COOH} \stackrel{\text { water }}{\rightleftharpoons} \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\)

⇒ \(\mathrm{H}_2 \mathrm{CO}_3 \stackrel{\text { water }}{\rightleftharpoons} 2 \mathrm{H}^{+}+\mathrm{CO}_3^{2-} \)

⇒ \(\mathrm{H}_3 \mathrm{PO}_4 \stackrel{\text { water }}{\rightleftharpoons}{=} 3 \mathrm{H}^{+}+\mathrm{PO}_4^{3-}\)

Bases like NaOH and KOH are strong bases because they dissociate completely.

⇒ \(\mathrm{NaOH} \stackrel{\text { water }}{\longrightarrow} \mathrm{Na}^{+}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{KOH} \stackrel{\text { water }}{\longrightarrow} \mathrm{K}^{+}+\mathrm{OH}^{-}\)

Substances like NH₄OH, Ca(OH)₂ and Al(OH)₃ dissociate to a small extent and are called weak bases.

⇒ \(\mathrm{NH}_4 \mathrm{OH} \stackrel{\text { water }}{\rightleftharpoons} \mathrm{NH}_4^{+}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{Ca}(\mathrm{OH})_2 \stackrel{\text { water }}{\rightleftharpoons} \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}\)

⇒ \(\mathrm{Al}(\mathrm{OH})_3 \stackrel{\text { water }}{\rightleftharpoons} \mathrm{Al}^{3+}+3 \mathrm{OH}^{-}\)

• According to Arrhenius, acids produce H⁺ ions and bases produce OH^– ions when dissolved in water.
• Actually, the H⁺ ion, formed by the loss of an electron from the hydrogen atom, is only an unshielded proton.
• This proton cannot exist independently since its charge density is very high.

In an aqueous solution, it binds itself with one of the two lone pairs available on the water molecule to form a hydronium ion (H₃O⁺).

⇒ \(\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}\)

For the sake of convenience, the hydrated proton (hydronium ion) is represented as H⁺ (aq). In an aqueous solution, the H⁺ and OH^– ions exist as hydrated ions.

Limitations of Arrhenius’s theory This concept of acids and bases could explain neutralisation, salt hydrolysis and the strength of adds and bases, but it had a few drawbacks.

1. This theory cannot explain the basic nature of substances like NH₃, Na₂CO₃ and CaO, which do not possess a hydroxyl (OH^–) group. Nor cannot it explain the basic nature of CO₂, SO₂, SO₃, etc., which do not contain hydrogen.
2. Another limitation of this theory is its inability to explain reactions between acidic and basic substances in the absence of water. Two examples follow.

⇒ \(\mathrm{SO}_3(\mathrm{~g})+\mathrm{CaO}(\mathrm{s}) \longrightarrow \mathrm{CaSO}_4(\mathrm{~s})\)

⇒ \(\mathrm{NH}_3(\mathrm{~g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{NH}_4 \mathrm{Cl}(\mathrm{s})\)

Bronsted-Lowry theory: In 1923, a Danish chemist, I N Bronsted, and a British chemist, T M Lowry, independently put forward a more general theory of acids and bases.

According to this theory, an acid is a proton donor and a base is a proton acceptor. This definition has some very interesting implications. To understand these, let us consider some examples.

⇒ \(\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}\)

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O}  \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{CH}_3 \mathrm{COO}^{-}\)

⇒ \(\mathrm{CO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HCO}_3^{-}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O}  \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{NH}_3\)

⇒ \(\mathrm{HCl}+\mathrm{NH}_3 \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{Cl}^{-}\)

Consider the first two reactions. HCl is an acid because it donates one proton to become Cl^– and H₂O is a base because it accepts a proton to become H₃O⁺.

• Similarly, in the second equation, CH₃COOH is an acid and H₂O is a base. Now consider the next two reactions. In these, water acts as an acid.
• It donates a proton to become OH^–. This means that the same substance can act as an acid or a base. Such substances which can act both as an acid and a base are called amphoteric.
• This definition of acids and bases does not restrict itself to neutral molecules. Even ions can act as acids or bases. For example, CO₃^2- in the third equation acts as a base by accepting a proton.

Nor is it necessary for an acid-base reaction to take place in an aqueous medium. For example, in the sixth equation HCI acts as an acid by donating a proton and NH₃ acts as a base by accepting a proton.

• The presence of the hydroxyl group (OH^–) is not necessary for a substance to act as a base. Similarly, a substance can act as an acid even if it does not contain hydrogen.
• For example, NH₃ acts as a base in the fourth reaction. All that is necessary for an acid-base reaction is that one substance should be able to donate a proton and the other should be able to accept the proton.
• A corollary is that a substance which acts as an acid in the presence of a proton-acceptor, will not act in the same way in the absence of a proton-acceptor.
• For example, though acetic acid acts as an acid in an aqueous solution (second equation), it does not do so in a benzene solution because benzene is not a proton-acceptor.

The Bronsted-Lowry concept of acids and bases has a very interesting consequence. When a substance acts as an acid it donates a proton. But on donating a proton, it becomes a base because it becomes capable of accepting a proton. Let us take the case of NH₃ and H₂O.

⇒ \(\mathrm{H}_2 \mathrm{O}+\mathrm{NH}_3 \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}\)

Here water acts as an acid by donating a proton and becomes OH^–. Now OH^– can accept a proton, so it can act as a base.

• Similarly, NH₃ acts as a base by accepting a proton and becomes NH⁺₄, and NH⁺₄ has a proton to donate, so it is an acid.
• Now do you see how every acid turns into a base by donating a proton, and every base turns into an add by accepting a proton.
• The base formed when an acid donates a proton is called the conjugate base of the add and the acid formed when a base accepts a proton is called the conjugate acid of the base.

In the reaction between NH₃ and H₂O, OH^– is the conjugate base of H₂O and NH₄ is the conjugate acid of NH₃.

Thus, there are two acid-base pairs in the reaction and these are called conjugate acid-base pairs. A few examples will make this clearer.

⇒ \(\text { Acid } 1+\text { base } 2 \rightleftharpoons \text { acid } 2+\text { base } 1\)

⇒ \(\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}\)

⇒ \(\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_3^{2-} \rightleftharpoons \mathrm{HCO}_3^{-}+\mathrm{OH}^{-}\)

⇒ \(\mathrm{HCO}_3^{-}+\mathrm{NH}_3 \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{CO}_3^{2-}\)

⇒ \(\mathrm{NH}_4^{+}+\mathrm{CH}_3 \mathrm{COO}^{-} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}+\mathrm{NH}_3\)

• A strong add has a strong tendency to donate a proton but its conjugate base does not have an equally strong tendency to accept a proton.
• The conjugate base of a strong acid is weak and the conjugate base of a weak acid is strong. Also, if two acids are mixed, the weaker acid acts as a base.

⇒ \(\underset{\text { (perchloric acid) }}{\mathrm{HClO}_4}+\mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{H}_3 \mathrm{SO}_4^{+}+\mathrm{ClO}_4^{-}\)

In this reaction, H₂SO₄ acts as a base with respect to the stronger acid, HClO₄. Thus, H₂SO₄ is amphoteric or amphiprotic. Certain acids can lose two or three protons. They are called diprotic (H₂SO₄) or triprotic (H₃PO₄) adds.

Strengths of acids and bases: The strength of an acid or base depends upon its readiness to lose or gain a proton. Experimentally, the strength of an acid or base is measured by its ionisation constant or dissociation constant.

• A strong acid like HCl, HNO₃, H₂So₄, HBr, HI and HClO₄ dissociates almost completely in water. This means the molar concentration of H₃O⁺ ions in an aqueous solution of a strong acid is almost the same as that of the acid itself.
• But this is not the case for weak acids and we can compare the strengths of such acids if we know the dissociation constants at a particular temperature.
• The equilibrium constant for the dissociation of an acid is called its acid dissociation constant or acid ionisation constant and is denoted by K_a.
• Similarly, the base dissociation or base ionisation constant is K_b.

Consider a weak acid HA and suppose its dissociation equilibrium can be represented as follows.

⇒ \(\mathrm{HA}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})\)

At equilibrium, \(K_{\mathrm{a}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]\left[\mathrm{H}_2 \mathrm{O}\right]}\). Here [H₂O] = 1 as it is in a very large quantity and remains almost unchanged.

Suppose the initial concentration of HA is C mol L^-1 and its degree of dissociation is α. Then

⇒ \(K_{\mathrm{a}}=\frac{(C \alpha)^2}{C(1-\alpha)}=C \alpha^2\) (neglecting a in comparison with 1)

or \(\alpha =\sqrt{\frac{K_{\mathrm{a}}}{C}}\).

Thus, the degree of dissociation is inversely proportional to the square root of the concentration. Suppose we consider two acids of equimolar concentration (i.e., C is the same for both) and degrees of dissociation α₁ and α₂.

Then, \(\frac{\alpha_1}{\alpha_2}=\sqrt{\frac{K_{a_1}}{K_{a_2}}}\)

However, the degree of dissociation of an acid is a measure of its strength.

Therefore, \(\frac{\text { strength of acid }}{\text { strength of } \text { acid }_2}=\sqrt{\frac{K_{\mathrm{a}_1}}{K_{\mathrm{a}_2}}} \text {. }\)

• So, we may say that at a given temperature, the ionisation or dissociation constant of an add is the measure of
  its strength.
• As you already know, the equilibrium constant is a dimensionless quantity so Ka is also without any unit.
• Concentrations of all the speeds at ionisation equilibrium are measured with respect to the standard state concentration of 1 M.
• The larger the value of K_a, the stronger is the acid. Weak acids, however, show a great variation in their strengths. For instance, both formic add and hydrocyanic add are weak but their ionisation constants are very different.

Example: The ionisation constants of formic acid (HCOOH) and hydrocyanic acid (HCN) are 1.77 x 10²⁴ and  4.9 x 10^-10 respectively. Compare their strengths.
Solution:

\(\frac{\text { Strength of formic acid }}{\text { Strength of hydrocyanic acid }}=\sqrt{\frac{K_{\mathrm{a}}(\mathrm{HCOOH})}{K_{\mathrm{a}}(\mathrm{HCN})}}=\sqrt{\frac{1.77 \times 10^{-4}}{4.9 \times 10^{-10}}}=601 \text {. }\)

Therefore, formic acid is 601 times stronger than hydrocyanic acid.

The relative strengths of bases can be determined in a similar manner. First, let us consider the equilibria involved in solutions of weak bases. If B is the base, the equilibrium may be represented as

⇒ \(\mathrm{B}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BH}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

The equilibrium constant is K_b and is called base dissociation constant. The higher the value of K_b, the stronger is the base.

⇒ \(K_{\mathrm{b}}=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}\)

If C is the molar concentration of a base and α is its degree of dissociation \(\alpha=\sqrt{\frac{K_b}{C}},\)

where K_b is the dissociation constant of the base.

If we take two bases b₁ and b₂ of equimolar concentrations and dissociation constants \(K_{b_1} \text { and } K_{b_2}\) then

∴ \(\frac{\text { strength of base }}{\text { strength of base }{ }_2}=\sqrt{\frac{K_{b_1}}{K_{b_2}}} \text {. }\)

Relation between K_a and K_b We have seen that the strength of an acid (or a base) is expressed by its K_a (or K_b). For a conjugate acid-base pair, the two equilibrium constants are related in a simple manner.

Let us consider the case of the acid-base pair NH₄ and NH₃. The equilibrium for both acid and base may be represented as

⇒ \(\mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{NH}_3(\mathrm{aq})\)

∴ \(K_{\mathrm{a}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{NH}_3\right]}{\left[\mathrm{NH}_4^{+}\right]}=5.6 \times 10^{-10}\)

∴ \(\mathrm{NH}_3(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

∴ \(K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_3\right]}=1.8 \times 10^{-5}\).

On adding the two, we get the net reaction as \(2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

The sum of the two reactions is simply the dissociation of water, and its equilibrium constant,

⇒ \(K_{\mathrm{w}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]^2}\)

or \(K_{\mathrm{w}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}\).

The equilibrium constant for the net reaction is equal to the product of the equilibrium constants for the individual reactions.

⇒ \(K_{\mathrm{a}} \times K_{\mathrm{b}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{NH}_3\right]}{\left[\mathrm{NH}_4^{+}\right]} \times \frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_3\right]}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=K_w,\)

i. e., (5.6 x 10^-10) x (18 x 10^-5) = 10 x 10^-14. The ionic product of water (K_w) is discussed later in the chapter.

Whenever an equation can be written as a sum of two or more equations, the equilibrium constant for the net reaction is the product of the equilibrium constants of all the individual reactions.

Di- and polybasic acids and di- and polyacidic bases Polybasic or polyprotic acids are those which have more than one ionisable proton. For example, sulphuric acid has two and phosphoric acid has three ionisable protons. The ionisation reactions for a dibasic acid like H₂SO₄ can written as

⇒ \(\mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HSO}_4^{-}(\mathrm{aq}) \)

⇒ \(\mathrm{HSO}_4^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})\)

The corresponding equilibrium constants for these reactions will be as follows.

⇒ \(K_{\mathrm{a}_1}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{HSO}_4^{-}\right]}{\left[\mathrm{H}_2 \mathrm{SO}_4\right]} \text { and } K_{\mathrm{a}_2}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \text {. }\)

• As you can see, the dibasic acid H₂SO₄ has two ionisation constants. Similarly, tribasic acids like phosphoric acid will have three ionisation constants.
• Like polybasic acids, polyacidic bases also have the ionisation constants corresponding to the number of ionisation steps.
• For example, a diacidic base like ethylenediamine has two ionisation constants \(K_{b_1} \text { and } K_{b_2}\). Table gives some values of the ionisation constants of polyprotic acids.
• As you can seen in Table among all the ionisations in an individual polybasic acid the first ionisation constant is the largest followed by the second and then the third.
• This is because after the first ionisation, we are left with a negatively charged ion and it is difficult to remove a proton from a negative ion.

Factors affecting acid strength: We have already seen how to compare the strengths of two acids. But why is one acid stronger than the other? Put simply, the acid strength is determined by the strength and polarity of the H—A bond.

• The strength of the bond depends on the enthalpy change associated with the dissociation of the HA molecule into H and A.
• The polarity of the H—A bond depends on the electronegativity of A and the ease with which electron transfer can occur from H to A resulting in H⁺ and A^– ions. The weaker and more polar the HA bond, the stronger is the acid.
• Let us consider the acids of the halogen series HF, HCl, HBr and HI.
• The variation in polarity is not very significant while we move from HF to Til but there is a considerable difference in the bond strength, which is the deciding factor here.

You already know that atomic size increases on moving down a group. Thus, the size increases from F to I and so the bond strength decreases, increasing the acidity from HF to HI.

• This trend is generally true of acids in the same group of the periodic table. When we go down the group from O to Te, the acid strength increases as H₂O<H₂S<H₂Se<H₂Te.
• For acids in the same row of the periodic table, the variation in bond strength is insignificant and the polarity of the bond is the deciding factor for acid strengths.
• Let us consider the hydrides of a few elements in the second period—CH₄, NH₃, H₂O and HF. As we move from C to F, electronegativity increases and hence the added strength also increases from CH₄ to HF.
• The electronegativities given are C, N, O and F respectively.

Oxoacids Oxoacids such as H₂SO₄, H₂CO₃ and HNO₃ have an O—H bond and the acid strength is dependent on the strength of this bond. Any factor that weakens the O—H bond or increases its polarity, increases the strength of the acid.

• In these acids, the nonmetallic atom is bonded on one side to the oxygen of O—H. The electronegativity of this atom and its oxidation number influences the acid strength.
• In the case of oxoacids containing the same number of O—H groups and the same number of O atoms, as the electronegativity of the nonmetallic atom increases, the acid strength increases.
• For example, consider HOCl, HOBr and HOI. The electronegativity decreases from Cl to I, and hence add strength increases from HOI to HOBr to HOCl.

With the increase in the electronegativity of the nonmetallic atom attached to the O—H bond, the bonded electron cloud is pulled more towards the atom, thus weakening the O—H bond, which then tends to break easily.

If the oxoacids contain the same nonmetallic atom but different numbers of oxygen atoms, then as the oxidation number of the atom increases, the acid strength increases. For example, among the oxoacids of chlorine, the strength of the acids increases as

Common-Ion effect in the ionisation of acids and bases: You are already aware how acetic acid ionises in water. You also know how to calculate the degree of dissociation of this acid. The equilibriun involved is \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}\)

∴ \(K_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} .\)

If we add sodium acetate (the salt of the conjugate base of acetic acid) to the acetic acid solution, the hydrogen ion concentration changes (it decreases).

• This is because the addition of acetate ions causes an increase in the concentration of the products and the equilibrium shifts to the left-hand side, reducing the dissociation of acetic acid.
• This is according to Le Chatelier’s principle. This is called the common ion effect, which is simply the shift in equilibrium on the addition of a substance that provides more of an ion already involved in the equilibrium. Therefore, the concentration of H₃O⁺ decreases.

Suppose acetate ions of concentration 0.1 M are added a 0.1-M acetic acid solution. Then the initial concentrations of the acetate ion and acetic acid solution are both 0.1 M.

Let the concentration of the acetic acid dissociated be x. Then the concentrations of the species present can be given as follows:

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})\)

Initial conc. \( 0.1 0.1 \sim 0\)

Change \(-x +x +x\)

Equilibrium conc. \(0.1-x 0.1+x x\)

⇒ \(K_{\mathrm{a}}=18 \times 10^{-5}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\)

= \(\frac{(0.1+x)(x)}{0.1-x}\).

Since K_a is small for a very weak acid, x is small compared to 0.1 and we can make the approximation that 0.1 +x ≈ 0.1 ≈ 0.1-x.

∴ \(K_{\mathrm{a}}=18 \times 10^{-5}=\frac{x(0.1)}{0.1}=x\)

or x = [H₃O⁺] = 1.8 x 10^-5 M.

Example: Calculate the hydronium iott concentration of a 0.1-M solution of acetic acid, given K_a = 1.8 x 10^-5.
Solution:

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})\)

⇒ \(K_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\right.}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\).

If x is the concentration of CH₃COOH that dissociates, then the concentrations for the various species in the solution are as follows.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}\)

Initial conc. \(0.1 0 0\)

Change \(-x +x +x\)

Equilibrium conc. \(0.1-x x x\)

Substituting the values at equilibrium in the expression for K_a,

⇒ \(K_{\mathrm{a}}=\frac{x \times x}{0.1-x}=\frac{x^2}{0.1-x}\)

This will result in a quadratic equation in x. However, it may be simplified by assuming that x << 0.1

Hence, \(K_{\mathrm{a}}=\frac{x^2}{0.1}\)

⇒ x² = 0.1 x K_a = 0.1 x 1.8 x 10^-5

x = 1.34 x 10^-3

∴ [H₃O⁺] = 1.34 x 10^-3

Limitations of Bronsted-Lowry theory The Bronsted-Lowry theory could explain many reactions which the Arrhenius theory could not. However, it still could not account for certain acid-base reactions which do not involve the transfer of a proton.

1. Reactions between acidic oxides like CO₂, SO₂ and SO₃ and basic oxides like CaO, BaO and MgO cannot be explained by the theory of proton transfer. Such reactions take place in the absence of a solvent.

⇒ \(\mathrm{CaO}+\mathrm{SO}_3 \longrightarrow \mathrm{CaSO}_4\)

⇒ \(\mathrm{MgO}+\mathrm{CO}_2 \longrightarrow \mathrm{MgCO}_3\)

2. The acidic behaviour of compounds like BF₃ and AlCl₃ cannot be explained by the Bronsted-Lowry theory.

Lewis theory: In 1923, G N Lewis, an American chemist, extended the concept of adds and bases further.

• According to the Lewis theory, an acid is a substance which can accept a pair of electrons and a base is a substance which can donate a pair of electrons.
• While this definition encompasses most add-base reactions, it has the drawback of being so general as to indude other types of reactions in its ambit.
• According to the Lewis theory, an acid-base reaction involves the donation of a pair of electrons by a base to an add, leading to the formation of a coordinate bond between them.

Lewis bases Lewis bases can be of two types,

1. Neutral molecules like H₂O, RNH₄, NH₃ and ROH, in which
   one atom has at least one unshared pair of electrons, can act as Lewis bases,
2. All negative ions (F^–, Cl^–, OH^–, CN^–, etc.) can behave as Lewis bases.

Lewis acids Species with vacant orbitals in the valence shell of one of the atoms can act as Lewis acids. Some examples of Lewis acids are as follows.

1. Simple cations like Ag⁺, Cu²⁺ and Fe³⁺ can accept a pair of electrons and act as Lewis acids.

2. Molecules which have an atom with an incomplete octet can behave as Lewis acids, for example, BF₃ and AlCl₃.

3. Molecules in which the central atom has vacant orbitals may acquire more than an octet of valence electrons, for example, SnCl₄, SiF₄ and PF₅, and can act as Lewis acids.

⇒ \(\underset{\substack{\text { Lewis acid } \\ \text { acid }}}{\mathrm{SiF}_4}+\underset{\text { Lewis base }}{2 \mathrm{~F}^{-}} \longrightarrow \mathrm{SiF}_6^{2-}\)

4. Molecules which have a multiple bond between two atoms of different electronegativities, for example, CO₂ and SO₂, also behave as Lewis acids.

In the CO₂ (O=C=O) molecule, carbon being less electronegative than oxygen, acquires a slight positive charge (because one n electron pair shifts more towards oxygen) and can, thus, accept a pair of electrons.

⇒ \(\underset{\substack{\text { Lewis } \\ \text { acid }}}{\mathrm{CO}_2}+\underset{\substack{\text { Lewis } \\ \text { base }}}{\mathrm{OH}^{-}} \longrightarrow \mathrm{HCO}_3^{-}\)

• While it is true that the definition of acids and bases proposed by Lewis is more general than the other definitions, it does not mean that it encompasses all acid-base reactions.
• There are substances, for instance, which are Bronsted acids but not Leivis acids. Bronsted acids like HCl and H₂SO₄ can donate a proton but are not capable of accepting a pair of electrons, so they are not Lewis acids.

However, all Bronsted bases are also Lewis bases because a species capable of donating an electron pair (Lewis base) also has the tendency to accept a proton (Bronsted base).

⇒ \(\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4^{+}+\mathrm{OH}^{-}\)

Here NH₃ is a Lewis base as well as n Bronsted base but H₂O is a Bronsted acid, but not a Lewis add.

Limitations of Lewis theory The Lewis theory of adds and bases can explain the acidic and basic nature of many substances which cannot be explained by either the Bronsted-Lowry theory or the Arrhenius theory. Still it is not a flawless theory.

1. This concept of adds and bases is so general that it labels all reactions leading to the formation of coordinate bonds as acid-base reactions.
2. It does not explain the addic behaviour of adds like HCl and H₂SO₃ which do not form coordinate bonds with bases.
3. The formation of coordination compounds is usually a slow process, so add-base reactions should be slow- according to the Lewis theory. But acid-base reactions are generally fast.
4. The Lewis theory cannot be used to ascertain the relative strengths of adds and bases.
5. The catalytic property of many adds is due to the H⁺(aq) ion. Since a Lewis add need not possess hydrogen, a Lewis acid need not have this property.

While dealing with acids and bases such as HCl, CH₃COOH, NaOH and NH₄OH in aqueous solutions, the Bronsted-LowTy concept is the most suitable and hence widely accepted.

Self Ionisation Of Water

So far we have considered water as a solvent and seen that it has both acidic and basic properties, i.e., it is amphoteric in nature. We will be using the Bronsted-Low’ry concept to understand the ionisation of water. In the presence of an add, it acts as a base while in the presence of a base, it acts as an acid.

In pure water, one molecule of water can donate a proton to another molecule of water in a reaction in which water acts as acid and base simultaneously. There are always some H₃O⁺ and OH^– ions present in pure water.

or \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-}\)

This is referred to as dissociation self-ionisation or auto-ionisation of water and is characterised by the equilibrium constant,

or, \(K_{\mathrm{c}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]\left[\mathrm{H}_2 \mathrm{O}\right]}\)

Multiplying the denominator of the above expression ([H₂O]²) by Kc (as H₂O is in excess and is essentially constant), we get K=[H₂O⁺][OH^–]

K_w is called the ionic product of water. It is a constant at a particular temperature. At 298 K its value is 1.0 x 10¹⁴ mol² L^-2. Such a small value of K_w indicates that the auto-ionisation of water does not occur to a large extent.

In fact, the reverse reaction, the formation of water by the reaction of H₃O⁺ and OH^– essentially goes to completion as its equilibrium constant is

⇒ \(\frac{1}{1 \times 10^{-14}}=1 \times 10^{14}\)(As stated earlier the units have been dropped.)

K_w increases with temperature because the degree of ionisation of water increases with temperature.

In pure water both H₃O⁺ and [OH^–] are equal.

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{1 \times 10^{-14}}=1 \times 10^{-7} \text { at } 298 \mathrm{~K} \text {. }\)

We can calculate the extent of dissociation of water molecules by finding the concentration of dissociated and undissociated water.

The concentration of dissociated water is simply the concentration of [H₃O⁺] or [OH^–]

Therefore concentration of dissociated water =1 x 10^-7 M at 298 K.

The molar concentration of pure water can be calculated from its density and volume.

Now, number of moles = \(\frac{\text { mass }}{\text { molar mass }}=\frac{D \times V}{\text { molar mass }}\) (since mass = density x volume)

Concentration of undissociated water =\(\left(\frac{1000 \mathrm{~g}}{\mathrm{~L}}\right)\left(\frac{1 \mathrm{~mol}}{18 \mathrm{~g}}\right)=55.4 \mathrm{~mol} \mathrm{~L}^{-1}\)

(The density of pure water is 1000 gL^-1 and its molar mass is 18.0 g mol^-1.)

Therefore degree of dissociation = \(\frac{\text { conc. of dissociated water }}{\text { conc. of undissociated water }}\)

= \(\frac{1 \times 10^{-7}}{55.4}=18 \times 10^{-9} \approx 2 \times 10^{-9} \text { or } \frac{2}{10^9} \text {. }\)

This means that out of 10⁹ molecules of water, two molecules are ionised.

Aqueous solutions of acids and bases: The concentrations of H₃O⁺ ions and OH^– ions are the same in pure water. When some acid or base is added to water, these concentrations do not remain equal, but the value of Kw remains the same (at a particular temperature).

• If some acid is added to water, the concentration of H₃O⁺ ions will increase.
• Then according to Le Chatelier’s principle, the system will try to counteract the change and the reverse reaction will be favoured, i.e., H₃O⁺ ions will combine with OH^– ions to form water molecules.
• In other words, the equilibrium will shift backwards, so that the value of Kw may remain the same.

At the new equilibrium, the concentration of OH^– ions will be less than the concentration of H₃O⁺ ions. The concentration of OH^– ions can be obtained from the following relation.

⇒ \(\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}\)

Similarly, the addition of a base will increase the concentration of OH^– ions and decrease the concentration of H₃O⁺ ions.

The concentration of H₃O⁺ ions in an aqueous solution of a base can be obtained from the following expression.

⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]} .\)

In pure water or neutral solutions [H₃O⁺] = [OH^–].

In acidic solutions [H₃O⁺]> [OH^–].

In basic solutions [H₃O⁺]<[OH^–].

According to Arrhenius, a strong acid or a strong base completely dissociates into hydrogen ions and the corresponding anions or hydroxyl ions and the corresponding cations (as the case may be).

Thus, a 0.1-M HCl solution will dissociate to give 0.1-M H⁺ ions and 0.1-M Cl^– ions.

Example: Calculate the hydronium ion and hydroxyl ion concentrations in

1. a 0.001-M HCl solution,
2. a 0.01-M HNO₃ solution,
3. a 0.01-M NaOH solution,
4. a 0.01-M Ba(OH)₂ solution,
5. a 0.1-M HCOOH which is 10% ionised and
6. a solution containing 3.65 x 10^-3 g of HCl per 100 mL.

Solution:

1. HCl is a strong acid which ionises completely in water, so[H₃O⁺] = 0.00 1 M = 1 x 10^-3 mol L^-1.

The equilibrium can be represented as follows.

⇒ \(\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}\)

⇒ \(K_{\mathrm{w}}=\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

⇒ \({\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}=\frac{1 \times 10^{-14}}{10^{-3}}=1 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1}}\)

2. The equilibrium in this case can be represented as \(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{NO}_3^{-}\)

Being a strong acid HN03 too dissociates completely and

⇒ \({\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{HNO}_3\right]=0.01 \mathrm{M}=1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} . }\)

∴ \(\quad{\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_0 \mathrm{O}^{+}\right.}=\frac{1 \times 10^{-14}}{10^{-2}}=1 \times 10^{-12} \mathrm{~mol} \mathrm{~L}^{-1}.}\)

3. Since NaOH is a strong base it ionises completely in water.

∴ \(\quad\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=0.01 \mathrm{M}=10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\)

∴ \(\quad\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1 \times 10^{-14}}{10^{-2}}=1 \times 10^{-12} \mathrm{~mol} \mathrm{~L}^{-1}\).

4. Since Ba(OH)₂ is completely ionised and its acidity is 2, i.e.,1 mol of Ba(OH)₂ gives 2 mol of OH^– ions,

⇒ \(\mathrm{Ba}(\mathrm{OH})_2 \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\)

⇒ \({\left[\mathrm{OH}^{-}\right]=2\left[\mathrm{Ba}(\mathrm{OH})_2\right]=2 \times 0.01=0.02 \mathrm{~mol} \mathrm{~L}^{-1} .}\)

∴ \({\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1 \times 10^{-14}}{2 \times 10^{-2}}=5 \times 10^{-13} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

5. HCOOH is a weak acid whose degree of dissociation (α) is 10% (given).

⇒ \(\underset{\mathrm{C}(1-\alpha)}{\mathrm{HCOOH}}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\mathrm{C} \alpha}{\mathrm{H}_3 \mathrm{O}^{+}}+\underset{\mathrm{C} \alpha}{\mathrm{HCOO}^{-}}\)

⇒ \({\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\mathrm{C} \alpha=0.1 \times 0.1=0.01=10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

⇒ \({\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}=\frac{1 \times 10^{-14}}{10^{-2}}=10^{-12} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

6. Molarity of a solution = no. of moles in1 L of solution.

The given solution contains 3.65 x 10^-3 g of HCl in 100 mL or 0.1 L.

∴ M = \(\frac{3.65 \times 10^{-3}}{36.5 \times 0.1}\) [mol. wt. of HCl = 36.5] = 1 x 10^-3

HCl is a strong acid.

∴ \({\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=[\mathrm{HCl}]=1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

∴ \({\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{10^{-3}}=1 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

The pH scale: From our discussions on the ionisation of water so far, it must have become clear that K_w (ionic product) remains constant and that for a neutral solution [H₃O⁺] is equal to [OH^–], for an acidic solution [H₃O⁺] > [OH^–] and for a basic solution [H₃O⁺]< [OH^–].

• Surely then if we know the H₃O⁺ ion concentration in a solution we would have a fair idea about its relative acidic or basic character.
• The hydrogen ion concentration may vary from, say, 1 M in a strong acid to 10^-14 M in a strongly basic solution. Expressing H₃O⁺ ion concentrations in this fashion would be rather cumbersome.
• In 1909, P L Sorensen, a Danish chemist, proposed a more convenient way of expressing the
• H₃O⁺
• ion concentration in a solution. The scale proposed by Sorensen is called the pH (power of hydrogen) scale, which may be defined as

1. The magnitude of the negative power to which 10 must be raised to express the hydronium ion concentration; or
2. The negative logarithm (base 10) of the hydronium ion concentration in mol L^-1.

• Say the hydronium ion concentration in a solution is 10^-x Then its pH would be x, according to the first definition. This is simple enough and the first definition of pH seems to be apt.
• But what if the H₃O⁺ ion concentration in some solution is 1.5 x 10^-4 M? The first definition would not be so convenient.
• This is why the second definition is preferred. Actually, the second definition can be derived from the first.

Suppose [H₃O⁺] = 10^-x

Then log[H₃O⁺] = log 10^-x = -x log 10 = -x.

∴ x = pH = -log[H₃O⁺].

If you happen to know the pH of a solution, you could easily calculate the concentration of H₃O⁺ ions in it. [H₃O⁺]= antilog (-pH).

The hydroxyl ion concentration in a solution can be expressed in terms of pOH.

pOH = – log[OH^–].

For any solution at 298 K

K_w = [H₃O⁺][OH^–] = 10-14

∴ log[H₃O⁺] + log[OH^–] = logK_w = log 10-14 =-14

or -log[H₃O⁺]-log[OH^–] = -logK_w =14

or pH + pOH = pK_w =14.

For a neutral solution, [H₃O⁺] = [OH^–] = 10^-7.

∴ pH = -log 10^-7 =7.

This brings us to something very useful.

For an acidic solution, pH < 7.

For a basic solution, pH > 7.

For a neutral solution, pH = 7.

For practical purposes, the pH range is taken as 0 to 14, though theoretically, it may be possible to have H₃O⁺ ion concentrations of more than 10° M (i.e., 1 M) or less than 10^-14 M.

The table gives the pH values of some common substances in everyday life.

Example 1. Calculate the pH value of a solution of

1. 0.01-M HCl and
2. 0.01-M NaOH.

Solution:

1. \(\mathrm{HCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

Since HCl is completely ionised,

[H₃O⁺] = [HCl] = 0.01 M = 1 x 10^-2 mol L^-1.

pH = – log[H₃O⁺] = – log 1 x 10^-2 = – [-2] = 2.

2. \(\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)

NaOH is a strong base which ionises completely.

[OH^–] = [NaOH] = 0.01 M = 10^-2 mol L^-1.

∴ \(K_{\mathrm{w}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\)

∴  \(\quad {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1 \times 10^{-14}}{1 \times 10^{-2}}=10^{-12} \mathrm{~mol} \mathrm{~L}^{-1} . }\)

∴ \(\quad \mathrm{pH}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log 10^{-12}=-[-12]=12\).

Example 2. 1000 mL of a solution contains 6.3 g of nitric acid. What is the pH of the solution if the acid is completely dissociated?
Solution:

The concentration of HNO₃ = 6.3 g L^-1.

Molecular weight of HNO₃ = 63.

∴ M = 6.3/6.3 = 0.1.

⇒ \(\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{NO}_3^{-}\)

Since the acid is completely dissociated

[H₃O⁺] = [HNO₃]=0.1M = 10^-1 M.

pH = – log[H₃O⁺] = – log 10^-1 = 1.

• The pH scale is very useful in getting an idea about the relative acidic (or basic) strengths of solutions. The higher the pH, the lower the concentration of hydronium ions in the solution.
• If the pH of a solution changes by one unit, there is a tenfold change in the molar concentration of that solution. Here it is important for you to know that pH is a measure of ‘hydrogen ion activity’.
• At the present level of learning, for the sake of simplicity, we have been using hydrogen ion concentration as a measure of pH since we are considering ideal (dilute) behavior of solutions.
• However, many solutions deviate from ideal behaviour with rise in concentration. To overcome this problem, we use the concept of activity.
• Activity is a physical quantity, a thermodynamic function that can be used in place of concentration. Now consider a concentrated solution, for example, 1 M HCl.

Let us use both its activity and molar concentration to determine its pH.

⇒ \(a_{\mathrm{H}_3 \mathrm{O}^{+}} \text {(activity) }=\gamma\left[\mathrm{H}_3 \mathrm{O}^{+} \right.\)., where y is the activity coefficient, a dimensionless quantity.

The activity being 0.81 the pH value of the solution turns out to be 0.092. On the other hand, if we use the concentration, pH = – log[1] = 0.

The ‘p’ in pH indicates (-log) so that px means -log(x). This is very useful in calculations involving exponential numbers so that it has been extended to other species. The most widely used are pK_w, pK_a, and pK_b.

PK_a = -logK_a; pK_b =-logK_b.

As you already know, for a conjugate acid-base pair, K_a x K_b = K_w.

Taking negative logarithm on both sides, (- log K_a) + (- log K_b ) = (- log K_w)

or PK_a+pK_b =pK_w.

Measuring pH The pH of a solution can be measured with the help of pH papers, indicators, or pH meters. While pH papers and indicators give an approximate pH value of the solution in question, pH meters give the exact value.

• An acid-base indicator is a substance that changes colour in a specific pH range of about 2 pH units. To determine the pH of a solution, a few drops of the indicator are added to the solution.
• To make the determination easier, nowadays a universal indicator is available, which is simply a mixture of several indicators, to make approximate measurements in the pH range 3-10.
• The color change on the addition of the indicator is noted and compared with a color chart to know the pH of the given solution.
• Nowadays, pH meters are available for commercial purposes. They measure pH with high precision.

A typical pH meter consists of a special measuring probe (a glass electrode) connected to an electronic meter that measures and displays the pH reading.

Hydrolysis Of Salts

The term hydrolysis refers to any reaction in which water is one of the reactants. Water, being an amphiprotic solvent, can act both as an acid or a base. When a salt is dissolved in water, it dissociates into its ions. If the salt is represented as MA, then we have

⇒ \(\mathrm{MA}(\mathrm{aq}) \longrightarrow \mathrm{M}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})\)

Now, water may interact with the ions in two ways:

⇒ \(\underset{\text { acid }}{\mathrm{M}^{+}(\mathrm{aq})}+\underset{\text { base }}{2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})} \underset{\text { base }}{\mathrm{MOH}(\mathrm{aq})}+\underset{\text { adid }}{\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})}\) ….(1)

⇒ \(\underset{\text { base }}{\mathrm{A}^{-}(\mathrm{aq})}+\underset{\text { acid }}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})} \underset{\text { add }}{\mathrm{HA}(\mathrm{aq})}+\underset{\text { base }}{\mathrm{OH}^{-}(\mathrm{aq})}\)…..(2)

As you can see in reaction (1), the base from which the cation is obtained is generated while in reaction (2), the acid from which the anion is obtained is generated. The salt MA is obtained by a neutralisation reaction between the base, MOH, and the acid, HA.

• A closer look at the reverse reactions shows that M⁺ is the conjugate acid of the base MOH while A^– is the conjugate base of the acid HA.
• On dissolving a particular salt in water, whether reaction (1) or (2) or both will occur and to what extent is dependent upon the strengths of the acid and the base from which the salt has been formed.
• The strength of an acid is inversely proportional to that of its conjugate form. If HA is a strong acid, its conjugate base, A^–, will be weak and hence reaction (2) will occur only insignificantly.
• Similarly, if the base MOH is strong, its conjugate acid M⁺ will be weak and reaction (1) will occur to a very small extent.
• If both M⁺ and A^– are obtained from a strong base and a strong acid respectively, hydrolysis occurs to a negligible extent or we may say that it does not occur at all. One such example is NaCl.
• Na⁺ is obtained from NaOH, a strong base, and Cl^– is obtained from HCl, a strong acid.
• A solution of NaCl will not show any hydrolysis and the species present in solution will be only Na⁺ (aq), Cl^–(aq), and H₂O. Hence the pH will be 7 and the solution is neutral.

When a salt dissociates into ions on dissolution in water, the ion which is the conjugate of a strong acid/base does not hydrolyse.

For example, sodium acetate, CH₃COONa (formed from a strong base and a weak acid), will dissociate in water as \(\mathrm{CH}_3 \mathrm{COONa}(\mathrm{aq}) \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}).\)

Here CH₃COO^– is the conjugate base of the weak acid CH₃COOH. Therefore it will hydrolyse forming OH ions in the solution.

⇒ \(\mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)….(3)

However, Na⁺ will not hydrolyse as it is obtained from a strong base, NaOH. Therefore, the solution will be basic due to the hydrolysis reaction (3), in which hydroxyl ions are produced.

Similarly, NH₄Cl (formed from a strong acid and a weak base) dissociates as \(\mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq}) \rightleftharpoons \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

In this case NH₄⁺ is the conjugate acid of the weak base NH₃ and hence undergoes hydrolysis as \(\underset{\text { acid }}{\mathrm{NH}_4^{+}(\mathrm{aq})}+\underset{\text { base }}{2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})} \rightleftharpoons \underset{\text { base }}{\mathrm{NH}_4 \mathrm{OH}(\mathrm{aq})}+\underset{\text { acid }}{\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})}\)…(4)

Cl^– will not undergo hydrolysis as it is obtained from HCl (strong acid). As a result, a solution of NH₄Cl will be acidic due to reaction (4), where H₃O⁺ is produced.

• Now let us see what happens when ammonium acetate, CH₃COONH₄, is dissolved in water.
• Since the salt is formed from the weak acid CH₃COOH and the weak base NH4OH, both the ions (CH₃COO^– and NH₄) formed will hydrolyse.
• Since both OH^– and H₃O⁺ are produced during hydrolysis, the pH of the solution will depend on the comparative strengths of the parent acid (K_a) and base (K_b) (from which the salt is formed).
• If K_a < K_b, the solution will be alkaline as then the conjugate base of the acid (A^–) is stronger and will hydrolyse more as compared to the conjugate acid of the base (M⁺).

If K_a >K_b, the solution will be acidic. However, if K_a = K_b, the solution will be neutral. It is interesting to note that the pH of such solutions is independent of the concentration of the salt and is given by

pH=7 + 1/2(pK_a-pK_b)

Example 1. The pH of a 0.1-M solution of aniline, a weak base, is 8.8. Find its pK_b.
Solution:

⇒ \(\underset{\text { anline }}(\mathrm{B})+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BH}^{+}+\mathrm{OH}^{-}\)

Initial conc.(M) C 0 0

Change -x +x +x

Equilibrium cone. C-x x x

Since \(K_{\mathrm{b}}=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}\),

∴ \(K_b=\frac{x^2}{C-x}\).

The pH of the solution = – log[H₃O⁺] = 8.8.

∴ [H₃O⁺] = antilog (-8.8) =16 x 10^-9 M.

Since we know that \(\left[\mathrm{OH}^{-}\right]=\frac{K_w}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]},\)

⇒ \(\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{16 \times 10^{-9}}=6.25 \times 10^{-6}=x\)

Substituting the values of x and C in the expression for K_b we get

∴\(K_{\mathrm{b}}=\frac{\left(6.25 \times 10^{-6}\right)^2}{0.1-6.25 \times 10^{-6}}\)

Since 6.25 x 10^-6 <<0.1, 0.1- 6.25 x 10^-6 = 0.1

∴ \(K_{\mathrm{b}}=\frac{\left(6.25 \times 10^{-6}\right)^2}{0.1}=3.9 \times 10^{-10}\).

pK_b =- log K_b =-log3.9 x 10^-10

pK_b =9.4

Example 2. The pK_a of hypobromous acid (HOBr) is 8.7. Calculate the pH of a 0.1-M solution of the acid.
Solution:

⇒ \(\mathrm{HA}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-}\)

Equilibrium cone. (M) C- x x x

Since \(K_4=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{x^2}{C-x}\),

pK_a =-log K_a =8.7.

K_a =antilog(-8.7) = 2 x 10^-9

Now substituting the value of K_a and C in the expression for K_a, we get

2 x 10^-9 = \(\frac{x^2}{0.1-x}\)

As K_a is small, 0.1 >> x.

∴ 2 x 10^-9 = \(\frac{x^2}{0.1}.\)

or x² =2×10^-9– x 0.1 = 2 x 10^-10

or x \(=\sqrt{2 \times 10^{-10}}=1.4 \times 10^{-5} \mathrm{M} .\)

This is the concentration of H₃O⁺ ions

[H₃O⁺]=x=1.4 X 10^-5 M.

pH =- log[H₃O⁺] = 4.8.

The pH of a 0.1-M solution of HOBr is 4.8.

Example 3. Calculate the degree of ionisation of 0.1-M acetic acid in the presence of 0.1-M HCl The pK_a of acetic acid is 4.74. Compare it with its ionisation in pure water.
Solution:

⇒ \(\mathrm{HAc}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Ac}^{-}\)

⇒ \(K_{\mathrm{a}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{Ac}^{-}\right]}{[\mathrm{HAc}]}\)

or \(\mathrm{p} K_{\mathrm{a}}=-\log K_{\mathrm{a}}=4.74\)

or K_a = antilog(-4.74) =1.8 x 10^-5.

Let us first calculate α in pure water.

If the equilibrium concentration of [H₃O⁺] = Cα then that of Ac’ is also Cα and the concentration of acetic acid that remains undissociated is [HAc]= C(1 – α).

∴ \(K_4=\frac{(\mathrm{C} \alpha)(\mathrm{C} \alpha)}{C(1-\alpha)}\)

Since acetic add is a weak acid, α << 1

Hence \(K_{\mathrm{a}} \cong \frac{C^2 \alpha^2}{C}=C \alpha^2\)

or \(\alpha=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{18 \times 10^{-5}}{0.1}}=0.013 .\)

Therefore the percentage dissociation is 1.3%. In the presence of 0.01-M HCl, the source of H₃O⁺ is acetic acid as well as hydrochloric acid. The equilibrium concentrations are as shown below.

⇒ \(\begin{array}{ll}
\mathrm{HAc}+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Ac}^{-} \\
\mathrm{C}(1-\alpha) & \mathrm{C} \alpha+0.01 \quad \mathrm{C} \alpha
\end{array}\)

Now, \(K_{\mathrm{a}}=\frac{(C \alpha+0.01)(C \alpha)}{C(1-\alpha)}\)

We may neglect α in comparison to 1.

∴ \(\quad K_a \cong \frac{C^2 \alpha^2+0.01 C \alpha}{C} \)

or \(K_a=C \alpha^2+0.01 \alpha\)

or \(C \alpha^2+0.01 \alpha-K_a=0\).

This is a quadratic equation and a can be calculated using the formula

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

∴\(\quad \alpha =\frac{-0.01 \pm \sqrt{(0.01)^2+4 \times C \times K_a}}{2 C}\).

Substituting for K_a and C, we get

α = \(\frac{-0.01 \pm \sqrt{(0.01)^2+4 \times 0.1 \times 18 \times 10^{-5}}}{2 \times 0.1}=\frac{-0.01 \pm 0.0104}{0.2} .\)

Discarding the negative root, a = 2 x 10^-3 = 0.002

Percentage dissociation = 0.002 x 100 = 0.2%

This is far less than that in pure water. This shows that the equilibrium is affected in the presence of a common ion.

Buffer Solutions

Suppose you have a solution of a particular pH or a solution which has a certain concentration of H₃O⁺ ions. If you add even a small amount of add or base to this solution, its pH will change—it will decrease if you add an add and increase if you add a base.

Both in nature and in industries, it is necessary to have solutions whose pH values do not change with the addition of a small amount of an add or a base.

The pH value of an ordinary solution may change even if it comes in contact with air. It can absorb carbon dioxide from the air and become more addic, for example.

• A solution which can resist change in pH when a small amount of an acid or a base is added to it (or when it is diluted) is called a buffer solution. The ability of such a solution to resist change in pH is called buffer action.
• A buffer solution can be acidic. Such a solution contains a weak acid and its salt with a strong base in equimolar quantities (a weak acid and a salt formed from its conjugate base), example, a solution of acetic acid and sodium acetate, which acts as a buffer solution around pH 4.75.
• A basic buffer solution, on the other hand, contains equimolar quantities of a weak base and its salt with a strong acid (a weak base and a salt formed from its conjugate base), example, NH₄OH and NH₄Cl, which acts as a buffer around pH 9.25.
• A buffer solution may also contain a single substance, generally the salt of a weak acid and a weak base, example, ammonium acetate (CH₃COONH₄).
• In order to understand how a buffer solution maintains a constant pH value, let us consider a solution containing equimolar quantities of CH₃COOH (weak acid) and CH₃COONa (a salt of acetic acid with a strong base).

In an aqueous solution acetic acid will be weakly ionised and sodium acetate will be mostly dissociated.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})\)

⇒ \(\mathrm{CH}_3 \mathrm{COONa}(\mathrm{aq}) \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq})\)

The solution will contain a large concentration of Na⁺ and CH₃COO^– ions, a very small concentration of H₃O⁺ ions, and a large amount of undissociated acetic acid molecules.

If a few drops of an acid are added to this solution, the H₃O⁺ ions provided by the acid disturb the equilibrium (between ionised and un-ionised acid), and CH₃COO^– ions in the solution combine with the H₃O⁺ ions to counter the change.

⇒ \(\underset{\text { from buffer }}{\mathrm{CH}_3 \mathrm{COO}^{-}(\mathrm{aq})}+\underset{\text { from acid }}{\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Thus, the H₃O⁺ ions provided by the acid are neutralised by CH₃COO ions from the buffer solution and there is no change in pH.

If a few drops of a base are added to the solution, the OH” ions from the base combine with the H₃O⁺ ions of the solution to form water.

⇒ \(\mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

• This disturbs the equilibrium (by reducing the concentration of H₃O⁺ ions) and in accordance with Le Chatelier’s principle, more CH₃COOH molecules ionise, restoring the original concentration of H₃O⁺ ions or the pH.
• Blood is a natural buffer solution whose pH is about 7.4. The pH of sea water is also almost constant.
• Buffer solutions are required for electroplating and in the manufacture of photographic material and dyes, among many other industrial processes.
• The pH of culture media in biological laboratories also need to be resistant to change. The buffer solution of desired pH can be prepared by taking appropriate ratios of salt and acid or salt and base and by knowing the respective pKa and pKb values.
• An equation which is used to calculate the pH of a buffer solution is as follows.

pH = \(\mathrm{p} K_{\mathrm{a}}+\log \frac{[\text { salt }]}{\text { [acid] }}\) (for acidic buffer)

or pH = \(\mathrm{p} K_{\mathrm{a}}+\log \frac{[\text { base }]}{[\text { salt }]}\)
(for basic buffer)

This is called the Henderson-Hasselbalch equation. In the equation [acid], [base], and [salt] denote the concentrations of the acid, base, and salt respectively used to prepare the buffer solution.

Note that, in both the equations, pK_a appears and it may be calculated for the conjugate acid of the base by using the expression pK_a =pK_w -pK_b.

Example 1. Calculate the pH of an acetic acid-sodium acetate buffer in which the concentrations of the acid and salt are 0.5 M and 0.25 M respectively. K_a = 1.8x 10^-5 for acetic acid.
Solution:

Using the Henderson-Hasselbalch equation,

pH = \(\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt] }}{\text { [acid] }}\)

pH = \(-\log \left(18 \times 10^{-5}\right)+\log \frac{0.25}{0.5}\) = 4.74 – 0.3010 = 4.44.

Example 2. Find the pH of an ammonia-ammonium chloride buffer solution in which the concentrations are \(C_{\mathrm{NH}_3}\)=0.2M and \(C_{\mathrm{NH}_4^{+}}\)= 0.3 M. Given that K_a for NH₃ = 1. 76 x 10^-5.
Solution:

pK_a + pK_b=pK_w

or pK_a=pK_w-pK_b

pK_a = 14 – (- log 1.76 x 10^-5) = 14-4.75 = 9.25.

Using this value of pK_a in the equation

pH = \(\mathrm{p} K_{\mathrm{a}}+\log \frac{[\text { base] }}{[\text { salt }]}\)

pH = \(9.25+\log \frac{[0.2]}{[0.3]}\) = 9.25-0.18 = 9.07.

Example 3. Determine the pH of a buffer solution containing 0.03-M boric acid and 0.043-M sodium borate, given that pK_a for B(OH)₃ =9.00.
Solution:

pH = \(p K_{\mathrm{a}}+\log \frac{\text { [salt] }}{\text { [acid] }}\)

= \(9+\log \frac{0.04}{0.03}\) = 9 + 0.12 = 9.12.

Solubility Equilibria

The solubility of different salts in water or any other solvent varies to a large extent. For instance, calcium chloride is hygroscopic (absorbs water vapour from the atmosphere) in nature, and, on the other hand, lithium fluoride is almost insoluble.

There are certain factors which influence the solubility of salts in solvents. Common salt dissolves readily in water because the energies of coulmbic or ionic interaction between the two ions of a salt are overcome due to the high dielectric constant of water.

• Thus, in water Na⁺ and C- exist freely— the two do not attract each other. The dissolution of a salt in water has the effect of reducing the force between the ions, which then separate as a consequence.
• Ion solvation (solvation enthalpy—energy released in the process of solvation) also affects tire solubility of a salt in water or any other solvent. An ion is solvated when surrounded by several molecules of solvent.
• Solvation enthalpy is more for polar solvents and less for nonpolar solvents. Thus, salts do not dissolve in nonpolar solvents as the solvation enthalpy is not enough to overcome lattice enthalpy. Different salts exhibit different solubilities at different temperatures.
• Salts are divided into three categories based on solubility. Salts of solubility 0.1 M or greater are considered soluble.
• Salts of solubility between 0.1 M and 0.01 M are considered arc considered slightly soluble. And salts of solubility less than 0.01 M, though called insoluble, are actually sparingly soluble.
• A solid dissolves in a solvent until the solution and the solid are in equilibrium. The solution, at equilibrium, is saturated and its molar concentration is the molar solubility of the solid.

Solubility product: BaSO₄ and AgCl arc sparingly soluble in water. When such a sparingly soluble salt is mixed with water, a small
amount of it dissolves and makes the solution saturated.

• The rest of it remains undissolved and an equilibrium is set up between the undissolved salt and the salt ions in solution.
• At equilibrium, the rate of dissolution of ions from the undissolved solid is equal to the rate of precipitation of ions from the saturated solution.
• You may have noticed that we are talking about an equilibrium between the undissolved solid and the ions in solution, and not the molecules in solution.

This is because the small amount of the salt that dissolves gets completely dissociated into ions. The equilibrium can be represented as follows.

⇒ \(\underset{\text { or electrolyte }}Undissolved salt \rightleftharpoons ions in solution\)

Let us consider solid AgCl in contact with its saturated aqueous solution. The dissolution of the salt may be represented as

⇒ \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

As in the other cases we have considered before, the equilibrium constant would be K= \(\frac{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]}{[\mathrm{AgCl}]}\)

The concentration of the undissolved salt is constant (it can be taken as 1) irrespective of the amount of the solid salt present. (This is true for any pure solid.)

Then K[AgCl] = [Ag⁺][C^–] =K_sp

This constant (K_sp) is called the solubility product. In general, if a sparingly soluble salt A_x B_y dissociates in water, the equilibrium thus set up can be represented as

⇒ \(\mathrm{A}_x \mathrm{~B}_y \rightleftharpoons x \mathrm{~A}^{y+}+y \mathrm{~B}^{x-}\).

The solubility product may be expressed as K_sp = [A^y+]^x[B^x-]^y

The solubility product of a sparingly soluble salt (or electrolyte) at a given temperature may be defined as the product of the molar concentrations of its ions in a saturated solution, with each concentration term raised to the power equal to the number of ions of that species produced by the dissociation of one molecule of the electrolyte.

A few examples will make this clear.

⇒ \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2: K_{\text {sp }}  =\left[\mathrm{Ca}^{2+}\right]^3\left[\mathrm{PO}_4^{3-}\right]^2\)

⇒ \(\mathrm{BaSO}_4: K_{\text {sp }} =\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]\)

⇒ \(\mathrm{Mg}\left(\mathrm{OH}_2\right): K_{\text {sp }}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2\)

⇒ \(\mathrm{Ag}_2 \mathrm{CrO}_4: K_{\text {sp }} =\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right]\)

• It is worthwhile to mention here that ionic product (or Q_sp) is the product of the concentration of ions in any solution raised to their respective stoichiometries.
• It may have struck you that both solubility product and ionic products are products of the concentrations of ions in a solution (not considering the powers of the concentration terms for the moment).
• However, the ionic product has a more general application. It applies to both saturated and unsaturated solutions, while the solubility product applies only to saturated solutions.

1. The ionic product is the same as the solubility product in a saturated solution.
2. If the solubility product is greater than the ionic product, the solution is unsaturated.
3. The higher the value of the solubility product the greater is the solubility of a salt.
4. The ionic product cannot be greater than the solubility product. When it tends to exceed the value of the solubility product, precipitation (or combination of ions) starts.

If you know the solubility of a sparingly soluble salt at a particular temperature, you can easily calculate its solubility product.

Example 1. Find the solubility product of AgCl at T°C if its solubility at this temperature is 1.08 x 10^-5 mol L^-1.
Solution:

⇒ \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

It is given that the solubility of AgCl is 1.08×10^-5 mol L^-1. AgCl is a sparingly soluble salt and it dissociates completely.

That is to say, 1 mol of AgCl in solution dissociates completely to produce 1 mol of Ag⁺ and 1 mol of Cl^– ions.

Therefore, 1.08 x 10^-5 mol of AgCl will give 108 x 10^-5 mol of Ag⁺ and 1.08 x 10^-5 mol of Cl^– ions,

or [Ag⁺] = 1.08 x 10^-5 mol L^-1

and [Cl^–] = 1.08x 10^-5 mol L^-1

= (1.08 x 10^-5 )(1.08 x 10^-5) = 1.16 x 10^-10.

Example 2. The solubility of PbCl₂ at a particular temperature is 2.2 x 10^-2 mol L^-1. Find its solubility product at this temperature.
Solution:

⇒ \(\mathrm{PbCl}_2 \rightleftharpoons \mathrm{Pb}^{2+}+2 \mathrm{Cl}^{-}\)

PbCl₂ is a sparingly soluble salt, so it is completely ionised in solution.

In other words, 1 mol of PbCl₂ in solution will dissociate to produce 1 mol of Pb⁺ and 2 mol of Cl^– ions. It is given that the solubility of PbCl₂ is 2.2 x 10^-2 mol L^-1.

[Pb²⁺] = 2.2 x 10^-2 mol L^-1

and [Cl^–] = 2 x 2.2 x 10^-2 mol L^-1.

∴ K_sp =[Pb²⁺][Cl^–]² = [2.2 x 10^-2][2 x 2.2 x 10^-2]² =4.3×10^-5.

Calculation of solubility: The solubility of a salt is defined as grams of solute that can be dissolved in 100 mL of the solvent. The solubility of a sparingly soluble salt can be calculated if its solubility product is known.

• However, since K_sp is obtained considering the concentration in moles per litre (M), solubility is also obtained in moles per litre.
• This is called molar solubility—the amount of n salt that dissolves in a requisite amount of water to produce one litre of a saturated solution.
• The following examples illustrate the calculation of molar solubility from solubility product.

Example 1. The solubility product of Ag₂CrO₄ at 298 K is 4 x 10^-12. Find its solubility at this temperature.
Solution:

At equilibrium \(\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{2-}\)

Suppose the solubility of Ag₂CrO₄ is x mol L^-1

Then [Ag⁺]=2x and [CrO₄^2-] = x. (Ag₂CrO₄ forms 2 moles of Ag⁺ and one mole of CrO₄^2-)

∴ \(K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right]=[2 x]^2[x]=4 x^3\)

Given that K_sp = 4 x 10^-12

or 4x³ =4 x 10^-12

∴ x = 1 x 10^-4

Therefore the solubility of Ag₂CrO₄ is 1 x 10^-4 mol L^-1.

Example 2. The solubility product of AgBr in water is 2.5 x 10^-13. Calculate its solubility in a 0.01-M NaBr solution.
Solution:

NaBr dissociates completely in solution.

∴ \(\left[\mathrm{Br}^{-}\right]=[\mathrm{NaBr}]_0=0.01 \mathrm{M}\) (The subscript ‘0’ denotes the initial concentration.)

Let the solubility of AgBr be x mol L^-1

Then [Ag⁺] = [Br^–] = x mol L^-1 (from the dissociation of AgBr)

∴ total[Br] = x + 0.01 M (x <<0.01 M; AgBr being sparingly soluble)^–

K_sp(AgBr) = [Ag⁺][B^–r] = x x 0.01

Given that K_sp(AgBr) = 2.5 x 10^-13

∴ x x 0.01 = 2.5 x 10^-13

or x = \(\frac{2.5 \times 10^{-13}}{0.01}=2.5 \times 10^{-11} \mathrm{~mol} \mathrm{~L}^{-1}\)

Factors affecting solubility: The principle that an equilibrium counteracts the change due to the addition of species to solutions is also applicable to solubility product constants.

Common-ion effect: The presence of a common ion can also affect the solubility equilibrium. Let us consider the dissolution of MgF₂ in water.

⇒ \(\mathrm{MgF}_2(\mathrm{~s}) \rightleftharpoons \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq})\)

Let x be the solubility of MgF₂ in moles per litre. Then the concentrations of Mg²⁺ and F^– are as follows:

[Mg²⁺] = x mol L^-1 (as one mole of MgF₂ gives one mole of Mg²⁺ and 2 moles of F^–).

[F^–] = 2 xmol L^-1.

∴ K_sp =[Mg²⁺][F^–]² = x x(2x)² =4x³.

But the solubility product of MgF₂ is 7.4 x 10^-11

∴ 7.4 x 10^-11 = 4x³

or \(x^3=\frac{7.4 \times 10^{-11}}{4}=1.8 \times 10^{-11} \Rightarrow x=\sqrt[3]{1.8 \times 10^{-11}}=2.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)

∴ \({\left[\mathrm{Mg}^{2+}\right]=2.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

Molar solubility = 2.6 x 10^-4 M.

∴ [F^–] = 2 x 2.6 x 10^-4 =5.2×10^-4M.

When NaF is added to a solution of MgF₂, F^– being the common ion, its concentration in solution increases.

In order to keep K_sp constant, [Mg²⁺] must become smaller, or in other words, MgF₂ is less soluble in an NaF solution than it is in pure water.

The presence of a common ion shifts the equilibrium to the left. This may be seen clearly from the following solved example.

Example: Calculate the molar solubility of MgF₂ in 0.05-M NaF at 298 K, given that K_spof MgF₂ is 7.4 x 10^-11.
Solution:

We have calculated the molar solubility of MgF₂ in pure water, which is 2.6×10^-4 M. Let us now calculate its solubility in NaF.

If we assume x to be the molar solubility of MgF₂ then one mole of MgF₂ gives 1 mol of Mg²⁺ ions and 2 moles of F^– ions.

Hence the concentrations of Mg²⁺ and F^– from MgF₂ alone are x and 2x respectively. There is another source of F^–, which is NaF. From 0.05 M of NaF, we get 0.05 M of F^– ions.

Equilibrium conc. \(\mathrm{MgF}_2(\mathrm{~s}) \rightleftharpoons \underset{x}{\mathrm{Mg}^{2+}(\mathrm{aq})}+\underset{2 x+0.05}{2 \mathrm{~F}^{-}(\mathrm{aq})}\)

Hence [F^–] = 2x+ 0.05 and [Mg²⁺] = x.

But K_sp =[Mg²⁺][F^–]² or 7.4 x 10^-11 = x x (2x + 0.05)²,

As K_sp is smqll 2x <<0.05.

Hence 7.4 x 10^-11 s x x (0.05)² or x = 7.4 x 10^-9 M.

The molar solubility of MgF₂ has decreased from 2.6 x 10^-4 M in pure water to 7.4 x 10^-9 M in the presence of NaF due to the common-ion effect.

pH of a solution: The solubility of salts of weak acids increases as the pH of the solution decreases or the acidity increases.

For example, the solubility of CaCO₃ increases with decreasing pH.

This is because the CO₃^2- ions react with H⁺ ions forming HCO₃^– ions react with H and thus are removed from the solution. As a result the reaction moves in the forward direction, dissolving more CaC03.

⇒ \(\begin{aligned}
\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{CO}_3^{2-}(\mathrm{aq}) \\
\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CO}_3^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{HCO}_3^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
\hline
\mathrm{CaCO}_3(\mathrm{~s})+\mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{HCO}_3^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
\hline
\end{aligned}\)

Salts containing anions such as \(\mathrm{PO}_4^{3-}, \mathrm{S}^{2-}, \mathrm{F}^{-} \text {and } \mathrm{CN}^{-}\) also behave in a similar fashion.

Predicting the precipitation of a salt: If the value of the ionic product exceeds that of the solubility product, precipitation of the excess ions occurs.

So, if you know the solubility product of a sparingly soluble salt, you will be able to predict whether or not mixing two solutions containing known concentrations of Us Ions will result in the precipitation of the salt.

Example 1. Predict whether precipitation will occur when equal volumes of a 0.02-M Na₂SO₄ solution and a 0.02-M BaC₂ solution are mixed.(K_sp of BaSO₄ = 1.5 x 10^-10)
Solution:

Let the volume of each solution = V mL

Total volume after mixing = 2V mL

Applying the relation \(\begin{aligned}
M_1 V_1=M_2 V_2 \\
0.02 \times V=M_2 \times 2 V
\end{aligned}\)

∴ after mixing [BaCl₂]₀ = 0.01 mol L^-1

Since BaCl₂, is completely ionised,

[Ba²⁺] =[BaCl₂]₀ = 0.01 mol L^-1

To determine the molarity of the Na₂SO₄ solution after mixing, we use the equation

M’₁V’₁(before mixing) = M’₂V’₂(after mixing)

∴ \(M_2^{\prime}=\frac{0.02 \times V}{2 V}=0.01\)

∴ after mixing[Na₂SO₄]₀ = 0.01 mol L^-1

Since Na₂SO₄is completely ionised,

[SO₄^2-] = [Na₂SO₄] = 0.01 mol L^-1

Ionic product of BaSO₄, LP. or Q_sp = [Ba²⁺][SO₄^2-] = 0.01 x 0.01 = 10^-4 mol L^-1

As Q_sp >  K_sp precipitation occurs.

Precipitation of soluble salts: The precipitation of soluble salts from their saturated solutions is called salting out This process is used to obtain pure sodium chloride from a saturated solution of impure sodium chloride.

• HCl gas is passed through a saturated solution of NaCL The addition of HCl (which ionises) increases the concentration of Cl^– ions.
• Consequently, the ink product exceeds the solubility product of NaCl and precipitation occurs. The salt which is precipitated is in the pure stale. The impurities remain in solution.
• Soap is salted out from its saturated solution by the addition of sodium chloride.
• Soaps are sodium salts of ratty acids. The addition of sodium chloride increases the concentration of Na⁺ ions in the solution, thus making the ionic product exceed the solubility product

Example 1. Calculate the solubility of BaSO₄ in pure water and in sea -water which contains 0.029-M SO₄^2- ions. [K_sp(BaSO₄) = 9.1x 10^-11n. ]
Solution:

⇒ \(\mathrm{BaSO}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})\)

Let x be the molar solubility of BaSO₄.

∴ \(\left[\mathrm{Ba}^{2-}\right]=x ;\left[\mathrm{SO}_4^{2-}\right]=x\)

⇒ [latexK_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]=x^2 .[/latex]

But \(9.1 \times 10^{-11}=x^2 \text { or } x=\sqrt{9.1 \times 10^{-11}}=9.5 \times 10^{-6}\) .

Hence the solubility of BaSO₄ in pure water = 9.5 x 10^-6 M

In sea water, [SO₄^2-] = 0.029 M.

Total concentration of SO₄^2- = 0.029 + x’, where x’ is the solubility of BaSO₄ in sea water.

Now, K_sp = (x’)(0.029 + x’)

We may assume that x'<< 0.029 as K_sp is small.

∴ K_sp =(x’)(0.029)

∴or 9.1 x 10^-11 = 0.29x’

or x’ = 3.1x 10^-10

Therefore, the solubility of BaSO₄ in sea water is 3.1 x 10^-10 M. This is far less than that in pure water.

Example 2. A carbonated drink was saturated with CO₂ at 0°C and 5 atm pressure. What was the concentration of CO₂ in the bottled drink if the Henry constant for an aqueous solution of CO₂ at 0°C is 7.5 x 10^-2 mol L^-1 atm^-1?
Solution:

According to Henry’s law, C = kp,

where C is the concentration of a gas in a solution, p is the partial pressure of the gas and k is the Henry constant.

∴ concentration of CO₂ in the drink = kp = 7.5 x 10^-2 x5

=37.5 x 10^-2 =0.38 mol L^-1.

Example 3. Calculate the equilibrium constantfor a reaction given that the rate constantsfor thefonvard and reverse reactions are 2.48 x 10^-4 and 8.25 x10^-5 respectively.
Solution:

Equilibrium constant K = \(\frac{k}{k^{\prime}}\),

where k and k’ are the rate constants for the forward and the reverse reaction respectively.

∴ K = \(=\frac{2.48 \times 10^{-4}}{8.25 \times 10^{-5}}=3\)

Example 4. Two moles of PCl₅ was heated to 327°C in a closed vessel of volume 2 L. When equilibrium was established, 40% of the PCl₅ had dissociated into PCI₃ and Cl₂ Calculate the equilibrium constantfor the reaction.
Solution:

The dissociation of PCl₅ into PCI₃ and Cl₂ can be represented as follows.

⇒ \(\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2\)

Initial concentration of PCl₅ = 2 mol.

Percentage dissociation at equilibrium = 40%.

PCI₅ dissociated at equilibrium = 40% of 2 mol = 0.8 mol.

∴ amount of PCI₅ at equilibrium = 2- 0.8 =12 mol

and amount of PCI₃ and Cl₂ at equilibrium = 0.8 mol.

Since the volume of the vessel is 2 L, at equilibrium

⇒ \(\left[\mathrm{PCl}_5\right]=\frac{1.2}{2} \mathrm{~mol} \mathrm{~L}^{-1}\)

⇒ \(\left[\mathrm{PCl}_3\right]=\frac{0.8}{2} \mathrm{~mol} \mathrm{~L}^{-1}\).

⇒ \(\left[\mathrm{Cl}_2\right]=\frac{0.8}{2} \mathrm{~mol} \mathrm{~L}^{-1}\).

∴ K=\(\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{0.4 \times 0.4}{0.6}=0.267\).

Example 5. 0.1 mol of PCl₅ is heated in a 1-L flask at 250°C. Calculate the equilibrium concentrations of PCl₅, PCl₃ and Cl₂, if the equilibrium constant for the dissociation of PCl₅ is 0.0414.
Solution:

PCI₅ dissociates according to the following reaction.

∴ \(\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2\)

Suppose at equilibrium x mol of PCl₅ dissociates to form x mol of PCl₃ and x mol of Cl₂. Since the volume of the vessel is 1 L, at equilibrium,

⇒ [PCl₅] = 0.1 – x mol L^-1.

⇒ [PCl₃] = xmolL

⇒ [Cl₂] = x mol L^-1.

∴ K = \(\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{x \times x}{0.1-x}\)

Given that K = 0.0414.

∴ 0.0414 = \(\frac{x^2}{0.1-x}\)

or x² +0.0414x-0.00414 = 0

This is a quadratic equation and x can be calculated using the formula

x =\(-\frac{b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-0.0414 \pm \sqrt{(0.0414)^2-4 \times 1 \times(-0.0414)}}{2 \times 1}\)

=\(\frac{-0.0414 \pm \sqrt{0.0017+0.1656}}{2}=\frac{-0.0414 \pm 0.135}{2}\) = 0.0882 and + 0.0468

The negative value of x is meaningless.

Thus, the concentrations of PCI₃, Cl₂ and PCl₅ at equilibrium are as follows.

[PCI₃] = [Cl₂] = x = 0.0468 mol LT^-1.

[PCI₅] = 0.1 – x = 0.0532 mol L^-1.

Example 6. The equilibrium constant for the reaction \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}\) is 4.0 at 298 K. Calculate the weight of ethyl acetate that will be obtained when 120 g of acetic acid reacts with 92 g of alcohol.
Solution:

Weight of ethyl alcohol = 92 g.

Weight of acetic acid = 120 g.

Number of moles = \(\frac{\text { weight of substance }}{\text { molecular weight }} \text {. }\)

∴ number of moles of ethyl alcohol = 92/46 = 2

and number of moles of acetic acid = 120/60 = 2.

According to the chemical equation that represents the reaction, 1 mol each of the reactants produce 1 mol each of the products. Therefore, if x mol of each of the reactants reacted, x mol each of the products would be formed at equilibrium.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}\)

Intial concentration \(\frac{2}{V} \frac{2}{V} \frac{0}{V} \frac{U}{V}\)

Conc. at equilibrium \(\frac{2-x}{V} \frac{2-x}{V} \frac{x}{V} \frac{x}{V}\)

Applying the law of equilibrium

K = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}\)

= \(\frac{[x / V][x / V]}{\left[\frac{2-x}{V}\right]\left[\frac{2-x}{V}\right]}=\frac{x^2}{(2-x)^2}\).

Given that K = 4.

∴ 4 = \(\frac{x^2}{(2-x)^2}\)

or 2 = \(\frac{x}{2-x}\)

or 4-2x = x

or x = 4/3 =1.33 mol.

Hence the amoimt of ethyl acetate formed = 1.33 mol

=1.33 x molecular weight of ethyl acetate

=1.33 x 88 =117.04 g

Example 7. The equilibrium constant at 298 K for \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) is 4.0 x 10¹⁵. In a solution in which copper has displaced some silver ions, the concentration of Cu²⁺ ions is 16 x 10^-2 mol L^-1 and the concentration of Ag⁺ ions is 2.0 x 10^-9 mol L^-1. Is this system at equilibrium?
Solution:

Applying the law of chemical equilibrium, assuming that the system is at equilibrium,

⇒ \(K_{\mathrm{c}}=\frac{\left[\mathrm{Cu}^{2+}(\mathrm{aq})\right][\mathrm{Ag}(\mathrm{s})]^2}{\left[\mathrm{Ag}^{+}(\mathrm{aq})\right]^2[\mathrm{Cu}(\mathrm{s})]} .\)

By convention, concentration of a solid = 1.

∴\(K_{\mathrm{c}}=\frac{\left[\mathrm{Cu}^{2+}(\mathrm{aq})\right]}{\left[\mathrm{Ag}^{+}(\mathrm{aq})\right]^2}\)

Substituting the concentrations of Cu²⁺ and Ag⁺ ions

∴ \(K_c=\frac{1.6 \times 10^{-2}}{\left(2 \times 10^{-9}\right)^2}=4 \times 10^{15}\)

This is the value of K_c for the reaction in equilibrium. Hence the given system is in equilibrium.

Example 8. The equilibrium constant for the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) is 50.5 at 718 K. Predict the direction in which the reaction will proceed to reach equilibrium at 718 K if we start with 2 x 10^-2 mol of HI,1.0 x 10^-2 mol of H₂ and 3.0 x 10^-2 mol of I₂ in a vessel of volume1 L

Solution:

Reaction quotient \(\left(\mathrm{Q}_c\right)=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

If the reaction quotient is equal to K_c, the system is at equilibrium, and if it is less than Kc, the reaction will proceed in the forward direction to attain equilibrium.

Since the volume of the vessel is 1 L,

⇒ \(Q_c=\frac{\left[2 \times 10^{-2}\right]^2}{\left[1.0 \times 10^{-2}\right]\left[3.0 \times 10^{-2}\right]}=1.33\)

Since this is less than K_c, the reaction will proceed in the forward direction.

Example 9. 1 mol of ammonia was injected into a 1-L flask at a certain temperature. The equilibrium mixture was found to contain 0.3 mol of H₂. Calculate the concentrations of N₂ and NH₃ at equilibrium, and the equilibrium constant.
Solution:

The equilibrium can be represented as follows.

⇒ \(2 \mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2+3 \mathrm{H}_2\)

The initial concentration of NH₃ =1 mol.

At equilibrium, 2 mol of ammonia produces 3 mol of hydrogen.

Therefore, 0.3 mol of H₂ will be obtained from 0.2 mol of NH₃.

Similarly, when 2 mol of NH₃ dissociates, 1 mol of N₂ is obtained.

Therefore, 0.2 mol of NH₃ will produce 0.1 mol of N₂.

Since the volume of the flask is1 L, at equilibrium,

[N₂] = 0.1 mol ^-1,

[H₂] = 0.3 mol L^-1,

[NH₃] =1-0.2 =0.8 mol L^-1.

∴ \(K_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2}=\frac{(0.1)(0.3)^3}{(0.8)^2}=0.004\)

Example 10. 1 mol of ammonium carbamate dissociates as shown below at 500 K. \(\mathrm{NH}_2 \cdot \mathrm{COONH}_4(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NH}_3+\mathrm{CO}_2\) If the pressure exerted by the released gases is 3.0 bar, calculate the value of K_p.
Solution:

Applying the law of equilibrium,

⇒ \(K_p=\frac{p_{\mathrm{NH}_3}^2 p_{\mathrm{CO}_2}}{p_{\mathrm{NH}_2 \cdot \mathrm{COONH}_4(s)}}\)

By convention  \(p_{\mathrm{NH}_2 \mathrm{COONH}}\) =1 as it is a solid.

Given that the pressure exerted by the gases = 3 bar; this is the total pressure.

(To obtain the partial pressures, the total pressure must be multiplied by the mole fractions of the respective gases.)

∴ \(p_{\mathrm{NH}_3}=3 \times \frac{2}{3}=2 \mathrm{bar}\).

and \(p_{\mathrm{CO}_3}=3 \times \frac{1}{3}=1 \mathrm{bar}\).

∴ \(K_p=(2)^2(1)=4 bar.\)

Example 11. The dissociation constant of a weak acid HA is 1.6 x 10^-5 at 298 K. Calculate the H₃O⁺ ion concentration in a 0.01-M solution of the acid.
Solution:

The ionisation of the acid may be represented as

⇒ \(\mathrm{HA}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})\)

⇒ \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\) .

If the concentration of HA that dissociates is x, then the concentration of H₃O⁺ and A^– is also x.

Then the concentration of the undissociated acid is (0.01- x).

Since HA is a weak acid, its degree of dissociation is small and at equilibrium, x may be neglected in comparison to 0.1.

[HA] = (0.1 -x) = 0.1 mol L^-1

[H₃O⁺] =[A^–] = X

∴ \(\quad K_a=\frac{x^2}{0.1}\)

or \(x^2=K_a \times 0.1=16 \times 10^{-5} \times 0.1\)

or \(x=\sqrt{16 \times 10^{-5} \times 0.1}=126 \times 10^{-3}\).

∴ \(\quad\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{A}^{-}\right]=1.26 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\).

Example 12. Calculate the degree of dissociation and the concentration of H₃O⁺ ions in a 0.1-M solution of formic acid, given that =2.1 x 10^-4 at 298 K.
Solution:

Formic acid dissociates in water according to the following equation.

⇒ \(\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{HCOO}^{-}\)

The initial concentration of the acid is given as 0.1 mol L^-1

Let the degree of dissociation be α.

Then at equilibrium

[H₃O⁺] = α x 0.1 mol L^-1

[HCOO^–] =α x 0.1 mol L^-1

[HCOOH] = 0.1(1- α) mol ^-1

∴\(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{HCOO}^{-}\right]}{[\mathrm{HCOOH}]}=\frac{(0.1 \alpha)(0.1 \alpha)}{0.1(1-\alpha)}\)

Since formic add is a weak acid, a is very small as compared to 1 and can be neglected in the denominator.

∴ \(K_a=\frac{(0.01 \alpha)^2}{0.1}=0.1 \times \alpha^2\)

Given that K_a = 2.1 x 10^-4.

or 2.1 x 10 = 0.1 x α²

or \(\frac{21 \times 10^{-4}}{0.1}=\alpha^2\)

or α² = 2.1 x 10^-3 = 21 x 10^-4

or α = 0.046

[H₃O⁺] = Cα = 0.1 x 0.046 = 4.6 x 10^-3 mol L^-1

Example 13. A solution of NaOH contains 4.0 g of the base per litre. Find the pH of the solution.
Solution:

Concentration of NaOH solution = 4 g L^-1; molecular weight of NaOH = 40.

∴ molarity of solution = 4/40 = 0.1

Since NaOH is completely dissociated, concentration of OH^– ions = 0.1 M = 10^-1 mol L^-1.

K_w=[H₃O⁺][OH^–]

or 1 x 10^-14 = [H₃O⁺][OH^–]

or \(\frac{1 \times 10^{-14}}{10^{-1}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)

or 1 x 10^-13 M = [H₃O⁺].

pH =- log[H₃O⁺] = -log 10^-13 = 13.

Example 14. Calculate the pH of a solution of a strong monobasic acid made by diluting 100 mL of a 0.01-M solution to 1 L
Solution:

M₁V₁ = M₂V₂

or 0.01 x100 = M₂ x 1000.

∴ \(M_2=\frac{0.01 \times 100}{1000}=10^{-3}\)

Assuming complete dissociation

[H₃O⁺] =10^-3. (it is a monobasic add, one mole of protons is released per mole of add.)

∴ pH =- log[H₃O⁺] =-[log10^-3] = 3.

Example 15. Calculate the pH of a

1. 0.02-M H₂SO₄ solution and
2. 0.01-M Ba(OH)₂ solution.

Solution:

1. The ionisation of H₂SO₄ can be represented by \(\mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{SO}_4^{2-}\)

Being a strong acid, H₂SO₄ is completely ionised.

1 mol of H₂SO₄ dissodates to give 2 mol of H₃O⁺.

[H₃O⁺] =2 x 0.02 = 0.04 mol L^-1.

∴ pH = -log 0.04 =-log 4 x 10^-2 = -(-1.3980) =1.398.

2. Ba(OH)₂ dissodates completely.

1 mol of Ba(OH)₂ dissociates to give 2 mol of OH^–.

∴ [OH^–] = 2[Ba(OH)₂] =2 x 0.01 = 0.02 M.

K_w =[OH^–][H₃O⁺]

∴ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1 \times 10^{-14}}{0.02}=5 \times 10^{-13} \mathrm{M}\)

∴ pH =-log 5 x 10^-13 =-(-12. 3010) =12.301

Example 16. How many grams of NaOH must be dissolved in 1 L of a solution of the base so that the pH of the solution is 13?
Solution:

pH = -log[H₃O⁺]

[H₃O⁺] = -antilog pH = -antilog13 =1 x 10^-13 mol L^-1.

K_w =[H₃O⁺][OH^–].

∴ \(\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}=\frac{1 \times 10^{-14}}{1 \times 10^{-13}}=1 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}\).

As NaOH is a strong electrolyte, [NaOH] =[OH^–] =1 x 10^-1 mol L^-1.

Amount of NaOH = number of moles x mol. wt. of NaOH =1 x10^-1 x 40 = 4 g.

Example 17. The dissociation constant of a weak acid HA is 1 x 10^-9. Find the pH of a 0.1-M solution of the acid.
Solution:

The dissociation of the acid in water may be represented as follows.

⇒ \(\mathrm{HA}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-}\)

If x mol dissociates at equilibrium, [HA] = 0.1- x.

∴ \({\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{A}^{-1}\right]=x .}\)

∴ \(\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{x \times x}{0.1-x}\).

Since x is negligible compared to 1,

∴ \(K_{\mathrm{a}}=\frac{x^2}{0.1}\)

∴ x = \(\sqrt{\mathrm{K}_{\mathrm{a}} \times 0.1}=\sqrt{1 \times 10^{-9} \times 0.1}=10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

∴ \({\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} .}\)

∴ \(\mathrm{pH}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log 10^{-5}=5\) .

Example 18. Calculate, the[H₃O⁺] of a solution of a monobasic acid whose pH is 4.35.
Solution:

pH = -log[H₃O⁺]

∴ log[H₃O⁺] = -pH = -4.35.

[H₃O⁺ ] = antilog(-4.35).

Since mantissas are always positive in logarithmic tables, add -1 to the characteristic and +1 to mantissa in order to make the mantissa positive.

[H₃O⁺] = antilog[-4-1 + (-0.35 +1)] = antilog 5.65 = 4.467 x 10^-5

Equilibrium  Multiple Choice Questions

Question 1. Which of the following is a characteristic of a reversible reaction?

1. The number of moles of the reactants is equal to that of the products.
2. It can be influenced by a catalyst.
3. It can never proceed to completion.
4. None of the above

Answer: 3. It can never proceed to completion.

Question 2. A chemical reaction A ⇔ B is said to be in equilibrium when

1. A has been completely converted to b
2. The conversion of a to b is 50%
3. Only 10% of the conversion of a to b has taken place
4. The rate of conversion of a to b is just equal to that of b to a

Answer: 4. The rate of conversion of a to b is just equal to that of b to a

Question 3. The equilibrium constants of these reactions are related as

⇒ latex]2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) K_1[/latex]

⇒ \(2 \mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) K_2\)

1. K₂ = K₁
2. \(K_2=K_1^2\)
3. \(K_2=\frac{1}{K_1^2}\)
4. \(K_2=\frac{1}{K_1}\)

Answer: 4. \(K_2=\frac{1}{K_1}\)

Question 4. In which of the following cases does the reaction proceed the fastest?

1. K =10²
2. K = 10^-2
3. K = 10
4. K = 1

Answer: 1. K =10²

Question 5. For the reaction \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\), the forward reaction at constant temperature is favoured by

1. The introduction of an inert gas at constant volume
2. The introduction of an inert gas at constant pressure
3. The introduction of Cl₂ at constant volume
4. An increase in pressure

Answer: 2. The introduction of an inert gas at constant pressure

Question 6. In the reaction \(2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g})+85 \mathrm{cal}\), which of the following conditions would give the best yield at equilibrium?

1. 500 atm, 100°C
2. 100 atm, 100°C
3. 500 atm, 500°C
4. 100 atm, 500°C

Answer: 1. 500 atm, 100°C

Question 7. Tire solubility of Agl in a solution of Nal is less than that in pure water because

1. Agl forms a complex with nal
2. Of the common-ion effect
3. The K_sp of AgI is less than that of nal
4. Of a decrease in temperature

Answer: 2. Of the common-ion effect

Question 8. Which of the following is not a Lewis acid?

1. BF₃
2. AlCl₃
3. BeCl₂
4. SnCl₄

Answer: 4. SnCl₄

Question 9. An acid solution of pH 6 is diluted a hundred times. The pH of the solution becomes

1. 6.95
2. 6
3. 4
4. 9

Answer: 1. 6.95

Question 10. If the solubility of a salt M₂X₃ is x mol L^-1, its solubility product is

1. x⁵
2. 76x²
3. 96x⁵
4. 108x²

Answer: 4. 108x²

Question 11. The solubility product of A₂B is 4 x 10^-9 mol L^-1 Its solubility is

1. 10^-3 M
2. 4^1/3 x 10^-3 M
3. 10^-4 M
4. 2 x 10^-5 M

Answer: 1. 10^-3 M

Question 12. Which of the following has the lowest value of K_sp at ordinary temperature?

1. Mg(OH)₂
2. Ba(OH)₂
3. Ca(OH)₂
4. Be(OH)₂

Answer: 4. Be(OH)₂

Question 13. Which of the following is the strongest Bronsted base?

1. ClO^–
2. ClO₂^–
3. ClO₃^–
4. ClO₄^–

Answer: 1. ClO^–

Question 14. The pH of a certain solution is 5.0. Just enough acid is added to decrease the pH to 2.0. The hydrogen ion concentration will

1. Increase 1000 times
2. Decrease 1000 times
3. Increase 100 times
4. Increase 10 times

Answer: 1. Increase 1000 times

Question 15. The K_sp of CaF₂ is 1.7 x 10^-10. The addition of what concentration of an equal volume of F^– ions to a solution containing 0.01-M Ca²⁺ ions will result in the precipitation of CaF₂?

1. 3.4×10^-8 M
2. 1.3×10^-4 M
3. 3.68 x 10^-4 M
4. 2.6×10^-4 M

Answer: 2. 1.3×10^-4 M

 

 

 

 

Thermodynamics

This chapter deals with energy changes that take place during a chemical reaction. In any chemical reaction, the atoms of the reactants are rearranged to form the products. This involves the breaking and forming of bonds. You know that energy is required to break bonds and that it is released when bonds are formed. It should then be easy to understand that a chemical reaction involving the dissociation and formation of bonds must be accompanied by energy changes.

The energy change accompanying a reaction may appear in different forms. When fuels are burnt, for instance, energy appears as heat and light. The chemical reaction in a battery produces electrical energy. When a grenade explodes, the chemical reaction produces heat, light, sound and kinetic energy. Consider the following reactions.

⇒ \(\mathrm{C}+\mathrm{O}_2\longrightarrow \mathrm{CO}_2 \text { + heat }\)

⇒ \(2 \mathrm{Mg}+\mathrm{O}_2\longrightarrow 2 \mathrm{MgO}+\text { heat }+ \text { light }\)

⇒ \(\mathrm{Zn}+\mathrm{CuSO}_4\longrightarrow \mathrm{ZnSO}_4+\mathrm{Cu}+\text { electrical energy }\)

In these reactions there is a net release of energy. There are other reactions in which energy is absorbed. Take the case of electrolysis, for example. Here electrical energy is absorbed and the electrolyte splits into its components. Photosynthesis, on which almost all living organisms depend directly or indirectly, is another example of a reaction in which energy is absorbed.

Thermodynamics Some Definitions

Before we start a systematic study of the energy changes associated with chemical reactions, let us define a few basic terms used in any discussion of energetics or chemical thermodynamics.

System: A system is that part of the universe which we are interested in investigating. For example, if we are studying a particular reaction in a vessel, then the vessel, the reactants, and the products constitute the system.

Surroundings: Everything other than the system or the part of the universe other than the system is called the surroundings. In the example we have just considered, everything other than the vessel in which the reaction is taking place is called the surroundings. A system is separated from the surroundings by boundaries which may be real or imaginary. Further, boundaries may be considered part of either the system or the surroundings, depending upon convenience.

There are three types of systems: open system, closed system, and isolated system.

Open system A system which can exchange matter and energy with the surroundings is called an open system. For example, an open test tube in which a reaction is taking place is an open system. It can exchange heat with the surroundings and gaseous products can escape into the surroundings.

Closed system A dosed system can exchange energy with the surroundings but not mass. If a reaction occurs in a sealed bulb, which can exchange heat with the surroundings but not matter, the bulb with the reactants and products constitutes a closed system.

Isolated system An isolated system can neither exchange matter nor energy with the surroundings. A thermos flask filled with hot tea is an isolated system.

Intensive and extensive properties: The properties of any substance can be classified as intensive or extensive. Intensive properties are those independent of the size of the substance. For example, by doubling the size of the given sample of a substance, the temperature and pressure of the substance do not double or do not change. These are called Intensive properties, other examples being viscosity, density, and all other molar properties.

If the value of the property depends on the size of the substance, it is called an extensive property. On doubling the size, internal energy doubles, and hence it is an extensive property. Other examples include mass, volume, heat capacity, enthalpy, entropy, and free energy.

State functions: State functions are those parameters or measurable macroscopic properties of a system that describe its state. The state of a system is defined by specifying the values of certain number of macroscopic properties. The number depends on the nature of the system.

• The values of state functions or state variables depend only on the state of the system and not on how that state is readied. Take a gaseous system, for example. Its state is described by the state functions: temperature (T), pressure (p), and volume (IQ. To take a more specific case, consider 1 mol of CO at stp. Its volume will be 22.7 L irrespective of the method by which it is obtained.
• All state functions or state variables are not independent because equations of state exist between different state functions. For example, consider the equation of state for an ideal gas, pV-nRT. Here out of the four variables (p. V. n and T), only three can be independently varied.
• Thus once the values of the minimum number of state functions are fixed, others have definite values. The equilibrium state of a system is characterised by the definite values of state functions which do not change with time. Internal energy, enthalpy, and entropy are the state functions you will learn about in this chapter.

The state of a system: As stated above, thermodynamic equilibrium exists in a system when the macroscopic properties or state functions do not change with time. If the equilibrium state is disturbed tire system again settles down to a new equilibrium state.

• If a particular state function does not have equal values in the system and its surroundings, exchange of matter or energy or both takes place between the thermodynamic system and its surroundings. After the interaction between system and its surroundings stops, the system attains a new equilibrium state with new values of state functions.
• The starting state of the system is referred to as the initial state and the state reached after interaction with the surroundings is the final state. When a thermodynamic system undergoes a change of state (from the initial to tire final), we say it has undergone a process. This implies that there will be a change in at least one of tire state functions of tire system during a process.
• A system can change from one state (initial state) to another (final state) through a number of paths. In other words, there can be a number of processes between an initial state and a final state. There are certain processes in which a particular state variable remains unchanged. Let us learn about these processes.

Isothermal If the temperature of the system remains constant during the change, the process is called isothermal.

Adiabatic If the system does not exchange heat with the surroundings, the process is called adiabatic.

Isobaric If the pressure of the system remains constant during the change, the process is called isobaric.

Isochoric If the volume of the system remains constant during the change, the process is called isochoric.

Reversible process In this process, the initial and the final states are connected through a succession of equilibrium states, i.e., at every state along the reversible path, there exists an equilibrium between the system and the surroundings. Such states are called quasi-equilibrium states.

Irreversible process The processes occurring in nature are generally irreversible. Their only difference from a reversible process is that equilibrium is not maintained during the transformation process.

Cyclic process The process that brings back a system to its original state after a series of changes is called a cyclic process.

Transference Of Energy

While describing open, closed, and isolated systems, we touched upon the idea of a system exchanging energy with the surroundings. There are two forms in which energy is exchanged between a system and the surroundings.

We often talk about heat flowing from a body at a higher temperature to one at a lower temperature. What we mean is that energy is exchanged between the two bodies because of a difference in temperature.

• So, one of the forms in which a system can exchange energy with the surroundings is heat. Energy is exchanged between a system and the surroundings in the form of heat when they are at different temperatures.
• The other mode of energy exchange between a system and the surroundings is energy. One example where energy is exchanged in this form is when the system and surroundings are at different pressures.
• Consider a gas enclosed in a cylinder that is fitted with a weightless, frictionless moving piston. If the gaseous system is at a higher pressure than the surroundings, the piston will move upwards (increasing the volume of the gas) until the pressure of the system and that of the surroundings become equal.

During the expansion of the gas, work is done by the system on the surroundings due to the pressure difference between the two. Here energy is transferred from the system to the surroundings as work.

• If, on the other hand, the pressure of the system is lower than that of the surroundings, the piston is pushed down until the pressures become the same. In this case the volume of the gas decreases and work is done by the surroundings on the system.
• Once again there is figure, when the gas enclosed in a cylinder transference of energy in the form of work, but from the fitted with a piston is at a higher pressure than the surroundings to the system. By convention, work done on the surroundings, the piston moves until the pressures system is positive, while work done by the system is negative. equalize.

Remember that neither heat, nor work is a property of a system. They are path functions as they depend on how a change is brought about. A system does not possess a particular amount of work or heat. It does, however, have a particular amount of energy under a particular set of conditions. The SI unit of heat is the same as that of work, i.e., the joule. However, heat is sometimes expressed in calories or kilocalories.

1 calorie = 4.184 J.

1 J = 10⁷ ergs.

Work

There are several kinds of work, such as mechanical work, electrical work, and chemical work. Work is said to be done if an object is moved by a force F through a distance d. The work done is given by

W = Fxd…….Equation 1

Some types of work encountered in chemistry are:

1. Expansion work (which we shall study in detail) in the case of expansion or compression of a gas against a force surface expansion in case of liquids—these are examples of mechanical work
2. When larger molecules are synthesized from smaller ones within living organisms—chemical work
3. An ion moving in the presence of an electric field—electrical work.

The most common form of mechanical work encountered in chemistry is the expansion work associated with the expansion of a gas against pressure. It is also called p-V or pressure-volume work.

To understand the concept further, let us once again consider a Figure but with a weight attached to the piston. When the gas expands, the piston moves up and the weight is lifted up. We say that work is done by the system.

Let us now calculate the work done in terms of pressure and volume. If an external pressure p_ex is applied on the piston and the piston moves inwards by a distance dl, then the change in volume of the gas is given by

dV=A • dl …….(1)

where A is the area of the cross-section of the piston. The force on the piston is given as

F = p_ex • A …..(2)

Now, the work done in moving the piston by a distance dl against an opposing force of magnitude F is given by

dW = -Fdl …….(3)

(the negative sign is due to the opposing nature of the force).

Substituting (2)  in (3)

dW = -p_ex • Adl

or dW = -p_ex dV

or simply W = -p Δ V …..Equation 2

• By the help of this equation, we can calculate the work done (or pressure/volume) during the expansion or compression of a gas. When expansion occurs ΔV in Equation 2 is positive, thereby resulting in a negative value for W.
• Similarly, when compression occurs, weight is lowered in the surroundings and we say that work is done on the system. Since AV is negative now, W is positive.
• Thus we can say that work is done when there is a change in height of a weight in the surroundings.

If due to a change in state of a system, a weight is lifted in the surroundings, work is done by the system and if a weight is lowered in the surroundings, work is done on the system. Now W can also be given as W = mgh,

where m is the mass lifted, g is the acceleration due to gravity and h is the height through which the mass is lifted.

Thus, during a change in state of the system, we can learn about any work done by measuring the relevant quantities only in the surroundings.

Reversible expansion or compression of a gas: Suppose a gas confined in a cylinder with a movable piston is in equilibrium. This means that there is no change in state of the system. This is possible when the external pressure p_ex (pressure that is applied) is equal to the internal pressure, p, of the gas. A small change in p_ex can disturb the equilibrium and the gas may expand or compress, resulting in a change in the volume and hence the state of the system.

• When p_ex is increased infinitesimally (infinitesimal means negligibly small) then compression occurs and the volume of the gas decreases. Equilibrium with the surroundings is also established almost immediately. The system can also be restored to its original state by reducing the external pressure to the original value.
• By continuously increasing the external pressure, infinitesimally, the gas can be made to undergo a finite amount of compression. This is achieved by keeping different weights on the piston one after the other.
• In each step, however, the change is infinitesimal and can be reversed by an infinitesimal change in external pressure. Again, in each step equilibrium is immediately established.
• Thus we have a process which is carried out in several infinitesimally small steps and throughout the process the system is in equilibrium with its surroundings. This is a reversible process.

If the difference between the internal pressure and the external pressure is large then there is a considerable disturbance of equilibrium and an infinitesimal change in p_ex in the opposite direction will not bring back the system to its original state. Such a process is termed an irreversible process.

Calculation of work done in various cases

1. Work done in free expansion When the external pressure is zero (p_ex = 0), there is no opposing force on the piston and the gas can expand freely. This is called free expansion. This occurs when a gas expands in vacuum. Both the work done on the system and by the system are zero in this case.

2. Work done in expansion against constant pressure If a constant external pressure is applied, the gas expands or compresses until the internal pressure becomes equal to the external pressure. The work done at every displacement dV is given by Equation 2. The total work done in the expansion from V₁ to V₂ is the sum of all such contributions. It is obtained by finding the area under the p-V plot, i.e., by integrating Equation.

∴ \(W=-\int_{V_1}^{V_2} p_{e x} d V\) ….. (1)

= \(-p_{e x} \int_{V_1}^{V_2} d V=-p_{e x}\left(V_2-V_1\right)\)

∴ \(W=-p_{e x} \Delta V\)

In case of expansion, ΔV is positive and W turns out to be negative. Therefore, work is done by the system.

If ΔV is negative as in case of compression, W is positive, i.e., work is done on the system.

3. Work done in reversible expansion In reversible expansion, the difference between external and internal pressure is infinitesimally small. With very small changes in pressure, the process of expansion and compression may be reversed. If the expansion occurs in several stages, at each stage, it is ensured that the external pressure is only infinitesimally different (less in case of expansion and more in case of compression) from the internal pressure of the gas.

Since \(p_{e x}=p_{\text {in }}\) we may replace p_ex in Equation 2 by p_in

dW = -p_in dV.

This is the work done at every stage of the expansion. The total work is obtained by summing up all such infinitesimal contributions (again) given by the area under the p-V plot.

⇒ \(\int d W=-\int p_{\mathrm{ma}} d V\)

∴ \(W_{\mathrm{rev}}=-\int p_{\mathrm{in}} d V\)

The subscript ‘rev’ indicates the reversible nature of the process.

If it is assumed that the gas behaves ideally then p_in be replaced by nRT/v using the ideal gas equation.

Therefore \(W_{\mathrm{rev}}=-\int_{V_1}^{V_2} \frac{n R T}{V} d V\)

If the process is carried out at a constant temperature (isothermally), T may be taken out of the integral sign.

⇒ \(W_{r e v}=-n R T \int_{V_1}^{V_1} \frac{d V}{V}\)

= \(-n R T|\ln V|_{V_1}^{V_2}\)

or, \(W_{n e v}=-n R T \ln \left(\frac{V_2}{V_1}\right)=-2303 n R T \log \frac{V_2}{V_1}\),

where V₁ and V₂ are the initial and final volumes respectively.

Internal Energy

The energy stored within a system (say chemical system) is its internal energy or intrinsic energy. Under a particular set of conditions, a thermodynamic system has a definite amount of internal energy.

• In other words, the internal energy of a system, generally represented as U, depends on the nature, amount, temperature, and pressure of the system. This energy possessed by a system is due to the different types of energy that its atoms or molecules have.
• It is, in fact, the sum of the energies of the elementary particles of the system, i.e., the sum of their potential and kinetic energies and the bond energy between constituting atoms.
• Instead of saying that a particular system under a particular set of conditions possesses a definite amount of internal energy, we could have said that internal energy is a state function, which means that it depends only on the initial and final state of the system and is independent of the way (path) the change takes place.

However, unlike other state functions, like pressure and temperature, it is not possible to determine the value of internal energy. This is because it is not possible to determine with exactness the values of the components of internal energy. Note that this matters little, because in the study of chemical processes, what we are really interested in knowing is the clumge in internal energy, ΔU (delta U).

ΔU = U₂-U₁

where U₂ is the internal energy of the system at the end of the process being studied and is the internal energy of the system at the start of the process. For a chemical reaction

ΔU = U_p-U_r,

where U₂ is the internal energy of the products and U_r is that of the reactants. Obviously, All is a state function since U_p and U_r are state functions.

So, how do we measure ΔU or the change in internal energy associated with a chemical process? Suppose we carry out a chemical reaction in such a way that the system remains at the same temperature, there is no work done on the system and it does no work on the surroundings.

• Whatever the change in the internal energy of the system in the course of the reaction must be equal to the energy exchanged by the system with the surroundings in the form of heat.
• Why? This in fact follows from the law of conservation of energy, which we will discuss subsequently, for the moment, imagine that the change in internal energy of the system is negative, i.e., U_p < U_r.
• What happens to this energy which the system has apparently lost? It cannot just vanish because that is against the law of conservation of energy. We have assumed that the system does not work and nor is any work done on it (i.e., volume remains constant), so the energy cannot be exchanged with the surroundings in the form of work.
• Since the temperature is constant, the energy must be exchanged with the surroundings in the form of heat.
• The change in the internal energy associated with a chemical reaction is, thus, determined by letting the reaction occur at constant temperature and constant volume and measuring the heat exchanged with the surroundings.

Law Of Conservation Of Energy

This is one of those common-sense laws which seem very obvious but have far-reaching implications. It states that energy can neither be created nor destroyed, though it may change from one form to another.

In the context of the system and the surroundings that we have been discussing, this means that the total energy of the system and the surroundings (i.e., the universe) remains constant.

• More specifically, it means that during a chemical reaction, energy may be absorbed or released, but the total energy of the reaction system and the surroundings remains constant.
• The law of conservation of energy is also known as the first law of thermodynamics. It is of relevance not only to chemists but also to physicists, engineers, and others who deal with the conversion of energy from one form into another.

The first law of thermodynamics being identical to the law of conservation of energy can be stated as:

1. Energy can neither be created nor destroyed although it can be changed from one form to another.
2. The energy of an isolated system is constant.

Consider the universe, which is an isolated system. Tire energy of the universe is conserved. Inside the universe, energy can be transferred from one part to another or it can be converted from one form to another, but it can neither be created nor destroyed.

Let us see if we can express this law mathematically. If the internal energy of a particular system is U₁ and it absorbs a certain amount of heat q, then its internal energy will become U₁ + q. Now suppose W amount of work is done on the system, then U₂, the final energy of the system is

U₂ =U₁ +q + W

or ΔU = U₂ – U₁ = q + W….. Equation 3

(If W is the work done on the system, -W is the work done by the system.)

• This means the change in the internal energy of the system is equal to the sum of the heat absorbed by the system and the work done on it.
• One can generalise this to say the change in the internal energy of the system is the sum of the heat exchanged by the system (with the surroundings) and the work done on or by the system. Equation 3 is the mathematical statement of the first law of thermodynamics.

If we now substitute for W in Equation 3, the expression for change in internal energy (when only pressure-volume type of work is done) becomes

ΔU = q – pΔV …. Equation 4

Now, if there is no change in the volume of the system then W = 0, i.e., W = -∫pdV = 0.

We can easily see that if there is no change in volume during a reaction, Equation 4 becomes ΔU = q_v,

where q_v is the heat exchanged (evolved or absorbed) by the reaction system at constant volume. When heat is absorbed by the system, q is positive and when heat is evolved by the system, q is negative. This is exactly what we said in the previous section.

The change in the internal energy of a system during a chemical reaction is equal to the heat exchanged with the surroundings, provided the volume and temperature of the system remain constant.

For any isothermal expansion of an ideal gas the total energy remains the same, and q = -W. In other words, the internal energy remains constant (ΔU = 0) when an ideal gas expands isothermally.

Thus for an isothermal irreversible change, q = -W = p_ex ΔV

and for an isothermal reversible change, q = \(-W=n R T \ln \frac{V_2}{V_1}\)

= \(-2.303 n R T \log \frac{V_2}{V_1}\),

where V₁ and V₂ are the initial and final volumes respectively.

For an adiabatic change, q = 0,

∴ ΔU = W_adiabstic  (from Equation 3).

Enthalpy

So far we have considered reactions taking place at constant volume and temperature, and the energy changes associated with such reactions. But the reactions we carry out in the laboratory normally occur in open vessels (beakers or test tubes, say).

Obviously, the volume of such a reaction system does not remain constant. However, the pressure does, since such a system, open to the atmosphere, is at atmospheric pressure and this pressure is more or less constant.

• The heat exchanged with the surroundings by a reaction system in the course of a reaction is equal to the change in the internal energy of the system if the temperature and volume are constant.
• What about the heat exchanged with the surroundings by a reaction system at constant pressure and not maintained at constant volume? It should be different, but why and how?
• Let us consider a reaction in which the volume of the system increases. An increase in volume can occur when the system does some work against the atmospheric pressure and energy is required for this work.
• Thus, in this case, the heat exchanged with the surroundings would be lower than that for a system at constant volume and temperature.
• On the other hand, if the reaction involves a decrease in the volume of the system, work would have to be done on the system by tire surroundings. The heat exchanged would then be greater than that exchanged by a system at constant volume.
• Thus, when a reaction proceeds at constant pressure and temperature, the heat exchanged by the system with tire surroundings is not equal to the change in the internal energy of the system.

In other words, the change in the internal energy of the system is not the only contributing factor to the total energy change associated with the reaction. Tire total energy change includes change in energy due to the pressure-volume type of work done by or on tire system.

Enthalpy or heat content is a property or state function introduced to take care of the energy changes associated with tire kind of system we have just been discussing. It is denoted by H and expressed mathematically as follows.

H = U + pV, …… Equation 5

where U is the internal energy, p the pressure, and V the volume of the system. The enthalpy of a system (substance) can be defined as the total energy associated with it, i.e., its internal energy and the energy due to factors such as pressure-volume conditions. The enthalpy of a substance depends on its state (temperature, pressure, etc.).

• Enthalpy changes are normally expressed with the substances in their standard states. The standard state of a pure substance is the pure form at a pressure of 1 bar and a specified temperature, the conventional temperature being 298.15 K.
• Enthalpy change in the standard state is called the standard enthalpy change. It is denoted by \(\Delta H^ominus\), where the superscript \(\ominus\) denotes standard conditions.

The molar enthalpy of a substance H_m = H/n (where n is the number of moles) is an intensive property and H_m = U_m + pV_m. When enthalpy changes have to be compared, the states of the systems must be identical.

Enthalpy change: As in the case of internal energy, the absolute value of the enthalpy of a system cannot be determined. What we are interested in knowing is the enthalpy change associated with a process. The total change in the energy of a system during a reaction at constant pressure is the enthalpy change, denoted by ΔH.

∴ \(\Delta H=H_{\text {products }}-H_{\text {reactants }}\)

= \(\left(U_p+p V_p\right)-\left(U_r+p V_r\right)\)

= \(\Delta U+\Delta p V\),

where U_p = internal energy of products,

V_p =volume of products,

U_r = internal energy of reactants, and V_t = volume of reactants.

Since p is constant, ΔH = ΔU + pΔV…….. Equation 6

The enthalpy change during a process is the sum of the change in internal energy and the pressure-volume work done.

Suppose the heat exchanged by a system with the surroundings during a reaction at constant pressure and temperature is q_p. Suppose also that the change in internal energy of the system is ΔU and the work done on the system is W. Then

q_p = ΔU-W = ΔU – (-pΔV)

= ΔU + pΔV

= ΔH. (ΔU + pΔV = ΔH)

• The heat exchanged by a system with the surroundings during a reaction at constant pressure and temperature is the enthalpy change associated with the reaction. In practice, the enthalpy change associated with a reaction is determined by insulating the system and allowing the heat of the reaction to alter its temperature.
• The amount of heat required to be supplied to or taken away from the system to let its temperature go back to the original value is then calculated to obtain the enthalpy change.

Significance of ΔU and ΔH: It is understood that the amount of heat exchanged with the surroundings for a reaction at a constant pressure (ΔH) is different from that exchanged at constant volume and temperature (ΔU).

The difference is not significant for solid or liquid systems but it does matter when gases are involved. Let us consider a gaseous reaction, where V_r is the total volume of the gaseous reactants V_p is the total volume of gaseous products, n_r, is the number of moles of reactants and n_r is the number of moles of products, all at constant pressure and temperature.

By the ideal gas law,

∴ \(\quad p V_t=n_r R T\)

and \(p V_p=n_p R T\)

Thus, \(p V_p-p V_r=\left(n_p-n_r\right) R T\)

or \(\quad p\left(V_p-V_r\right)=\left(n_p-n_r\right) R T\)

or, \(\quad p \Delta V=\Delta n_g R T\).

Here, Δn_g is the difference between the number of moles of the gaseous products and that of the gaseous reactants.

From the definition of enthalpy, it follows that ΔH = ΔU + pΔV.

Therefore, ΔH = ΔU + Δn_gRT.

The above equation is useful in situations where one of the two quantities ΔH and ΔU is known and the other has to be calculated. ΔH and ΔU differ significantly for processes involving gases. For solids and liquids, pV_p is only slightly different from pV_x.

Source of enthalpy change: You may be wondering what exactly leads to enthalpy change during a reaction. Any reaction involves the breaking and forming of bonds.

• Energy is released when bonds are broken and required for the formation of bonds. Simply put, the net change in energy due to the breaking and forming of bonds is the enthalpy change of the reaction.
• Of course, for the net change in energy to be termed enthalpy change, the reaction must take place at constant pressure and temperature. The simplest case is that of a gaseous system in which gas A reacts with gas B to form gas C.

If the reaction involves solutions, the interactions (hence energy changes) between the solvents and the reactants and products have to be considered. For liquid or solid reactants, we have to also consider the interactions between the molecules.

Enthalpy change(for a gaseous system at constant pressure) = (energy required to break bonds) – (energy released in formation of bonds).

Let us consider the reaction between hydrogen gas and chlorine gas to form HCl gas. Energy is required to break tire H—H and Cl—Cl bonds and released when H—Cl bonds are formed.

∴ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g})\)

The energy required to break H—H bonds = 436 kJ mol^-1.

Energy required to break Cl—Cl bonds = 242 kJ mol^-1.

Energy released in the formation of two H—Cl bonds = 2 x 431 kj mol^-1.

Enthalpy change (ΔH) = 436 + 242 – 2 x 431 = -184 kJ.

The negative value of ΔH indicates a decrease in the enthalpy of the system.

The heat evolved in this reaction is 184 kJ mol^-1.

Exothermic And Endothermic Reactions

In our discussion of the energy changes associated with chemical reactions so far, we have only mentioned that energy is released in some reactions and absorbed in others. Reactions in which energy is released are called exothermic while those in which energy is absorbed are called endothermic.

Exothermic reactions: While writing an exothermic reaction, the heat evolved is indicated on the right side, after the products.

∴ \(\mathrm{N}_2+3 \mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3+93.7 \mathrm{~kJ}\)

• If an exothermic reaction is carried out at constant volume and temperature, the heat evolved (q_ν) is (numerically) equal to the change in internal energy (ΔU). In such a reaction the internal energy of the products (U_p) is less than the internal energy of the reactants (U_r) and ΔU is negative.
• If an exothermic reaction occurs at constant pressure and temperature, the heat evolved is (numerically) equal to the change in enthalpy (ΔH). The enthalpy of the products is less than the enthalpy of the reactants, i.e., ΔH is negative.

Endothermic reactions: In the case of an endothermic reaction, the heat absorbed can be indicated along with reactants or with the products. Obviously, if it is shown on the left side of the equation, it will have a positive sign, and if it is shown on the right side, it will carry a negative sign.

∴ \(\mathrm{N}_2+\mathrm{O}_2+180.5 \mathrm{~kJ} \longrightarrow 2 \mathrm{NO}\)

or, \(\mathrm{N}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{NO}-180.5 \mathrm{~kJ}\)

When an endothermic reaction occurs at constant temperature and constant volume, the heat absorbed is (numerically) equal to the change in the internal energy of the system. The internal energy of the products (U_p) is greater than the internal energy of the reactants (U_r) and ΔU is positive.

When an endothermic reaction proceeds at constant pressure and temperature, the heat absorbed is (numerically) equal to the change in enthalpy of the system. The enthalpy of the products (H_p) is greater than the enthalpy of the reactants (H_r) and ΔH is positive.

Thermochemical Equations

You already know how to write a chemical equation. When a chemical equation not only indicates the quantities and physical states of the reactants and products involved, but also the change in enthalpy during a reaction, it is called a thermochemical equation. Fractional coefficients may also be used in such an equation.

⇒ \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \quad \Delta H=572 \mathrm{~kJ}\)

Certain conventions must be followed while writing a thermochemical equation. These conventions are listed below.

1. The heat evolved or absorbed is indicated as in the equation above, in terms of ΔH. Remember that ΔH is negative for exothermic reactions and positive for endothermic reactions.

2. The numerical value of ΔH corresponds to the reaction as written. In the absence of any information, it is assumed that a certain value of ΔH is due to the number of moles of reactants that combine as indicated by the chemical equation. Thus ΔH is expressed in kJ mol^-1.

3. The value of standard enthalpy (\(\Delta H^{\ominus}\)) in a thermochemical equation corresponds to the standard state of the substances involved in the reaction. The term ‘standard state of a substance’ refers to the pure substance at exactly 1 bar pressure.

4. The coefficients of the various substances represent the number of moles of the substances involved in the reaction and the value of ΔH corresponds to these coefficients. It stands to reason, therefore, that if the coefficients are multiplied or divided by some factor, so must ΔH be. Consider the following example.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \quad \Delta H=-242 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

If the coefficients are multiplied by 2, ΔH must also be multiplied by 2.

⇒ \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \quad \Delta \mathrm{H}=-(242 \times 2) \mathrm{kJ} \mathrm{mol}^{-1}\)

When the reaction is reversed, the sign of AH is reversed but its magnitude remains the same.

\(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\Delta H=+53.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(2 \mathrm{HI}(\mathrm{g}) \longrightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \Delta H=-53.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

5. The physical states of the reactants and products have to be indicated because ΔH changes with the physical state. The following example will make this clear.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\Delta H=-242 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})\Delta H=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Enthalpy Of Reactions

The enthalpy change accompanying a reaction is called reaction enthalpy and is denoted by Δ_rH. The standard enthalpy of reaction (Δ_rH^Θ) is the change in enthalpy per unit amount of the reaction when the reactants in their standard states change to products in their standard states.

By standard states we mean reference state, i.e., the most stable state of aggregation. For example the reference state of H₂ is pure gas at 1 bar and that of CaCO₃ is pure solid at 1 bar. By convention the standard states are reported at 298 K. There are several factors that determine the value of enthalpy of a reaction.

Quantities of reactants The amount of heat absorbed or released as also the enthalpy of a reaction depends on the quantities of reactants involved. This should be pretty easy to understand. If you bum 1 kg of coal it will produce less heat than if you bum 10 kg of coal.

The standard enthalpy change for the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) has the units of kJ mol^-1. Per mole refers to the reaction as written above.

Let us take another example.

⇒ \(2 \mathrm{CH}_3 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

∴ \(\Delta_{\mathrm{r}} H^{\ominus}\) for this reaction is -1276.9 kj mol^-1.

However, if the reaction is written with different stoichiometric coefficients, i.e., \(\mathrm{CH}_3 \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

∴ \(\Delta_{\mathrm{r}} H^{\ominus}\) is -638.4 kJ mol^-1, ‘per mole’ referring to the reaction as written here.

Physical State of Reactants and products The change of state of a substance involves heat changes, so the enthalpy of a reaction depends on the physical states of the reactants and products. We have already discussed the case of the formation of liquid water and gaseous water vapour from gaseous hydrogen and oxygen.

Allotropic forms The enthalpy of a reaction also depends on the allotropic forms of the substances involved in the reaction.

⇒ \(\mathrm{C} \text { (diamond) }+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta H^{\ominus}=-395.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta H^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Temperature The enthalpy of a reaction depends on the temperature at which the reaction proceeds. The value of enthalpy indicated in a thermochemical equation is generally the value at 298 K.

Constant pressure or volume As we have discussed already, the enthalpy of a reaction at a particular temperature depends on whether the reaction occurs at constant pressure or constant volume. From our discussion so far

ΔH = ΔU + pΔV.

But ΔH = q_p and ΔU = q_v.

∴ q_p = q_v + pΔV

Hess’s Law

The enthalpy of a substance is a stale function, independent of the method by which the substance is made. The enthalpy change in a reaction is the difference between the enthalpies of the products and the reactants.

• Both of these are consequences of the law of conservation of energy. Another consequence of the law is what is known as
  Hess’s law of constant heat summation, which states that the enthalpy change in a chemical or physical process is independent of the path taken or the manner in which the change is brought about.
• In the context of a chemical reaction, the law can be stated as follows. “The amount of heat absorbed or released during a reaction is the same whether the reaction proceeds in a single step or through several steps.”

Though G H Hess (a Russian chemist) came up with this law as a result of experimental observations, if you think a little you will realise that it follows from, or is a corollary of, the law of conservation of energy. How? Suppose a reactant A is converted into a product D and the heat evolved in the process is q.

• Now suppose the same reactant A changes first to B, then B is converted to C and finally, C changes to D and the heat evolved in the three steps is q₁, q₂, and q₃. Let q₁ + q₂ + q₃ = q′.  According to Hess’s law q = q’. If Hess’s law were not true, either q> q’ or q < q’. Let us consider the possibility that q >q’.
• Then the heat evolved in converting A to D in a single step would be greater than the heat absorbed when D is converted to A in three steps. This would lead to a situation in which (q – q‘) of heat would be ‘created’ after the completion of the cyclic process and the law of conservation of energy would be violated.

Hess’s law can be stated as follows: The enthalpy change for a reaction that is the sum of two or more other reactions is equal to the sum of the enthalpy changes of the constituent reactions.

Let us consider two ways in which carbon dioxide may be produced from carbon and oxygen. Either carbon dioxide can be produced directly by the combustion of carbon, or carbon can first be converted to carbon monoxide, which can then be converted to carbon dioxide.

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta_{\mathrm{r}} H^{\ominus}=-394 \mathrm{~kJ}\) …. (1)

⇒ \(\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \Delta_{\mathrm{r}} H_1^{\ominus}=-110.5 \mathrm{~kJ}\) …. (2)

⇒ \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta_{\mathrm{r}} H_2^{\ominus}=-283.5 \mathrm{~kJ}\) …..(3)

Adding the enthalpy changes in (2) and (3) you will get -394 kJ mol^-1, which is the same as the enthalpy change in (1). Do you see now why the law is called Hess’s law of constant heat summation?

Application of Hess’s law: Hess’s law is very useful in determining enthalpy changes in reactions for which it is not possible to experimentally determine enthalpy changes. It follows from Hess’s law that thermochemical equations can be added, subtracted, multiplied, or divided like algebraic equations.

Let us again consider the reaction between carbon and oxygen. The principal product of this reaction is CO₂ but CO may also be produced due to insufficient oxygen and this may then form CO₂. It is difficult to measure directly the enthalpy of combustion of carbon to carbon monoxide, CO. Let us assume it is unknown and determine its values by applying Hess’s law.

Reaction for which enthalpy is to be found: \(2 \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})\) …(1)

Reactions for which data is available: \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})\) \(\Delta_{\mathrm{r}} H^\ominus=-393.5 \mathrm{~kJ}\) …. (2)

⇒ \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})\) \(\Delta_{\mathrm{r}} H^{\ominus}=-566.0 \mathrm{~kJ}\) ….. (3)

How do we use these equations (with data) to get equation (1)? We need C and O₂, on the LHS and CO on the RHS. In Equation (2) C and O₂ are already on the LHS but the coefficient of C is 1 while we require 2 (from Equation (1)).

Hence, by multiplying Equation (2) by 2, we get

⇒ \(2 \mathrm{C}(\mathrm{s})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\ominus}=2 \times(-393.5) \mathrm{kJ}\) ….(4)

[If the equation is multiplied by 2, \(\Delta_r H^{\ominus}\)must also be multiplied by 2.]

Next CO must appear on the RHS. Reversing Equation (3) to get CO on RHS, we get

⇒ \(2 \mathrm{CO}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\ominus}=-(-566)=+566 \mathrm{~kJ}\) …. (5)

[If the reaction is reversed, the sign of ΔH must also be reversed.]

Now adding reactions (4) and (5)

∴ \(2 \mathrm{C}(\mathrm{s})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\ominus}=-787.0 \mathrm{~kJ} \\
\begin{array}{lll}
2 \mathrm{CO}_2(\mathrm{~g}) & \longrightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g})  \Delta_{\mathrm{r}} H^{\ominus}=566 \mathrm{~kJ} \\
\hline 2 \mathrm{C}+\mathrm{O}_2 \longrightarrow 2 \mathrm{CO} & \Delta H^{\ominus}=-787+566=-221 \mathrm{~kJ}
\end{array}\)

Enthalpy Of Formation

The enthalpy or heat of the formation of a substance is the heat change accompanying the formation of 1 mol of the substance from its constituent elements and is usually written as Δ_tH. If all the substances involved in the reaction are in the standard state, the heat change is referred to as the standard heat of formation and is denoted by \(\quad \Delta_{\mathrm{f}} H^{\ominus}\) Consider the following equations.

∴ \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta_{\mathrm{f}} H^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

This means the heat liberated in the formation of 1 mol of gaseous carbon dioxide from its constituent elements is 393.5 kJ or that the enthalpy of formation of carbon dioxide is 393.5 kJ mol^-1.

∴ \(6 \mathrm{C}(\mathrm{s})+6 \mathrm{H}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s}) \quad \Delta_{\mathrm{f}} H^{\ominus}=-1169 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

When 1 mol of glucose is formed from its constituent elements in their standard states, the heat liberated is 1169 kJ, or the standard molar enthalpy of formation of glucose is 1169 kJ mol^-1 (‘molar’ referring to the formation of one mole of the substance).

By convention, the standard enthalpy of formation of elements is zero at all temperatures. However, when an element changes its state, the standard enthalpy of formation is not zero.

Usefulness of \(\Delta_{\mathrm{f}} H^{\ominus}\).

If you know the standard enthalpies of formation of the reactants and products of a chemical reaction, you can easily calculate the standard enthalpy change for that reaction. The difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants is the enthalpy change associated with the reaction.

Standard enthalpy = \(\left(\begin{array}{c}
\text { sum of standard heats of } \\
\text { formation of products }
\end{array}\right)-\left(\begin{array}{c}
\text { sum of standard heats of } \\
\text { formation of reactants }
\end{array}\right)\)

∴ \(\Delta_r H^{\ominus}=\Sigma \Delta_f H^{\ominus} \text { (products) }-\Sigma \Delta_f H^{\ominus} \text { (reactants) }\)

Example 1. Calculate the enthalpy change for the following reaction. \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Given that \(\Delta_t H^{\ominus}\) for CO₂(g), CO(g) and H₂O(g) are -393.5 kJ mol^-1, -111.3 kJ mol^-1, and -241.8 kJ mol^-1 respectively.

Solution:

⇒ \(\Delta_{\mathrm{r}} H^{\ominus}=\Sigma \Delta_{\mathrm{f}} H^{\ominus} \text { (products) }-\Delta_{\mathrm{f}} H^{\ominus} \text { (reactants) }\)

= \(\left[\Delta_f H^\ominus(\mathrm{CO})+\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{H}_2 \mathrm{O}\right)\right]-\left[\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{CO}_2\right)+\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{H}_2\right)\right]\)

=[(-1113) + (-2418)]-[(-393.5) + (0)] =- 353.1 + 393.5 = 40.4 kJ mol^-1.

Example 2. Calculate the enthalpy change for the following reaction.
\(\mathrm{CCl}_4(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})_{;} \Delta_{\mathrm{r}} H^{\ominus}=-41.4 \mathrm{kcal}\)

Given that,

⇒ \(\Delta_{\mathrm{f}} \mathrm{H}^{\ominus} \text { for } \mathrm{CCl}_4(\mathrm{~g}), \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text { and } \mathrm{CO}_2(\mathrm{~g}) \text { are }\) -25.5 k cal mol^-1,-57.8 k cal mol^-1 and -94.1k cal mol^-1 respectively.

Solution:

⇒ \(\Delta_{\mathrm{r}} H^{\ominus} =\left[\Sigma \Delta_{\mathrm{f}} H^{\ominus}(\text { products })\right]-\left[\Sigma \Delta_{\mathrm{f}} H^{\ominus}(\text { reactants })\right]\)

= \(\left[\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{CO}_2\right)+4 \times \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{HCl})\right]-\left[\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{CCl}_4\right)+2 \times \Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{H}_2 \mathrm{O}\right)\right]\)

-41.4 = \(\left[(-94.1)+4 \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{HCl})\right]-[(-25.5)+(2 \times-57.8)]\)

or -41.4 = \(-94.1+4 \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{HCl})+141.1\)

∴ \(4 \times \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{HCl})=-41.4-141.1+94.1=-88.4\).

∴ \(\Delta_f H^{\ominus}(\mathrm{HCl})=\frac{-88.4}{4}=-22.1 \mathrm{kcal} \mathrm{mol}^{-1}\).

In our discussion so far, we have not really distinguished between various types of reactions or processes such as combustion, neutralization, and change of state. When we have used the term enthalpy of a reaction, we have used it to refer to any chemical process or reaction. The enthalpies associated with different types of reactions or processes actually go by different names.

Enthalpy of combustion: The enthalpy or heat of combustion of a substance is the heat change accompanying the complete combustion of 1 mol of the substance in (excess) oxygen or air. The heat of combustion of carbon, for instance, is -393.5 kJ mol^-1.

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) ; \quad \Delta_{\mathrm{c}} H^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Standard enthalpy of combustion is defined as the enthalpy per mol of a substance, when all the reactants and products are in their standard states at the specified temperature.

Remember that the heat of combustion of a substance is the heat evolved when 1 mol of the substance is completely burnt or oxidised. Carbon is converted to carbon monoxide according to the following reaction.

⇒ \(\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g}) ; \quad \Delta_c H^{\ominus}=-110.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

But -110.5 kJ mol^-1 is not the heat of combustion of carbon.

Remember also that combustion reactions are always accompanied by the evolution of heat, so the heat of combustion is always negative.

Calorific value Combustion reactions are very important for us. We bum fuels to meet our energy requirements. We use the energy released by the oxidation of food to survive. The calorific value of a fuel or food is the amount of heat (in calories) released when 1 mol of the fuel or food is burnt or oxidised completely.

Example 1. A cylinder of cooking gas contains 11 kg of butane. The thermochemical equation for the combustion of butane is \(\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(\mathrm{l}); \quad \Delta_{\mathrm{c}} \mathrm{H}=-2658 \mathrm{~kJ}\)

If a family’s energy requirement per day for cooking is 12000 kJ, how long would a cylinder last?
Solution:

Molar mass of butane = 58 g.

58 g of butane produces = 2658 of heat.

Daily requirement of energy = 12000 kJ.

∴ a cyclinder would last = \(\frac{2658 \times 11 \times 10^3}{58 \times 12,000}=\) = 42 days.

Example 2. Calculate the heat of combustion of glucose from the following data.

1. \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}); \Delta_{\mathrm{r}} \mathrm{H}_1^{\ominus}=-395.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{r}} \mathrm{H}_2^{\ominus}=-269.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
3. \(6 \mathrm{C}(\mathrm{s})+6 \mathrm{H}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s}) ; \Delta_{\mathrm{r}} H_3^{\ominus}=-1169.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Solution: The required equation for the heat of combustion of glucose is

∴ \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \longrightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O} \quad \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}=?\)

Applying Hess’s law, the required equation can be obtained by multiplying (1) by 6 and (2) by 6, adding the products and subtracting (3) from the sum.

∴ \(\begin{aligned}
& 6 \mathrm{C}(\mathrm{s})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_2(\mathrm{~g}) \\
& \frac{6 \mathrm{H}_2(\mathrm{~g})+3\mathrm{O}_2(\mathrm{~g})}{} \longrightarrow 6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
& \hline 6 \mathrm{H}_2(\mathrm{~g})+6 \mathrm{C}(\mathrm{s})+9 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
&-\left[6 \mathrm{H}_2(\mathrm{~g})+6 \mathrm{C}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g})\right.\left.\longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})\right] \\
& \hline \mathrm{C}_6 \mathrm{H}_1 \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})
\end{aligned}\)

Then \(\Delta_{\mathrm{r}} H^{\ominus}\) = \(\Delta_{\mathrm{r}} H^{\ominus}=6 \Delta_{\mathrm{r}} H_1^{\ominus}+6 \Delta_{\mathrm{r}} H_2^{\ominus}-\Delta_{\mathrm{r}} H_3^{\ominus}\)

= 6(-395) + 6(-269.4)-(-1169.8)=-2816.6 kJ.

Enthalpy of neutralisation: The enthalpy change associated with the neutralisation of one gram equivalent of an acid by a base (or vice versa) in a dilute aqueous solution is called the enthalpy of neutralisation. For instance, the enthalpy of neutralisation of NaOH by HCl or HCl by NaOH is -57.1 kJ mol^-1. In other words, the enthalpy change associated with the neutralisation of one gram equivalent of HCl by NaOH (or vice versa) is -57.1 kJ mol^-1.

∴ \(\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} ; \quad \Delta_{\mathrm{n}} H^{\ominus}=-57.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The enthalpy of neutralisation of any strong acid by a strong base (or vice versa) is always -57.1 kJ mol^-1. This is because neutralisation is actually a reaction between the H⁺ ions produced by an acid and the OH^– ions produced by a base. Let us see how this is so in the case of HCl and NaOH.

∴ \(\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

or \(\mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}\)

• When 1 mol of OH^– ions combines with 1 mol of H⁺ ions to produce 1 mol of water, the heat liberated is 57.1 kJ mol^-1. Strong acids and bases are completely ionised in dilute aqueous solutions and the amount of H⁺ ions or OH^– ions produced by one gram equivalent of an acid or a base is always the same, i.e., 1 mol.
• If either the acid or the base, or both, are weak, the enthalpy of neutralisation is usually less than 57.1 kJ mol^-1 in magnitude. This is because weak acids and weak bases do not dissociate completely in an aqueous solution and a part of tire energy liberated during the combination of H⁺ ions and OH^– ions is utilised to ionise the weak acid or weak base.
• The heat used for ionising the weak acid (or base) is called the heat of ionisation or dissociation and the net heat of neutralisation is the difference between 57.1 kJ mol^-1 and the heat of dissociation.

∴ \(\underset{\text { Weak acdd }}{\mathrm{CH}_3 \mathrm{COOH}}+\underset{\text { Strong base }}{\mathrm{NaOH}} leftrightharpoons \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} ; \quad \Delta \mathrm{H}=-55.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

To calculate the enthalpy of ionisation, let us write down the ionisation reaction and the neutralisation reaction separately.

Ionisation \(\mathrm{CH}_3 \mathrm{COOH} \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} ; \quad \Delta_{\text {ion }} H^{\ominus}=?\)

Neutralisation \(\mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O} ; \quad \Delta_{\mathrm{n}} H^{\ominus}=-57.1 \mathrm{~kJ}\)

On adding the two equations, we get, \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}; \Delta_{\text {net }} H^{\ominus}=-55.4 \mathrm{~kJ}\)

The net result is that \(\quad \Delta_{\text {net }} H^{\ominus}=\Delta_{\text {ion }} H^{\ominus}+\Delta_{\mathrm{n}} H^{\ominus} \text { (strong acid-strong base) }\)

or \(\quad-55.4=\Delta_{\text {ion }} H^{\ominus}-57.1 \)

∴\(\quad \Delta_{\text {ion }} H^{\ominus}=-55.4+57.1=2.1 \mathrm{~kJ}\)

In general \(\Delta_{\text {ion }} H^{\ominus}=\left(\Delta_{\text {net }} H^{\ominus}+57.1\right) \mathrm{kJ}\) .

Example: What would be the enthalpy change when

1. 0.25 mol of HCl in solution is neutralised by 0.25 mol of a NaOH solution?
2. 0.5 mol of nitric acid in solution is mixed with a solution containing 0.2 mol of a potassium hydroxide solution?
3. 200 cm³ of a 0.2-M HCl solution is mixed with 300 cm3 of a 0.1-M NaOH solution?

Solution:

1. 0.25 mol of HC1 s 0.25 mol of H⁺.

0. 25 mol of NaOH = 0.25 mol of OH^–.

⇒ \(\mathrm{H}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O} \quad \Delta_n H^{\ominus}=-57.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Enthalpy change during the formation of 1 mol of H₂O = 57.1 kJ.

∴ enthalpy change when 0.25 mol of water is formed = 57.1 x 0.25 = 14.27 kJ.

2. 0.5 mol of HNO₃ = 0.5 mol of H⁺

0. 2 mol of KOH = 0.2 mol of OH^–

0. 5 mol of H⁺ ions will react with 0.2 mol of OH^– ions to produce 0.2 mol of H₂O.

Enthalpy change = 0.2 x 57.1 = 1142 kj.

3. 200 cm³ of 0.2-M HCl = 0.04 mol of H⁺.

300 cm³ of 0.1-M NaOH = 0.03 mol of OH^–.

0.03 moles of OH^– will react with 0.04 mol of H⁺ to produce 0.03 mol of H₂O.

Enthalpy change = 57.1 x 0.03 = 1.71 kj.

Enthalpies of phase change: A phase of a system is a homogeneous part of it throughout which all physical and chemical properties are the same. You will study phases and phase changes in detail later. A couple of examples will give you a basic idea.

• If you have a beaker full of water, the system you are considering consists only of liquid water, so it has one phase. If you drop a few cubes of ice in the water, your system now contains water and ice, so it is a two-phase system.
• When a system changes from one phase to another the process is referred to as a phase change or phase transition. Thus, the conversion of a solid into a liquid (fusion) is a phase change, as also the conversion of a liquid into a gas (vaporisation).
• There are other kinds of phase changes, but we are not concerned with those here. You already know that energy is needed to change a solid to a liquid and a liquid to a gas. The enthalpy changes associated with phase transitions go by different names. Remember, during a phase change, the temperature remains constant.

Enthalpy of fusion The enthalpy of fusion of a substance is the enthalpy change accompanying the conversion of 1 mol of the substance in its solid state into its liquid state at its melting point. For example, the standard enthalpy of fusion of ice is 6.0 kJ mol^-1.

∴ \(\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \quad \Delta_{\text {fus }} H^{\ominus}=+6.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

∴ \(\Delta_{\text {fus }} H^{\ominus}\) is the enthalpy of fusion in the standard state. Freezing is the reverse process of fusion, in that case an equal amount of heat is given off to the surroundings.

∴ \(\Delta_{\text {freex }} H^{\ominus}=-\Delta_{\text {fus }} H^{\ominus}\)

Enthalpy of vaporisation The enthalpy of vaporisation of a substance is the enthalpy change accompanying the conversion of 1 mol of the substance in its liquid state into its vapour state at the boiling point of the liquid. The standard enthalpy of vaporisation of water is 40.7 kj mol-1.

∴ \(\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \quad \Delta_{\text {vap }} H^{\ominus}=40.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Enthalpy of sublimation The enthalpy of sublimation of a substance is the heat change associated with the conversion of 1 mol of it directly from its solid to its gaseous state at a temperature below its melting point. For instance, the standard enthalpy of sublimation of iodine is 62.39 kJ mol^-1.

∴ \(\mathrm{I}_2(\mathrm{~s}) \longrightarrow \mathrm{I}_2(\mathrm{~g}) ; \quad \Delta_{\mathrm{sub}} H^{\ominus}=61.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The enthalpy change of n reverse transition is the negative of the enthalpy change of the forward transition (under the same conditions). This further proves enthalpy to be a state property. Therefore, the change in enthalpy would be same in both the cases when a solid is directly or indirectly converted to vapour. Thus, the enthalpy of sublimation can be expressed as:

⇒ \(\Delta_{\text {sub }} H^{\ominus}=\Delta_{\text {fus }} H^\ominus+\Delta_{\text {vap }} H^{\ominus}\)

• The striking differences in the magnitude of the enthalpy change for various substances is attributed to the intermolecular interaction in the substances. The enthalpy of vaporisation of water at its boiling point is 40.79 kJ mol^-1.
• This signifies that water molecules are held together more tightly than any other liquid with low enthalpy of vaporisation, for instance, acetone. The high enthalpy of vaporisation of water is partly responsible for low humidity in the atmosphere.

Enthalpy of atomisation: The enthalpy change on breaking a molecule completely into its gaseous atoms is called enthalpy of atomisation \(\Delta_{\mathrm{a}} H\). In case of a metal, the enthalpy of atomisation is tine same as the enthalpy of sublimation.

⇒ \(\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{C}(\mathrm{g}) \Delta_{\mathrm{a}} H^{\ominus}=716.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\mathrm{Na}(\mathrm{s}) \longrightarrow \mathrm{Na}(\mathrm{g}) \Delta_{\mathrm{a}} H^{\ominus}=108.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The enthalpy of atomisation of diatomic gases is half the bond dissociation enthalpy. For example, the standard bond dissociation enthalpy of O₂ is 497 kJ mol^-1 and the standard enthalpy of atomisation pertaining to the reaction is 249.2 kJ mol^-1.

⇒ \(\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{O}(\mathrm{g})\)

Enthalpy of allotropic transformation: You know that elements like carbon, sulfur, and phosphorus can exist in different allotropic forms and that such elements can change from one allotropic form into another. Allotropic transformations involve enthalpy changes.

• The enthalpy of the allotropic transformation of one allotropic form of a substance into another is the heat change accompanying the transformation per mole of the substance. The standard enthalpy of transformation of rhombic sulphur to monoclinic sulfur, for example, is 1.3 kJ mol^-1.
• It is not easy to determine enthalpy changes for allotropic transformations experimentally because such processes are rather slow and the enthalpy changes associated with them are small. Hess’s law can be used conveniently to determine the enthalpy changes accompanying allotropic transformations.

Example: Calculate the enthalpy of the allotropic transformation of rhombic sulphur to monoclinic sulphur from the following thermochemical equations.

1. \(
   \mathrm{S}\left(\text { rhombic) }+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g})\right. \Delta_{\mathrm{c}} H_1^{\ominus}=-295.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) …(1)
2. \(\mathrm{~S}(\text { monoclinic })+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g}) \Delta_{\mathrm{c}} H_2^{\ominus}=-296.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)……(2)

Solution:

The required equation is S(rhombic) → S(monoclinic) ΔH =?

Applying Hess’s law and subtracting equation (2) from (1)

∴ \(\begin{array}{rc}
\mathrm{S}(\mathrm{r})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g}) & \Delta_{\mathrm{c}} H_1=-295.1 \mathrm{~kJ} \\
\mathrm{~S}(\mathrm{~m}) \pm \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{SO}_2(\mathrm{~g}) & \Delta_{\mathrm{c}} H_2=-296.4 \mathrm{~kJ} \\
\hline \mathrm{S}(\mathrm{r}) \longrightarrow \mathrm{S}(\mathrm{m}) & \Delta H=13 \mathrm{~kJ}
\end{array}\)

Therefore, the enthalpy of the allotropic transformation of rhombic sulphur to monoclinic sulphur is 1.3 kJ mol^-1.

Enthalpy of solution: The enthalpy of reaction for dissolving one mole of a solute in n moles of a solvent is known as the integral enthalpy of the solution or the integral heat of the solution. Let us look at some such values for a solution of HNO₃ in water at 298 K.

1. \(\left.\mathrm{HNO}_3(\mathrm{l})+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HNO}_3 \text { (in } 1 \mathrm{H}_2 \mathrm{O}\right) \Delta_{\text {sol }} H^{\ominus}=-187.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ….(1)

2. \(\left.\mathrm{HNO}_3(\mathrm{l})+10 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HNO}_3 \text { (in } 10 \mathrm{H}_2 \mathrm{O}\right) \Delta_{\text {sol }} H^{\ominus}=-205.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) …. (2)

3. \(\mathrm{HNO}_3(\mathrm{l})+100 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HNO}_3 \text { (in } 100 \mathrm{H}_2 \mathrm{O} \text { ) } \Delta_{\text {sol }} H^{\ominus}=-206.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)…. (3)

4. \(\mathrm{HNO}_3(\mathrm{l})+500 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HNO}_3 \text { (in } 500 \mathrm{H}_2 \mathrm{O} \text { ) } \Delta_{\text {sol }} H^{\ominus}=-206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\)…..(4)

5. \(\left.\mathrm{HNO}_3(\mathrm{l})+1000 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HNO}_3 \text { (in } 1000 \mathrm{H}_2 \mathrm{O}\right) \Delta_{\text {sol }} H^{\ominus}=-206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\)….(5)

6. \(\mathrm{HNO}_3(\mathrm{l})+\mathrm{aq} \longrightarrow \mathrm{HNO}_3 \text { (aq) } \Delta_{\text {sol }} H^{\ominus}=-207.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) …..(6)

As you can see from Equations (1)-(6), the heat of solution varies with concentration or decreases with increase in the amount of the solvent and finally stabilises at infinite dilution.

The enthalpy of solution at infinite dilution is the enthalpy change when 1 mol of the substance dissolves in such a large amount of solvent that the interaction between the solute molecules is negligible. In case of water as a solvent, infinite dilution is denoted by ‘aq’ (aqueous).

For HNO₃, the enthalpy of solution at infinite dilution refers to Equation (6) and is -207.36 kJ mol^-1.

Some more values of enthalpy of solution at infinite dilution are given below.

⇒ \(\mathrm{KCl}(\mathrm{s})+\mathrm{aq} \longrightarrow \mathrm{KCl}(\mathrm{aq}) ; \Delta_{\mathrm{sol}} H^{\ominus}=+18.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\mathrm{CuSO}_4(\mathrm{~s})+\mathrm{aq} \longrightarrow \mathrm{CuSO}_4(\mathrm{aq}) ;  \Delta_{\mathrm{sol}} H^{\ominus}=-66.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}+\mathrm{aq} \longrightarrow \mathrm{CuSO}_4(\mathrm{aq}) ; \Delta_{\text {sol}} H^{\ominus}=+11.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

• In the first and third cases \(\Delta_{\text {sol}}\) is positive (endothermic), while in the second, it is negative (exothermic). In general, salts which do not form hydrates dissolve in water with the absorption of heat. Hydrated salts also dissolve in water with the absorption of heat. The process is exothermic only in the case of anhydrous salts which form hydrates.
• When a substance dissolves in water, it either absorbs (positive enthalpy) or releases energy (negative enthalpy). Ammonium nitrate has a positive enthalpy of solution. Tire cold pack used for minor injuries to sportsmen contains NH₄NO₃ and water in separate compartments.
• It is activated on punching the partition between the two upon which the salt dissolves and the temperature of water falls. Calcium chloride, on the other hand, has a negative enthalpy of solution, and this phenomenon is used to dissolve ice on sidewalks in winters.
• Let us see what happens when NaCl dissolves in water. First, the three-dimensional network of Na⁺ and Cl^– breaks into individual ions. For this purpose the lattice energy which is responsible for holding the ions together must be overcome. The separated Na⁺ and Cl^– ions get stabilised in tire solution by their interaction with the water molecules.

This process involves breaking some of the hydrogen bonds between the water molecules. The ions are then said to be hydrated. The process of dissolution of an ionic compound in a solvent (water) involves a complex interaction between the solute and the solvent species. However, it can be broadly taken to be happening in two steps:

1. First the ionic lattice breaks: \(\mathrm{NaCl}(\mathrm{s}) \longrightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g})\)

Energy is required for this process and therefore the reaction is endothermic. The enthalpy change associated with it is called lattice enthalpy it may be defined as the heat change when one mole of an ionic solid separates into its constituent gaseous atoms.

2. Next the gaseous Na⁺ and Cl^– ions interact with water molecules and become hydrated:

⇒ \(\mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g}) \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

The enthalpy change associated with this process is called the enthalpy of solvation or more precisely the enthalpy of hydration as water is the solvent.

∴ \(\Delta_{\text {sol }} H=\Delta_{\text {lattlee }} H+\Delta_{\text {hyd }} H\)

Lattice enthalpy: It is not possible to determine lattice energy directly by experiments. Hence, it is determined on the basis of the Born-Haber cycle. The Bom-Haber cycle is a closed sequence of different processes which are involved in the breaking and finally in the making of an ionic crystal. To understand the cycle, let us take the example of NaCl(s). The steps involved in the formation of NaCl(s) are as follows:

1. Sublimation of Na(s) to Na(g): \(\mathrm{Na}(\mathrm{s}) \longrightarrow \mathrm{Na}(\mathrm{g}) \quad \Delta H=\Delta_{\mathrm{a}} H^{\ominus}[\mathrm{Na}(\mathrm{s})]=108.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

2. Dissociation of Cl₂(g) to give Cl(g): \(\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{Cl}(\mathrm{g}) \quad \Delta H=\frac{1}{2} \Delta_{\mathrm{Cl}-\mathrm{C}} H^{\ominus}=\frac{1}{2} \times 242 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

3. Ionisation of Na(g): \(\mathrm{Na}(\mathrm{g}) \longrightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-} \quad \Delta H=\Delta_{\text {ien }} H^{\ominus}=495.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Ionisation enthalpy of sodium = 495.8 kJ mol^-1. Since sodium is in the gaseous state, the enthalpy change is given by

∴ \(\Delta_{\text {ion }} H=\Delta_{\text {ion }} U+\Delta n_g R T\)

∴ \(\Delta_{\text {ion }} H=495.8+(1) \times R T\)

(It may be observed subsequently that RT cancels out with -RT in the next step.)

4. The electron released by the sodium atom is gained by chlorine to give Cl^–.

⇒ \(\mathrm{Cl}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(\mathrm{g})\)

The electron affinity (E_A) of Cl is 348.6 kJ mol^-1. Since electron affinity is defined as the energy released in the above process, from thermodynamic considerations we must take a negative value.

Hence energy involved (-E_A) =-348.6 kJ mol^-1. Here again, we need the electron gain enthalpy (\(\Delta_{\mathrm{eg}} H)\), which is -E_A – RT (since Δn_g = -1)

∴ \(\Delta_{\mathrm{eg}} H)\) = -348.6 + (-1)RT = -348.6 – RT.

5. Na⁺(g) and Cl^– (g) (steps 3 and 4) combine to give NaCl(s).

⇒ \(\mathrm{Na}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g}) \longrightarrow \mathrm{NaCl}(\mathrm{s})\)

Enthalpy change for this process = –\(\Delta_{\text {Laltice }} H\)(The process involves the formation of the lattice.) These steps are depicted as the sequence of steps in the Bom-Haber cycle.

The basis of calculations using the above cycle is that enthalpy being a state function, the sum of enthalpy changes around the cycle (clockwise direction) is zero.

∴ \(-\Delta_{\mathrm{f}} H[\mathrm{NaCl}(\mathrm{s})]+\Delta_{\text {sub }} H[\mathrm{Na}(\mathrm{s})]\)+\(\frac{1}{2} \Delta_{\text {diss }} H\left[\mathrm{Cl}_2(\mathrm{~g})\right]+I \cdot E .+R T-E . A-R T-\Delta_{\text {lattice }}\) H =0

or \(\Delta_{\text {lattice }} H=-\Delta_{\mathrm{f}} H+\Delta_{\text {sub }} H+\frac{1}{2} \Delta_{\text {diss }} H+I \cdot E .-E . A .\)

The \(\Delta_{\mathrm{f}}H\) of NaCl(s) is -411.2 kJ.

Substituting the appropriate values on the RHS, we get

∴ \(\Delta_{\text {Lattice }} H =411.2+108.4+\frac{1}{2} \times 242+495.8-348.6\) =1136.4-348.6

∴ \(\Delta_{\text {Lattice }} H =787.8 \mathrm{~kJ}\).

Bond energy: By now you know what enthalpies of formation of compounds are. You also know that compounds are formed by the breaking and making of bonds. To have a precise knowledge of the enthalpy change for a process, the bond dissociation enthalpy or the bond dissociation energy [ΔH(A-B)] is considered.

For example, when the dissociation, or breaking, of a chemical bond occurs as in the following process,the corresponding molar enthalpy change is called the bond dissociation enthalpy. Thus, the bond dissociation enthalpy is tire enthalpy associated with the breaking of 1 mol of a substance in the gaseous state completely into its gaseous atoms.

∴ AB(g) → A(g) + B(g)

The values of bond enthalpies depend on the bonding present between two atoms in the molecule. Even in the same molecule, the values may differ. For example, to break the first O—H bond in water, the enthalpy change \(\Delta H^{\ominus}(\mathrm{HO}-\mathrm{H})\)= 492 kJ mol^-1 while for breaking the second bond, \(\Delta H^{\ominus}(\mathrm{O}-\mathrm{H})\) is 428 kJ mol^-1.

• In methane four hydrogen atoms are attached to a carbon atom and the four C—H bonds are equal in energy and bond length. In this case the total enthalpy is expected to be four times the C—H bond enthalpy.
• However, this is not the case, the reason being different dissociation steps. The first step involves the breaking of a C—H bond in the CH₄ molecule but in the next step the C—H bond is broken in a CH₃ radical, and so on.

The energies required to break the individual C—H bonds in different dissociation steps are as follows.

∴ \(\mathrm{CH}_4(\mathrm{~g}) \longrightarrow \mathrm{CH}_3(\mathrm{~g})+\mathrm{H}(\mathrm{g}) ; \Delta_{\text {bond }} H^{\ominus}=427 \mathrm{~kJ} \mathrm{~mol}^{-1},\)

∴ \(\mathrm{CH}_3(\mathrm{~g}) \longrightarrow \mathrm{CH}_2(\mathrm{~g})+\mathrm{H}(\mathrm{g}) ; \Delta_{\text {bond }} H^{\ominus}=439 \mathrm{~kJ} \mathrm{~mol}^{-1},\)

∴ \(\mathrm{CH}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}(\mathrm{g})+\mathrm{H}(\mathrm{g}) ; \Delta_{\text {bond }} H^{\ominus}=452 \mathrm{~kJ} \mathrm{~mol}^{-1},\)

∴ \(\mathrm{CH}(\mathrm{g}) \longrightarrow \mathrm{C}(\mathrm{g})+\mathrm{H}(\mathrm{g}) ; \Delta_{\text {bond }} H^{\ominus}=347 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

The total enthalpy change for the atomisation of methane \(\left[\mathrm{CH}_4(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})\right] \text { is } \Delta_{\mathrm{a}} H^{\ominus}\) = 166 x 103 kJ mol^-1.

In case of reactions for which experimental data are not available for successive steps, mean bond enthalpy is used. It is the value of bond dissociation energy of a bond A—B averaged over a series of related compounds. For example, the bond energy for the O—H bond, \(\Delta H^{\ominus}\)(O—H) (averaged between H₂O and alcohols) is 463 kJ mol^-1.

Similarly, in case of CH₄, all the bond enthalpies added together come to 1665 kJ mol^-1. The mean bond enthalpy of C—H bond in CH₄ is

⇒ \(\frac{1}{4} \Delta_a H^{\ominus}=\frac{1}{4} \times 1665=416 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Now consider the atomisation of a diatomic molecule: \(\mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{g}) ; \quad \Delta_a H^{\ominus}=435 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Since two atoms are formed, the energy value is the enthalpy of atomisation of H₂ at 298 K.

The corresponding bond energy \(\Delta_a U^{\ominus}\) for the same reaction (with a little difference from the enthalpy) is 430.8 kJ mol^-1. That is why the two quantities are often used interchangeably. But the correct way would be to convert from ΔU to ΔH.

We know that ΔH = ΔU + pΔV

Some important bond enthalpies (kJ mol^-1) of multiple bonds are as follows.

C—C = 612 N=N = 418 O=O= 497

C≡C = 837 N≡N = 946

C=O = 741 C=N = 615

• For gases, pΔV (at constant pressure) may be replaced by RT. In case of a diatomic molecule, the bond energy is equal to the bond dissociation energy.
• The dissociation energy may here be defined as the enthalpy change involved in breaking the bond between atoms of a gaseous diatomic molecule. Also, it can be seen that bond enthalpy in homonuclear diatomic molecules is twice the enthalpy of formation for the atom in the gaseous state.

Bond energies can be used to calculate the enthalpy of a reaction. The standard reaction enthalpy, \(\Delta_r H^{\ominus}\) is the difference between the sum of the standard enthalpies of the reactants and products.

∴ \(\Delta_r H^{\ominus}=\Sigma \text { bond enthalpies }{ }_{\text {reactants }}-\Sigma \text { bond enthalpies products }\)

This relationship is valid when all the substances in a chemical reaction are in the gaseous state.

Example 1. Calculate the enthalpy change for the following reaction. \(2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

The average bond energies of the C—H, C≡C, O=O, C—O, and O—H bonds are 414, Hit), 499, 724, and 460 kJ mol^-1 respectively.
Solution:

ΔH = sum of bond energies of reactants – sum of bond energies of products.

The given reaction can be written as 2H—C≡C—H(g) + 5O=O(g) → 4O=C=)(g) + 2H—O—H(g)

∴ \(\Delta H=\left[2 \Delta H_{\mathrm{C} \equiv \mathrm{C}}+4 \Delta H_{\mathrm{C}-\mathrm{H}}+5 \Delta H_{\mathrm{O}=\mathrm{O}}\right]-\left[8 \Delta H_{\mathrm{C}=\mathrm{O}}+4 \Delta H_{O-H}\right]\)

= (2 x 810 + 4 x 414 + 5 x 499) – (8 x 724 + 4 x 460)

= 5771 – 7632 = -1861 kJ mol^-1.

Example 2. Calculate the bond energy of the C—H bond in CH₄ front the following data.

1. \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_4(\mathrm{~g}) \quad \Delta H_1^{\ominus}=-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ….(1)
2. \(\mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{g}) \quad \Delta \mathrm{H}_2^{\ominus}=435.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) …..(2)
3. \(\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{C}(\mathrm{g}) \quad \Delta H_3^{\ominus}=718.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)….(3)

Solution:

To find the bond energy of the C—H bond in CH₄, let us first write the equation for the dissociation of CH₄ into C and H, The enthalpy change for this reaction would be four times the bond energy of the C—H bond since the dissociation of the CH₄ molecule involves the breaking of four C—H bonds.

CH₄ (g) → C(g) + 4H(g) ΔH = ?

The enthalpy change for this reaction can be calculated from the given data by applying Hess’s law. The equation for this reaction can be obtained by multiplying Equation (2) by 2, adding the product to Equation (3), and then subtracting Equation (1) from the sum.

∴ \(\begin{aligned}
\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g}) & \longrightarrow 4 \mathrm{H}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \\
\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g}) & \longrightarrow \mathrm{CH}_4(\mathrm{~g}) \\
\hline \mathrm{CH}_4(\mathrm{~g}) & \longrightarrow 4 \mathrm{H}(\mathrm{g})+\mathrm{C}(\mathrm{g})
\end{aligned}\)

∴ ΔH = 2 x ΔH₂ + ΔH₃ – ΔH₁

= 2 x 435.4 + 718.4-(-748)

= 870.8 + 718.4 + 748 = 16640 kJ mol^-1.

Thus, the energy required to break four C—H bonds = 1664.0 kJ mol^-1.

Therefore, the bond energy of the C—H bond = 1664/4 = 416 kJ mol^-1.

Alternatively, the enthalpy of the formation of the reactants and products can also be used to find the enthalpy change for the required reaction.

ΔH = ∑ΔH_f (products) -∑ΔH_f (reactants)

ΔH = [ΔH₃ +2ΔH₂]-ΔH₁

= [718.4 + 2 x 435.4] – [-748]

= 718.4 + 870.4 + 74.8 = 1664.0 kJ mol^-1,

∴bond energy of C—H bond = 1664/4 = 416 kJ mol^-1.

The Measurement Of Heat

In the laboratory, heat changes in physical and chemical processes are measured with a calorimeter. Calorimetry is the measurement of heat changes associated with a process. The study of calorimetry involves the concept of specific heat and heat capacity. So let us define these first

The heal capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of it by 1°C Heat capacity is an extensive property. It is more convenient to use an intensive property. Therefore, the molar heat capacity, C_m (C_m = C/n in JK^-1 mol^-1 where n is the amount of the substance), is used. The specific heat capacity C_s (C_s = C/m in JK^-1 g^-1, where m is the mass of the substance) is also an intensive property.

For water, the specific heat is \(1 \mathrm{cal} \mathrm{g}^{-1} \mathrm{C}^{-1} \text { or } 4.18 \mathrm{~J} \mathrm{~g}^{-10} \mathrm{C}^{-1}\)

The molar heat capacity of a substance is the heat required to raise the temperature of 1 mol of the substance by one degree.

The heat capacity at constant volume (C_v) is given by \(C_v=\frac{q_v}{\Delta T} .\)

From the first law of thermodynamics, we have q = ΔU ÷ pΔV.

At constant volume, ΔV = 0. q_v = AU.

∴ heat capacity at constant volume is given by \(C_V=\left(\frac{\Delta U}{\Delta T}\right)_V\)

Thus, the heat capacity at constant volume is defined as the rate of change of internal energy with temperature.

The heat capacity at constant pressure is given by \(C_p=\frac{q_p}{\Delta T}.\)

We have already learned that, at constant pressure, ΔH = q_p.

Therefore, heat capacity at constant pressure is given by \(C_p=\left(\frac{\Delta H}{\Delta T}\right)_p\)

Thus, the heat capacity at constant pressure is defined as the rate of change of enthalpy with temperature. Relationship between C_p and C_v We know that, at constant pressure (Equation 6),

ΔH = ΔU + pΔV.

For 1 mole of an ideal gas, pΔV = RΔT (ideal gas equation).

Thus, Equation 6 can be written as ΔH = ΔU + RΔT

On dividing the equation by ΔT, we get \(\frac{\Delta H}{\Delta T}=\frac{\Delta U}{\Delta T}+R\)

or C_p = C_v + R

or C_p – C_v = R

C_p is always greater than C_v for any gas, since in a constant-pressure process a system has to do work against the surroundings.

Coming back to the measurement of energy changes associated with chemical or physical processes, measurements are made under two different conditions:

1. at constant volume, ΔU -q_v, and
2. at constant pressure, ΔH = q_p.

Constant-volume calorimetry: The reactions that can be most easily studied under constant- volume conditions are combustion reactions. The apparatus used for this purpose is called the bomb calorimeter or adiabatic bomb calorimeter. It consists of a sealed constant-volume steel container (known as a bomb) in which the reaction occurs. The bomb can withstand high pressures.

• A known mass of a combustible substance is placed in the bomb, which is filled with oxygen at 30 atm pressure. The sealed bomb is then immersed in a known amount of water contained in an insulated container. The whole set-up is called the calorimeter. Since the outer covering is insulated, there is no heat exchange with the surroundings and hence the reaction process is adiabatic.
• The sample is ignited electrically and the heat produced by the combustion reaction is absorbed by the water surrounding the bomb and the bomb itself. The temperature of the water is monitored and the heat produced is calculated using the equation

∴ \(q=C_{\text {cal }} \Delta T,\) ….(1)

where C_cal is the heat capacity of the calorimeter. C_cal is determined separately by igniting a known amount of a substance whose heat of combustion is already known, in the same calorimeter.

• Thus the heat produced in the reaction may be determined. We already know that at constant volume q_v = ΔU. In constant- volume calorimetry, the heat change observed is simply the change in internal energy.
• This value may be modified to get the enthalpy, but the pressure changes are usually small and may be neglected. The heat changes may be simply assumed to correspond to enthalpy changes.

Constant-pressure calorimetry: To measure the heat changes at constant pressure, a device comparable to a thermos flask is used. The calorimeter is just a vessel with a stirrer and a thermometer.

The system is insulated from the surroundings and hence acts as an isolated system. The heat changes within the vessel are measured in terms of the change in temperature of the system. Constant monitoring of the temperature before and after the reaction helps in finding q and hence AH, as we know that, ΔH = q_p

The amount of heat exchanged (q_v or q_p) in any process is given by, q_p =msΔT……(2)

where m is the mass of the substance, s is its specific heat and ΔT is the change in the temperature.

You already know that q = CΔT =C_p ΔT (at constant pressure),

where C = ms, the heat capacity of the system.

• Thus to determine q, C and ΔT have to be known first. The experimental technique—calorimetry—actually involves two steps, the first being the determination of the heat capacity of the calorimeter.
• The second step involves the determination of the change in temperature during the completion of the reaction. Since it is an insulated system, no heat is lost to the surroundings as in the case of constant-volume calorimetry.

As you already know, the enthalpy of the reaction is simply the heat change at constant pressure,  ΔH = q_p (at constant pressure).

Enthalpy changes are commonly tabulated at 25°C and it is often required to know their values at other temperatures. They can be calculated from the heat capacities of the reactants and products.

You know that ΔH = H(products) – H(reactants).

At constant pressure ΔH /ΔT = C_p.

Therefore, for small changes in temperature, \(\frac{\Delta(\Delta H)}{\Delta T}=C_p \text { (products) }-C_p \text { (reactants) }=\Delta C_p \Delta T\)

∴ \(\Delta(\Delta H)=\Delta C_p\left(T_2-T_1\right)\)

or \(\Delta H_2-\Delta H_1=\Delta C_p\left(T_2-T_1\right)\),

where ΔH₂ is the enthalpy of the reaction at T₂ and ΔH₁ is that at T₁.

Example 1. Calculate the enthalpy change on the freezing of 1 mol of zoater at -5°C to ice at -5°C. Given that

⇒ \(\Delta_{\text {tom }} H^{\oplus}\left(\mathrm{H}_2 \mathrm{O}\right)=6.03 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { at } 0^{\circ} \mathrm{C} \text { and } C_p\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]\)

= \(75.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, C_{,},\left[\mathrm{H}_2 \mathrm{O}(\mathrm{s})\right]=36.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

Solution:

To find the enthalpy change at -5°C, we use the equation: \(\Delta H_2-\Delta H_1=\Delta C_p\left(T_2-T_1\right) .\)

First, we will find Δ_fus H at -5°C and then convert it to ΔH (freezing) by prefixing a (-) sign to Δ_fusH.

Also, ΔH is in kJ and C_p in J. Converting, Δ_fusH = 6.03 kJ mol^-1 = 6.03 x 1000 J mol^-1.

∴ \(\Delta H_2-6.03 \times 1000= (75.3-36.8)\times(-5-0) \)

= \(C_p(1)-C_p(\mathrm{~s})=-192.5\)

∴ \(\Delta H_2=-192.5+6030=5837.5\)

Thus \(\Delta_{\text {fus }} H at -5^{\circ} \mathrm{C}=5837.5 \mathrm{~J} \mathrm{~mol}^{-1}=5.84 \mathrm{~kJ} \mathrm{~mol}^{-1} and \Delta H (freezing) at -5^{\circ} \mathrm{C}\)

= \(-5.84 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Example 2. Find the heat required to raise the temperature of 27.9 g of Fe from 30 to 40°C. The molar heat capacity of Fe(s) =25.10 J K^-1mol^-1.
Solution:

The molar heat capacity is the heat required to raise the heat of 1 mol of the substance, i.e., 55.8 g (molar mass of Fe) by 1°C.

The heat required to raise the temperature of 55.8 g of Fe by 1°C = 25.1 J.

The heat required to raise the temperature of 27.9 g of Fe by 1°C = \(\frac{25.1}{55.8}\) x 27.9 J.

∴ the heat required to raise the temperature of 27.9 g of Fe by 10°C = \(\frac{25.1}{55.1}\)x 27.9 x 10 = 125.5 J.

Example 3. 1.922 g of methanol (CH₃OH) was burnt in a constant-volume bomb calorimeter. The temperature of water rose by 4.2°C. If the heat capacity of the calorimeter and its contents is 10.4 kJ °C^-1, calculate the molar heat of combustion of methanol.
Solution:

If qcal is the quantity of heat involved in the reaction and C„ is the heat capacity of the calorimeter, then

∴ \(q_{\text {eat }}=C_V \Delta T\)

But \(q_{\text {cal }}=-q_r\)

∴ \(\quad q_r=-C_v \Delta T\)

= \(-10.4 \times 4.2\left[because \text { the temperature rose by } 4.2^{\circ} \mathrm{C}, \Delta T \text { is positive }\right]\) = \(-43.68 \mathrm{~kJ}\).

The minus sign indicates exothermic nature of the reaction.

Since the pressure changes are small and can be neglected, q_r = ΔH_r.

Tim ΔH for burning 1.922 g of methanol is -43.68 kJ.

The ΔH for burning 1 mol or 32 g of methanol is obtained as

∴ \(\Delta H=\frac{-43.68 \mathrm{~kJ}}{1.922 \mathrm{~g}} \times 32 \mathrm{~g} \mathrm{~mol}^{-1}\) = -727.24 kJ mol^-1.

The molar enthalpy or molar heat of combustion is therefore -727.24 kJ mol^-1.

Spontaneous Processes

Generally speaking, when we say something has happened spontaneously, we mean that it has happened of its own accord. In chemistry, the meaning is a little different. It would be easy enough for you to associate the word spontaneous with a process like the dissolving of sugar in water, or the evaporation of water from an open vessel.

1. But the burning of coal in air, or the burning of ethane to produce carbon dioxide and water are also spontaneous processes, even though these processes have to be initiated by ignition.
2. Any process which can occur under a given set of conditions is called a spontaneous process, irrespective of whether it occurs on its own, or needs to be initiated. In other words, a spontaneous process occurs on its own or has a tendency to occur. A simple way of saying this is that a spontaneous process is a process which is possible.
3. It is not possible, for example, to dissolve sand in water, so this is not a spontaneous process. Water cannot flow uphill, so this is not a spontaneous process either. Some nonspontaneous processes can, however, be made to occur by providing energy from an external source.
4. For instance, water can be made to flow uphill by using a pump. Similarly, though the electrolysis of water is not a spontaneous process, it can be made to occur by supplying electrical energy.

You may wonder about the difference between a nonspontaneous process which can be made to occur by supplying energy and a spontaneous process which requires initiation. The combination of gaseous hydrogen and oxygen to form water is a spontaneous reaction which has to be initiated by an electric spark.

∴ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

• The electrolysis of water is a nonspontaneous process which requires electrical energy. The difference is that once a spontaneous reaction has been initiated, it proceeds on its own, while a nonspontaneous process stops as soon as the supply of external energy stops.
• Once the electric spark starts the reaction between hydrogen and oxygen, the process continues on its own. But if the supply of electric current is stopped during the electrolysis of water, the process stops.

Criterion for spontaneity: What makes certain processes possible or spontaneous, and certain others not possible or nonspontaneous? What is the driving force behind the occurrence of different processes? If we know this we can predict whether a certain process would be possible.

• You have already read in several contexts that every system seeks to have the minimum possible energy in order to acquire the maximum stability. Water flows down a hill to have the minimum possible potential energy. A wound spring unwinds itself to minimise energy. Atoms form bonds with each other in order to have less energy.
• This may lead you to the conclusion that the criterion for a process is the reduction of energy, or that a process is possible if it causes a reduction in the energy of the system. Let us consider some chemical reactions and see if this is right. In exothermic reactions, the energy of the products is less than the energy of the reactants and the reaction is accompanied by the evolution of heat.

⇒ \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta H =-394 \mathrm{~kJ}\)

⇒ \(\mathrm{~N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \Delta H =-92 \mathrm{~kJ}\)

In such cases at least we may be right in saying that the tendency to attain a state of minimum energy (i.e., negative enthalpy change) is the driving force behind the occurrence of a reaction. But what about endothermic reactions which are spontaneous?

⇒ \(\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta H=5.86 \mathrm{~kJ}\)

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(\mathrm{s})+\mathrm{aq} \longrightarrow \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \Delta H=177.8 \mathrm{~kJ}\)

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_3(\mathrm{~g}) \Delta H=15.1 \mathrm{~kJ}\)

• These three reactions involve the absorption of heat, which means that the reactions lead to an increase in the energy of the system, so reduction of energy cannot be the only criterion determining the spontaneity of a reaction.
• Reversible reactions also prove that decrease in energy or negative enthalpy change cannot be the only determining factor for the feasibility of a reaction. Suppose the forward reaction is exothermic then the backward reaction has to be endothermic and it would not occur if the only criterion for spontaneity were reduction of energy.

The tendency towards maximum disorder: To investigate what other factors could possibly determine the spontaneity of a process, let us consider a process for which ΔH = 0, since for such a process the energy factor could not be the driving force. The mixing of air and bromine vapour is just such a process.

• Suppose vessel A contains bromine vapour and vessel B contains air and there is a movable partition between the two. Now suppose that the partition is removed. The two gases will mix completely. What makes this possible? When the samples of air and bromine vapour are in two separate vessels, there is more order and when they diffuse into each other, there is more disorder.
• This tendency of a system to move towards greater disorder or to maximise randomness is the other driving force which makes reactions possible. It explains why endothermic reactions are possible despite the fact that ΔH is positive for such reactions. Take the case of the dissolution of ammonium chloride in water.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(\mathrm{s})+\mathrm{aq} \longrightarrow \mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\)

• This is an endothermic process, but it is possible because the randomness of the system increases. How? When the ions are held together in the crystal lattice there is more order and when they go into the aqueous solution, they are more free to move about, hence there is more disorder.
• Just as the energy factor (minimising energy) cannot be the only criterion which determines the spontaneity of a process, nor can maximising randomness be the sole determining factor. If this were the case, the liquefaction of a gas (which increases order or decreases randomness) would not have been possible. But we will come to that later.

Entropy: The randomness or disorder of a system is measured in terms of a function called entropy. It is a state function, represented by S and measured in J K^-1.

Change in entropy (ΔS) = S_{final state} – Sintial state

or ∑ S_products – ∑ S_reatants

When ΔS for a process is positive, the process leads to an increase in randomness or disorder.

In the examples discussed earlier, the processes are accompanied by an increase in entropy. We may again tend to conclude that for spontaneous processes there must be an increase in entropy. However, there are cases where entropy decreases. For example, the spontaneous condensation of steam to liquid water.

⇒ \(\underset{s^{\ominus}=188.7 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}{\mathrm{H}_2 \mathrm{O}(\mathrm{g})} \longrightarrow \underset{s^{\ominus}=70.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})}\)

• The standard molar entropy values for liquid water and water vapour are given and show that there is a decrease in entropy. Standard entropy may be defined in a manner similar to standard enthalpy.
• The standard molar entropy of a substance is the entropy of one mole of the pure substance at 1 bar pressure and a specified temperature, usually 298 K. Let us not be in a hurry to make conclusions. When steam condenses, energy is given out, \(\Delta H^\ominus\) being -44.1 kJ mol^-1 for the reaction.
• Where does this energy go? If the water condenses on a window, the energy is passed on to the window and when it condenses in the air, energy is taken up by the air molecules. Therefore, the entropy of the surroundings increases.
• The tendency towards disorder is summed up in the second law of thermodynamics, which states that spontaneous reactions are accompanied by a net increase in the entropy of the universe. Here the universe means the system and surroundings taken together,

⇒ \(\Delta S_{\text {univ }}>0 \) (2nd law thermodynamics).

⇒ \(\Delta S_{\text {uriv }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\)

In the case of condensation of water, \(\Delta S_{\text {univ }}=\Delta S_{\text {water }}+\Delta S_{\text {surr }}\)

• The positive ΔS_surr outweighs the negative ΔS_water and the total entropy increases.
• For all spontaneous processes, a stage is reached when the system attains equilibrium. A glass of hot water cools till it attains room temperature. It does not cool below room temperature. At this point, it is said to have attained equilibrium or, to be more precise, thermal equilibrium.

A gas expands till it fills the whole of the available space uniformly. Then no further expansion occurs. Again this is a state of equilibrium. At the point of equilibrium,

ΔS_univ = 0 (at equilibrium).

Having understood that entropy is a measure of disorder or randomness of a system, it may be easy to imagine that at absolute zero, the entropy of a perfectly crystalline substance would be zero. At any temperature above this, the entropy is positive and it increases with an increase in temperature.

For any substance S_solid < S_liquid < S_gas

For any system the heat changes are reflected in the entropy of that system. Hence, let us find an expression relating entropy and heat. Consider an exothermic reaction. The heat released in an exothermic reaction flows to the surroundings and the heat lost by the system is equal to the heat gained by the surroundings.

q_surr = -q_sys

This equation is valid even for an endothermic reaction. In the case of an exothermic reaction, the flow of heat into the surroundings increases its entropy, i.e.,

ΔS_surr>0.

The actual magnitude of ΔS_surr however, depends on the amount of heat released as well as the temperature. In fact, the effect of q on entropy is related to temperature. The addition of heat to colder surroundings has a greater effect on the entropy as compared to the effect produced on addition of heat to hotter surroundings. The effect of q is greater if the temperature is lower. This can be mathematically expressed as

⇒ \(\Delta S_{\text {surt }}=\frac{q_{\text {surr }}}{T}=-\frac{q_{\text {sys }}}{T} .\)….(1)

We know that at equilibrium \(\Delta S_{\text {undv }}=0=\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \text {. }\)….. (2)

Therefore, form equations (1) and (2) we get \(\Delta S_{s y s}=\frac{+q_{s y s}}{T}\)

If the process is reversible, the change in entropy of the system is given by \(\Delta S_{r v}=\frac{q_{r v v}}{T}.\)

• You may wonder that if no real processes take place reversibly, what is the use of the above equation. Although transfer of heat between system and surroundings is almost impossible to achieve in a reversible manner, this idealised path is important for the definition of ΔS.
• Entropy is a state function and the value of ΔS is the same for a change from say any state A to any other state B, irrespective of the path taken. Whether the change is carried out in a reversible or an irreversible manner, ΔS is the same.
• Thus whatever may be the actual path taken, ΔS is calculated using q_rev only. q_rev is the heat associated with the process had it traversed a reversible path.

The overall criterion for spontaneity: To get back to what we were discussing earlier, the overall criterion for the spontaneity of a process is the resultant of two tendencies—

1. The tendency towards minimum energy and
2. The tendency towards maximum entropy. These two tendencies act independently of each other and may work in the same or opposite directions.

The resultant of the two tendencies is expressed by another thermodynamic function called Gibbs free energy (G). The free energy of a system is a measure of its capacity to do useful work. The enthalpy (H), entropy (S), and free energy (G) of a system are related by the following expression.

G = H-TS (here T is the absolute temperature).

At constant temperature and pressure, the change in free energy (AG) during a process can be obtained by the following equation.

ΔG = ΔH -TΔS…… Equation 7

1. The change in free energy takes into account both the change in enthalpy and the change in entropy, and is the factor which determines the spontaneity of a process. You know that a process is possible if it is accompanied by a decrease in energy (ΔH negative) and an increase in entropy (ΔS positive).
2. Some processes for which ΔH is positive are possible because they are accompanied by an increase in entropy (ΔS positive). Some processes for which ΔS is negative are possible because they are accompanied by a decrease in energy (ΔH negative).
3. Obviously, the conditions that are most favourable for the occurrence of a process are a decrease in energy and an increase in entropy, i.e., when ΔH is negative and ΔS is positive.

This makes ΔG negative. Basically, the criterion for a spontaneous reaction or feasibility of a reaction is that ΔG should be negative. A spontaneous reaction is always accompanied by a decrease in free energy, i.e.,

ΔG < 0 for a spontaneous process.

The tendency of the system to undergo a change in the direction of decreasing free energy makes it reach a state when ΔG = 0 at some point. This is the equilibrium state.

Thus, ΔG > 0 for a nonspontaneous process.

• However, the reverse of a nonspontaneous process (for which ΔG is positive) is spontaneous.
• In order to understand the meaning of free energy for a process being zero, let us consider the simple case of water. Water freezes spontaneously at T<0°C and icc melts spontaneously at T>0°C. At T = 0°C and 1 atm pressure, the ΔG for melting of ice and freezing of water is zero.
• At this temperature, theoretically, ice should not melt and water should not freeze. This means the two phases, liquid and solid, may coexist in equilibrium with each other.
• We may also have chemical reactions for which ΔG = 0. As a spontaneous chemical reaction proceeds (ΔG < 0), reactants turn into products and ΔG becomes less negative. At a point ΔG becomes zero—no more products are formed nor are any reactants consumed. This is a state of equilibrium or in fact chemical equilibrium.
• Reactions with negative ΔG are called exergonic and those with positive ΔG are called endergonic reactions.

When is ΔG negative? It is worth finding an answer to this question because it will help us determine which processes are feasible and under what conditions. The figure represents graphically the relation between ΔH, ΔS, and ΔG.

1. Obviously, when ΔH is negative and ΔS is positive, ΔG will be negative (line ‘a’)
2. When ΔH is negative and ΔS, too, is negative, ΔG will be negative if the magnitude of ΔH is greater than that of TΔS (line ‘b’), i.e., below temperature T_b.
3. When both ΔH and ΔS are positive, AG will be negative if the magnitude of TΔS is greater than that of ΔH (line ‘c’, i.e., above temperature T_c).
4. When ΔH is positive, while ΔS is negative, ΔG will be positive and the reaction will not be spontaneous.

Effect of temperature on spontaneity If you think about the last two conditions for ΔG to be negative, it will strike you that temperature must be an important factor in determining the spontaneity of a reaction. Say, both ΔH and ΔS are negative for a process. If tire magnitude of ΔH is greater than that of TΔS, ΔG will be negative and the process will be feasible.

• It could be possible that this is true up to a certain value of T, beyond which TΔS becomes greater in magnitude than ΔH. Similarly, if both ΔH and ΔS are positive for a particular process, there could be a temperature below which the magnitude of TΔS is less than that of ΔH, and the process is not feasible.
• For an endothermic process, ΔH is positive, i.e., the energy factor opposes the process. For such a process to occur, TΔS must be positive and its magnitude must be greater than that of ΔH.

• For such a process, a higher temperature is more favorable as tills increases the magnitude of TΔS. In fact, there may be a temperature below which the magnitude of TΔS becomes less than that of ΔH and the process becomes unfeasible.
• For an exothermic process, ΔH is negative and ΔG is negative at all temperatures if ΔS is positive. However, if ΔS is negative, the process is feasible only if the magnitude of TΔS is less than that of ΔH.
• Obviously, such a reaction would be more feasible at a low temperature and may even become nonspontaneous above a particular temperature.

If ΔH is the major factor contributing to negative ΔG value, the reaction is called ‘enthalpy driven’ and if entropy is the factor behind it, it is called an ‘entropy driven’ reaction. A few values of \(\Delta H^{\ominus}, \Delta G^{\ominus} \text { and } \Delta S^\ominus\) are given in Table. Let us now summarise whatever we have discussed so far pertaining to the feasibility of a reaction in Table.

Units of \(\Delta G^{\ominus}\): cal mol^-1 in the CGS system and J mol^-1 or kJ mol^-1 in the SI system.

Standard free energy change: The standard free energy change of a reaction is defined as the free energy change at a specified temperature when the reactants in their standard states are converted to products in their standard states. It is denoted by \(\Delta G^{\ominus}\).

Calculation of \(\Delta G^{\ominus}\)° of a reaction Free energy is a state function. Hence, similar to \(\Delta H^{\ominus}\), \(\Delta G^{\ominus}\) for a reaction can be calculated from the free energy of formation of reactants and products.

The standard free energy of formation of an element in its standard state is taken as zero. Table lists values of standard enthalpy of formation, the standard free energy of formation, and absolute standard entropies of some compounds.

Tire standard free energy change of a reaction is obtained by subtracting the sum of the standard free energy of formation of the reactants from that of the products.

∴ \(\Delta_r G^{\ominus}=\Sigma v \Delta_f G^{\ominus} \text { (products) }-\Sigma v \Delta_f G^{\ominus} \text { (reactants) }\)

where v’s are the stoichiometric coefficients of the products and reactants respectively.

∴ \(\Delta G^{\ominus}\) may also be calculated from \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}\) of the reaction using an equation similar to Equation 7.

∴ \(\Delta_f G^{\ominus}\) = \(\Delta_f G^{\ominus}\) at a particular temperature.

We have seen how to calculate \(\Delta G^{\ominus}\) for a reaction. However, all chemical reactions do not occur with the reactants and products in their standard states. We can determine AG for a reaction of this kind using \(\Delta G^{\ominus}\) of the same reaction at standard conditions and the reaction quotient of the reaction. Consider the reaction

αA + βB → γC + δD

A, B, C, and D are the reactants and products while α,β,γ, and δ represent the respective stoichiometries.

The reaction quotient, Q_c, of this reaction is given by \(Q_c=\frac{[C]^\gamma[D]^8}{[A]^\alpha[B]^\beta}\)

The square brackets indicate the concentrations of the reactant/product. Note that the products appear in the numerator and reactants appear in the denominator. The concentration terms of the species are raised to their respective stoichiometric coefficients.

ΔG at any temperature T is related to \(\Delta G^{\ominus}\) as follows:

ΔG = \(\Delta G^{\ominus}\) + RT In Q_c ……. Equation 8

where R is the gas constant.

If the species involved in the reaction are gaseous in nature, concentration is replaced by partial pressure, and the reaction quotient is represented as Q_p.

At equilibrium, Q = K, the equilibrium constant. (Q_c = K_c or Q_p = K_p, as the case may be.)

The Equilibrium constant K of the reaction αA + βB → γC + δD  is given by

K = \(\frac{[\mathrm{C}]_e^r[\mathrm{D}]_e^5}{[\mathrm{~A}]_e^\alpha[\mathrm{B}]_e^\beta}\)

where the subscripts e denote equilibrium concentrations. Also, at equilibrium, ΔG = 0.

Hence, \(\Delta G^{\ominus}\) = -RT In K = -2.303 RT log K.

This gives yet another method of determining \(\Delta G^{\ominus}\) of a reaction. The pressure should be expressed in bar and the concentration in mol L^-1 while determining K_p and K_c respectively.

Example 1. Calculate the entropy change AS per mole for the following reactions.

1. Combustion of hydrogen at 298 K \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

⇒ \(\Delta H=-241.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\Delta G=-228.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

2. Vaporisation of methanol at its normal boiling point

Methanol (l) → Methanol (g), Δ_vap H = 23.9 kJ mol^-1, Boiling point = 338 K
Solution:

1. ΔG = ΔH – TΔS.

∴ \(\Delta S=\frac{\Delta H-\Delta G}{T}=\frac{-2416-(-228.4)}{298}=-0.044 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} .\)

2. ΔG = ΔH -TΔS.

ΔG = 0.

∴ \(\Delta S=\frac{\Delta_{\mathrm{vap}} H}{T}=\frac{23.9}{338}=0.071 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

Example 2. Calculate the free energy change per mole for the following reactions and predict the feasibility of the reactions.

1. \(\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \quad \text { at } 298 \mathrm{~K}\)

⇒ \(\Delta H=177.9 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad \Delta S=160.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

2. \(2 \mathrm{NO}_2(\mathrm{~g}) \longrightarrow \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \quad \text { at } 298 \mathrm{~K}\)

⇒ \(\Delta \mathrm{S}=175.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \quad \Delta H=-57.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Solution:

1. ΔG =ΔH – TΔS.

∴ ΔG = 177.9 x 10³ – 298 x 160.4 = 130100.8 J mol^-1 = 130.1 kJ mol^-1.

ΔG is positive. Therefore, the reaction is nonspontaneous at this temperature.

2. ΔG = -57.2×10³ – (298 x 175.6) =-109528.8 J mol-1 =-109.5 kj mol”1.

ΔG is negative. Therefore, this reaction is spontaneous at 298 K.

Why Should There Be An Energy Crisis

While reading about the law of conservation of energy, it may have struck you as strange that there should be an energy crisis in the world when energy is not really lost in any process, it is merely converted from one form into another. It is true that energy is merely converted from one form into another, but it is useful to us only in certain forms.

• In most processes that we encounter, a portion of energy is converted into forms we cannot use. Let us look at it in another way. Most of our energy requirements are met by the burning of fossil fuels. The combustion of fuels produces energy which we use. But it also produces carbon dioxide and water, which escape into the air.
• The combustion of fuels is a spontaneous process, but the reverse is not a spontaneous process. So, the combustion of fuels is a one-sided process, which increases disorder. It is this tendency of most processes to increase disorder that wastes energy in our terms.
• Even the energy released during the combustion of fuels cannot be utilised fully to perform useful work. A part of it goes to increase disorder or is wasted according to us. This is why the efficiency of an engine is always less than 1.

∴ \(\frac{\text { work done by engine }}{\text { energy fed into engine }}<1\)

This follows from the equation

ΔG = ΔH-TΔS

or ΔH = ΔG + TΔS.

• If ΔH is the total energy fed into an engine then some of it, i.e., TΔS, will be used to increase disorder and only a part of it (ΔG) will be available for useful work. This is why, though the total energy of the world remains constant, every living process converts a part of it into a form that cannot be used.
• This is the part that goes to increase disorder in the universe. Since AG is the useful part of energy that can be harnessed, it is a measure of the maximum amount of energy that is ‘free’ to do useful work. Hence its name.

Example 1. Calculate the standard enthalpy of formation of ethyl alcohol from the following data.

1. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta_{\mathrm{f}} \mathrm{H}_1^{\ominus}=-1368 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ….. (1)
2. \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta_{\mathrm{f}} H_2^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) …..(2)
3. \(\mathrm{H}_2(\mathrm{~g})+\mathrm{S}_2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta_{\mathrm{f}} H_3^{\ominus}=-286.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\)…..(3)

Solution: The required equation is

∴ \(2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l}) \quad \Delta_{\mathrm{f}} \mathrm{H}^{\ominus}=\)

To obtain this equation, first multiply Equation (2) by 2 and Equation (3) by 3 and add the products to get Equation (4). Then subtract Equation (1) from Equation (4),

∴ \(\Delta_{\mathrm{f}} H^{\ominus}=\left[2 \times \Delta_f H_2^{\ominus}+3 \times \Delta_{\mathrm{f}} H_3^{\ominus}\right]-\Delta_{\mathrm{f}} H_1^{\ominus}\)

= [(2 x – 393.5) + (3 x – 286.0)] – [-1368] = -1645 +1368 = -277 kJ mol^-1.

Example 2. Calculate the heat of formation of methane, given that the heats of combustion of methane, graphite and hydrogen are – 890, – 394, – 286 kJ mol^-1 respectively.
Solution:

The required equation is \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_4(\mathrm{~g}) \quad \Delta_{\mathrm{f}} \mathrm{H}=\text { ? }\)

The equation for the combustion of methane is \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \quad \Delta_{\mathrm{c}} \mathrm{H}_1=-890 \mathrm{~kJ} \mathrm{~mol}^{-1}\) ….(1)

The equation for the combustion of graphite is \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta_{\mathrm{c}} \mathrm{H}_2=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}\) …..(2)

The equation for the combustion of hydrogen is \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \quad \Delta_{\mathrm{c}} \mathrm{H}_3=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\)…. (3)

Multiply Equation (3) by 2 and add the product to Equation (2).

This is the required equation.

Then Δ_fH =[Δ_cH₂ + (2 X Δ_cH₃)]-[Δ_cH₁]

= ((-394) + (2 x – 286)] – [-890] = -394-572 + 890 = -76 kJ mol^-1.

Example 3. Calculate the heat of reaction of \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

Given that

1. \(\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \Delta_f H_2=-110.35 \mathrm{~kJ}\)
2. \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta_f H_2=-393.35 \mathrm{~kJ}\)
3. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta_f H_3=-241.6 \mathrm{~kJ}\)

Solution:

This problem can be solved by two methods.

1. Applying Hess’s law:

Add Equation (1) and Equation (3) and then subtract Equation (2) from the resultant equation.

⇒ \(\begin{aligned}
\mathrm{C}(\mathrm{s})+\mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) & \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \\
-\mathrm{C}\left(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g})\right. & \left.\longrightarrow \mathrm{CO}_2(\mathrm{~g})\right) \\
\hline \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) \longrightarrow & \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})
\end{aligned}\)

This is the required equation.

∴ Δ_rH = Δ_fH₁ – Δ_fH₃ – Δ_fH₂ =(-110.35) + (- 2416)- (-39335) = + 414 kj.

2. Δ_fH = ∑Δ_fH(products) – ∑Δ_fH(reactants)

= [Δ_fH(CO) + Δ_fH(H₂O)]- Δ_fH(CO₂)

= -110.35 + (- 2416)- (-39335) = + 414 kj.

Example 4. Just before an athletic event, a participant is given 100 g of glucose (C₆H₁₂O₆), whose energy equivalent is 1560 kj. He utilises 50% of this gained energy in the event. Calculate the weight of water he would need to perspire in order to avoid storing extra energy in the body, ifthe enthalpy ofevaporation ofwater is 44 kj mol^-1.
Solution:

Energy from 100 g of glucose = 1560 kJ.

Energy utilised = 50% of1560 = 780 kJ.

Energy not used = 1560- 780 = 780 kJ.

Enthalpy of evaporation of water = 44 kJ mol^-1.

44 kJ of energy evaporates1 mol of water.

780 kJ of energy would evaporate \(\frac{1 \times 780}{44}\) =17.73 mol of water.

17.73 mol of water = 17.73 x18 g = 319.14 g of water.

Example 5. Calculate the bond energy of the Cl—Cl bondfrom the equation \(\mathrm{CH}_4(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_3 \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \Delta \mathrm{H}=-100.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

given that the bond energies ofC—H, C—Cl, H—Cl bonds are 413, 326 and 431 kJ mol^-1 respectively.

Solution: The given equation can be written as

∴ \(\Delta H=\left[4 \Delta H_{\mathrm{C}-\mathrm{H}}+\Delta H_{\mathrm{C}-\mathrm{a}}\right]-\left[3 \Delta H_{\mathrm{C}-\mathrm{H}}+\Delta H_{\mathrm{C}-\mathrm{a}}+\Delta H_{\mathrm{H}-\mathrm{a}}\right]\)

or \(-100.3=\left[(4 \times 413)+\Delta H_{\mathrm{a}-\mathrm{a}}\right]-[(3 \times 413)+326+431]\)

∴ \(\quad \Delta H_{\mathrm{a}-\mathrm{C}}=243.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Example 6. Calculate the enthalpy change for the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HBr}(\mathrm{g})\) given that the bond enthalpies of H—H, Br—Br and H—Br are 435,192 and 364 kj mol^-1 respectively.
Solution:

ΔH =∑ bond enthalpies of reactants – ∑ bond enthalpies of products

= [ΔH_H-H + ΔH_Br-Br] – [2 x ΔH_H-Br]

= 435 + 192-(364 x 2)

= 627-728 =-101 kj mol^-1.

Example 7. Calculate the Gibbs energy change for the reaction \(4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})\) at 25°C when O₂ is at a partial pressure of 1.5 bar. \(\Delta_r G^e\) =-1484.4 kJ mol^-1

Solution:

The reaction quotient Q = \(\frac{\left[\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})\right]^2}{[\mathrm{Fe}(\mathrm{s})]^4\left[p_{\mathrm{O}_2}\right]^3}\).

Since the concentrations of pure solids and liquids are constant,

Q = \(\frac{1}{\left(p_{\mathrm{O}_2}\right)^3}=\frac{1}{(1.5)^3}=0.2963\)

Now, \(\Delta_r G=\Delta_r G^\ominus\) + RT ln Q

= -1484.4 x 10³ + 8.314 x 298 x ln (0.2963) (R = 8.314 JK^-1 mol^-1)

= – 1,484,400-3013.7 =-1,487,413.7 J mol^-1 = -1487.4 kJ mol^-1.

Example 8. Arrange the following systems in order of increasing entropy (compare one mole of each).

1. \(\mathrm{H}_2 \mathrm{O}$ (l), \mathrm{H}_2 \mathrm{O} (s), \mathrm{H}_2 \mathrm{O} (g)\)

2. \(\mathrm{H}_2(\mathrm{~g}), \mathrm{HBrO}_4(\mathrm{~g}), \mathrm{HBr}(\mathrm{g})\)

Solution:

Entropy varies as

1. \(\mathrm{H}_2 \mathrm{O}(\mathrm{s})<\mathrm{H}_2 \mathrm{O}(\mathrm{l})<\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

2. \(\mathrm{H}_2(\mathrm{~g})<\mathrm{HBr}(\mathrm{g})<\mathrm{HBrO}_4(\mathrm{~g})\)

If the physical state is the same, the more complicated the molecule, the more is the entropy.

Example 9. For the reaction \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_3(\mathrm{~g})\) find ΔG at 25°C. The partial pressures of SO₂, O₂ and SO₃ are 1.0 bar, 0.5 bar and 0.1 bar respectively. \(\Delta G^{\ominus}\) = -141.8kj mol^-1.
Solution:

⇒ \(\Delta G=\Delta G^{\ominus}\) + RT In Qf; where Qf, is the reaction quotient.

⇒ \(Q_p=\frac{\left(p_{\mathrm{SO}_3}\right)^2}{\left(p_{\mathrm{SO}_2}\right)^2\left(p_{\mathrm{O}_2}\right)}=\frac{(0.1)^2}{(1)^2(0.5)}=0.02\)

∴ \(\Delta G=\Delta G^{\ominus}+R T \ln Q\)

= -1418 x 10³ + 8314 x 298 x In 0.02

=-151,4923 J mol^-1

=-151.5 kj mol^-1

Example 10. The \(\Delta S_{\text {system }} \text { and } \Delta_t H^{\ominus}\) for the following reaction at 25 C are 220.5 J k^-1 mol^-1 and 90.7 kJ mol^-1. Find S and say in which direction the reaction is not spontaneous.

⇒ \(\mathrm{CH}_3 \mathrm{OH}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{~g})\)

Solution:

⇒ \(\Delta S_{\text {total }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \text {. }\)

⇒ \(\Delta S_{\text {sys }}=220.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

⇒ \(\Delta S_{\text {surr }}=\frac{-\Delta H^{\ominus}}{T}\)

= \(-\frac{\left(90.7 \times 10^3\right)}{298} \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

= \(-304.4 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

\(\Delta S_{\text {total }}=220.8-304.4=-83.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

As the entropy decreases (ΔS = -ve) the reaction is not spontaneous in the forward direction but the reverse reaction is spontaneous.

Example 11. Considering only entropy,find whether thefollowing reaction is spontaneous or not at 25°C

⇒ \(\mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2 \mathrm{H}_4(\mathrm{l})\)

⇒ \(\mathrm{N}_2(\mathrm{~g}) \quad \mathrm{H}_2(\mathrm{~g}) \quad \mathrm{N}_2 \mathrm{H}_4(\mathrm{l})\)

Given \(\begin{array}{lccc}
\Delta H^{\ominus}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) & 0 & 0 & 50.6 \\
\Delta S^{\ominus}\left(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) & 191.5 & 130.6 & 121.2
\end{array}\)

Solution: To predict whether or not a reaction is spontaneous, we must find the sign of \(\Delta S_{\text {total }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}\)

The entropy change of the system is equal to the standard entropy of the reaction and

⇒ \(\Delta S_{\text {surr }}=-\Delta_r H^{\ominus} / T \text {. }\)

⇒ \(\Delta S_{\mathrm{sys}}=S^{\ominus}\left(\mathrm{N}_2 \mathrm{H}_4(\mathrm{l})\right)-S^{\ominus}\left(\mathrm{N}_2(\mathrm{~g})\right)-2 S^{\ominus}\left(\mathrm{H}_2(\mathrm{~g})\right)\)

= \(121.2-191.5-(2 \times 130.6)\)

= \(-3315 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

⇒ \(\Delta_{\mathrm{r}} H^{\ominus}=\Delta_f H^{\ominus}\left(\mathrm{N}_2 \mathrm{H}_4(\mathrm{l})\right)-\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{N}_2(\mathrm{~g})\right)-2 \Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{H}_2(\mathrm{~g})\right)\)

= \(50.6-0-0=50.6 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1} \text {. }\)

⇒ \(\Delta S_{\text {surr }}=-\frac{\Delta_r H^{\ominus}}{T}=\frac{-50.6}{25+273}=-0.1698 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}=-169.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

⇒ \(\Delta S_{\text {total }}=\Delta S_{\text {sys }}+\Delta S_{\text {surt }}\)

= \(-331.5+(-169.8)\)

= \(-5013 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }\)

As \(\Delta S_{\text {total }}\) is negative, the reaction is not spontaneous.

Example 12. The values of equilibrium constant for the following reaction at two different temperatures are 168 x 10^-5 and 0.0123 respectively. If the first temperature is 1000°C, find the second temperature.

⇒ \(\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g})\)

Solution:

⇒ \(\Delta G^{\ominus}=-R T_1 \ln K_{p_1}\) …. (1)

⇒ \(\Delta G^{\ominus}=-R T_2 \ln K_{p_2}\)…. (2)

where T₁ and T₂ are the two different temperatures with the corresponding equilibrium constants \(K_{p_1}\) and \(K_{p_2}\) . Equating RHS of (1) and (2),

∴ \(R T_1 \ln K_{p_1}=R T_2 \ln K_{p_2}\)

or \(T_1 \ln K_{p_1}=T_2 \ln K_{p_2}\)

or \(T_2=T_1 \cdot \frac{\ln K_{p_1}}{\ln K_{p_2}}=(1273) \times \frac{\ln \left(168 \times 10^{-5}\right)}{\ln (0.0123)}=3182 \mathrm{~K}\) .

Example 13. Calculate Kp for thefollowing reactions at 298 K

1. \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_2(\mathrm{~g}) \Delta G^{\ominus}=-142 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(\mathrm{SO}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_2 \mathrm{SO}_4(\mathrm{l}) \Delta G^{\ominus}=-81.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Solution:

1. \(\Delta G^{\ominus}=-R^{\prime} T \ln K_r \) =\(-2.303 R T \log K_p\)

⇒ \(\log K_p=-\frac{\Delta G^{\ominus}}{2.303 R T}\)

⇒  \(\log K_p=-\frac{\Delta G^{\ominus}}{2.303 R T}\)

or \(K_p=\operatorname{Antilog}\left(\frac{-\Delta G^{\ominus}}{2.303 R T}\right)\)

= \(\text { Antilog }\left(\frac{142,000}{2.303 \times 8.314 \times 2.98}\right)=7.7 \times 10^{24}\).

2. \(K_p= Antilog \left(\frac{81,700}{2.303 \times 8.314 \times 298}\right)=2.08 \times 10^4\).

Example 14. Calculate \(\Delta G^{\ominus}\) for thefollowing reactions

1. \(\mathrm{CaCO},(\mathrm{s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

⇒  \(K_p=1.06 \text { at } 500^{\circ} \mathrm{C}\)

2. \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

⇒  \(\mathrm{K}_p=98.9 \text { at } 300^{\circ} \mathrm{C}\).

3. \(\mathrm{CoO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Co}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

⇒  \(K_p=1.09 \times 10^4 \text { at } 277^{\circ} \mathrm{C}\) .

Solution: Since \(\Delta G^{\ominus}\), substituting the values of R, T and Kp we get \(\Delta G^{\ominus}\) for each reaction.

1. \(\Delta G^{\ominus}\) = -8.314 x (500 + 273) x ln(1.06) = -374.4 J mol^-1.

2. \(\Delta G^{\ominus}\) = -8.314 x (300 + 273) x ln(98.9) = -21,885.9 J mol^-1 = -21.88 kj mol^-1.

3. \(\Delta G^{\ominus}\) = -8.314 x (277 + 273) x ln(1.09 x 10⁴ )\(\Delta G^{\ominus}\)= -42310.2 J mol^-1 = -42.51 kJ mol^-1.

Example 15. Calculate \(\Delta G^{\ominus}\) for the following reactions which are carried out under standard state conditions.

1. \(2 \mathrm{LiOH}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \longrightarrow \mathrm{Li}_2 \mathrm{CO}_3(\mathrm{~s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)
2. \(\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{I}) \longrightarrow 2 \mathrm{HBr}(\mathrm{g})\)
3. \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s})\)
4. \(2 \mathrm{HgO}(\mathrm{s}, \text { red }) \rightarrow 2 \mathrm{Hg}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})\)

The standardfree energies of the compounds are given below.

⇒ \(\Delta_i G\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)\)

⇒  \(\mathrm{LiOH}(\mathrm{s}) -443.9 \)

⇒  \(\mathrm{CO}_2(\mathrm{~g}) -394.4\)

⇒ \(\mathrm{Li}_2 \mathrm{CO}_3(\mathrm{~s}) -1132.4\)

⇒ \(\mathrm{H}_2 \mathrm{O}(\mathrm{g}) -228.6\)

⇒  \(\mathrm{H}_2(\mathrm{~g}) 0\)

⇒ \(\mathrm{Br}_2(\mathrm{l}) 0\)

⇒ \(\mathrm{HBr}(\mathrm{g}) -53.43\)

⇒ \(\mathrm{CaO}(\mathrm{s}) -604.2\)

⇒ \(\mathrm{H}_2 \mathrm{O}(\mathrm{l}) -237.2\)

⇒ \(\mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s}) -896.8\)

⇒ \(\mathrm{HgO}(\mathrm{s}, \text { red) } -58.55\)

⇒ \(\mathrm{Hg}(\mathrm{l}) 0\)

⇒ \(\mathrm{O}_2(\mathrm{~g}) 0\)

⇒ \(\Delta_1 \mathrm{H}(\mathrm{Hg}(\mathrm{s}, \text { red }))=-90.83 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

⇒ \(\text { Substance } \mathrm{S}^{\ominus}\left(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\)

⇒ \(\mathrm{Hg}_{\mathrm{g}} 38.01\)

⇒ \(\mathrm{O}_2(\mathrm{~g}) 205.04\)

⇒ \(\mathrm{HgO}(\mathrm{s}, \text { red) } 70.29\)

Which of these reactions islare spontaneous? For those reactions that are not spontaneous,find the temperature at which they will become spontaneous assuming ΔH and εS to be independent of temperature.

Solution:

1. \(\Delta G^{\ominus}=\Delta_f G^{\ominus}\left(\mathrm{Li}_2 \mathrm{CO}_3(\mathrm{~s})\right)+\Delta_{\mathrm{f}} G^{\ominus}\left(\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right)-2 \times \Delta_f G^{\ominus}(\mathrm{LiOH}(\mathrm{s}))-\Delta_f G^{\ominus}\left(\mathrm{CO}_2(\mathrm{~g})\right)\)

= -1132.4 + (-228.6)- 2 x (-443.9)- (-3944)

= -78.8 kj Spontaneous reaction

2. \(\Delta G^{\ominus}=2 \times \Delta_f G^{\ominus}(\operatorname{HBr}(\mathrm{g}))-\Delta_f G^{\ominus}\left(\mathrm{H}_2(\mathrm{~g})\right)-\Delta_f G^{\ominus}\left(\mathrm{Br}_2(\mathrm{l})\right)\)

= 2 x -53.43- 0- 0

= -106.86 kj Spontaneous reaction

3. \(\Delta G^{\ominus}=\Delta_f G^{\ominus}\left(\mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s})\right)-\Delta_{\mathrm{f}} G^{\ominus}(\mathrm{CaO}(\mathrm{s}))-\Delta_f G^{\ominus}\left(\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right)\)

-896.8- (-6042)- (-237.2)

= -55.4 kjSpontaneous reaction

4. \(\Delta G^{\ominus}=2 \times \Delta_f G^{\ominus}(\mathrm{Hg}(\mathrm{l}))+\Delta_f G^{\ominus}\left(\mathrm{O}_2(\mathrm{~g})\right)-2 \times \Delta_f \mathrm{G}^{\ominus}(\mathrm{HgO}(\mathrm{s} ; \text { red }))\)

= 0 + 0 -(-585) = +58.5 kj.

As \(\Delta G^{\ominus}\) is positive, this reaction is not spontaneous. Now let us find \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}\).

⇒ \(\Delta H^{\ominus}=2 \times \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{Hg}(\mathrm{l}))+\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{O}_2(\mathrm{~g})\right)-2 \times \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{HgO}(\mathrm{s}, \mathrm{red}))\)

= 0 + 0- (-90.83) = 90.83 kj.

⇒ \(\Delta S^{\ominus}=2 \times S^{\ominus}(\mathrm{Hg}(\mathrm{l}))+S^{\ominus}\left(\mathrm{O}_2(\mathrm{~g})\right)-2 \times S^{\ominus}(\mathrm{HgO}(\mathrm{s}, \text { red }))\)

= (2 x 38.01) + 205.04- 2 x 70.29 = 140.47 JK^-1

For the reaction to be spontaneous T\(\Delta S^{\ominus}\) should exceed \(\Delta H^{\ominus}\). Let us find the temperature at which T\(\Delta S^{\ominus}\) is equal to \(\Delta H^{\ominus}\). Assuming that \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}\) are independent of temperature.

⇒ T\(\Delta S^{\ominus}\) = \(\Delta H^{\ominus}\)

or T = \(\frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}=\frac{90.83 \times 10^3}{140.47}\) = 646.6 K

Above 646.6 K, the reaction will be spontaneous because then T\(\Delta S^{\ominus}\) > \(\Delta H^{\ominus}\) and \(\Delta G^{\ominus}\) will be negative.

Example 16. Calculate \(\Delta S_295^{\ominus}\) for thefollowing phase changes.

1. \(\mathrm{C}_6 \mathrm{H}_6(\mathrm{l}) \longrightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})\)
2. \(\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}(\mathrm{g})\)
3. \(\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

The standard entropies of C₆H₆(l), C₆H₆(g), Cu(s), Cu(g), H₂O(l) and H₂O(g) are 172.8, 269.2, 33.15, 1663, 69.91 and 188.71 JK^-1 mol^-1 respectively.

Solution:

1. \(\Delta S_{\text {vap }}^{\ominus} =S^{\ominus}\left(\mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})\right)-S^{\ominus}\left(\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})\right)\)

=269.2-172.8

∴ \(\Delta S_{\text {vap }}^{\ominus} =96.4 \mathrm{JK}^{-1}\)

2. \(\Delta S_{\text {sub }}^{\ominus} =S^{\ominus}(\mathrm{Cu}(\mathrm{g}))-S^{\ominus}(\mathrm{Cu}(\mathrm{s}))\)

=166.3-33.15

∴ \(\Delta S_{\text {sub }}^{\ominus} =133.15 \mathrm{JK}^{-1}\)

3. \(\Delta S_{\text {vap }}^{\ominus} \left.=S^{\ominus}\left(\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right)-\mathrm{S}^{\ominus} \mathrm{H}_2 \mathrm{O}(\mathrm{l})\right)\)

=188.71-69.91

∴ \(\Delta S_{\text {vap }}^{\ominus} =118.8 \mathrm{JK}^{-1}\) .

Example 17. Calculate the standaril entropy changefor the combustion ofliquid ethanol, C₂H₅OH to gaseous CO₂ and liquid water. The standard entropies of CO₂(g), H₂O(l), C₂H₅OH(l) and O₂(g) are 213.6 K^-1 mol^-1, 161JK^-1 mol^-1 and 205.03 JK^-1 mol^-1 respectively.
Solution:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)

⇒  \(\Delta S^{\ominus}=2 \times S^{\ominus}\left(\mathrm{CO}_2(\mathrm{~g})\right)+3 \times S^{\ominus}\left(\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right)-S^{\ominus}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)-3 \times\left(S^{\ominus}\left(\mathrm{O}_2(\mathrm{~g})\right)\right.\)

= 2 x 213.4 +3 x 69.91- 161-3 x 205.3 = -139.97 J K^-1mol^-1.

Example 18. Find the equilibrium constant of the reaction \(\mathrm{CH}_4(\mathrm{~g})+4 \mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{Cl}_4(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})\)

The standard heats offormation of CH₄ (g), CCl₄ (g) and HCl(g) are -74.81 kJ mol^-1,-102.9 kJ mol^-1 and -92.31 kJ mol⁷ respectively. The standard molar entropies of CH₄ (g), C₂(g), CCl₄(g), and HCl(g) are 186.26, 223.07, 309.7 and 186.91 J K^-1 mol^-1 respectively at 298 K.

Solution:

⇒ \(\Delta_{\mathrm{r}} H^{ominus}=\underset{\text { Products }}{\Sigma \Delta_{\mathrm{f}} H^{ominus}}-\underset{\text { Reactants }}{\Sigma \Delta_f H^{ominus}}\)

= \(\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{CCl}_4(\mathrm{~g})\right)+4 \Delta_{\mathrm{f}} H^{\ominus}(\mathrm{HCl}(\mathrm{g}))-\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{CH}_4(\mathrm{~g})\right)-4 \Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{Cl}_2(\mathrm{~g})\right)\)

= -102. 9 + (4 x -92.31)- (-74.81)

Since the standard enthalpies of formation of elements in their reference states are zero by definition

⇒ \(\Delta_{\mathrm{f}} H^{\ominus}\left(\mathrm{Cl}_2(\mathrm{~g})\right) \text { is } 0\)

⇒ \(\Delta_{\mathrm{r}} H^{\ominus} =-397.33 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

∴ \(\Delta_{\mathrm{r} S^{\ominus}} =\underset{\text { Reaction }}{\Sigma S^{\ominus}}-\underset{\text { Products }}{\Sigma S^{\ominus}}\)

= \(S^{\ominus}\left(\mathrm{CCl}_4(\mathrm{~g})\right)+4 S^{\ominus}(\mathrm{HCl}(\mathrm{g}))-S^{\ominus}\left(\mathrm{CH}_4(\mathrm{~g})\right)-4 S^{\ominus}\left(\mathrm{Cl}_2(\mathrm{~g})\right)\)

= 309.7 + (4 x186.91)- 186.26- (4 x 223.07) = -21.2 JK^-1 mol^-1

Substituting this value of \(\Delta_{\mathrm{r}} S \text { in } \Delta G^{\ominus}=\Delta H^{\ominus}-T \Delta S^{\ominus}\)

⇒ \(\Delta G^{\ominus}\) = -39733 x 10³ -298(-21.2) = -391.01 kJ mol^-1 = -391.01 kJ mol^-1

⇒ \(\Delta G^{\ominus}\) = -RT In K = -2.303 RT log K.

log K = \(-\frac{\Delta G^{\ominus}}{2303 R T}\)

∴ K = \({antilog}\left(\frac{-\Delta G^{\ominus}}{2.303 R T}\right)\)

= \({antilog}\left(\frac{391.01 \times 10^3}{2.303 \times 8.314 \times 298}\right)\)

= antilog (68.53) = 3.37×10⁶⁸.

Example 19. The standard enthalpy changefor the reaction \(\mathrm{NH}_3(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq})\) at 298 K is -52.22 kJ. The standard Gibbs energy change is -52.81 kJ. Find the standard molar entropy of NH₃(aq) if that of HCl(aq) and NH₄Cl(aq) are 56.5 JK^-1 mol^-1 and 169.9 J K^-1 mol^-1 respectively.

Solution:

⇒ \(\Delta G^{\ominus}\)=\(\Delta H^{\ominus}-T \Delta S^{\ominus}\)

∴ \(T \Delta S^\ominus=\Delta H^\ominus-\Delta G^\ominus\)

or, \(\Delta S^{\ominus}=\frac{\Delta H^{\ominus}-\Delta G^{\ominus}}{T}\)

= \(\frac{-52220 \mathrm{~J}-(-52810 \mathrm{~J})}{298}=\frac{590}{298}=1.98 \mathrm{~J} \mathrm{~K}^{-1} \text {. }\)

Now \(\Delta S^{\ominus}=S_{\mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq})}^{\ominus}-\left(S_{\mathrm{NH}_3(\mathrm{aq})}^{\ominus}+S_{\mathrm{HCl}(\mathrm{aq})}^{\ominus}\right)\).

Substituting the values, we get

1.098= \(169.9-\left(S_{\mathrm{NH}_3(q)}^{\ominus}+56.5\right)\)

∴ \(S_{\mathrm{NH}_3}^{\ominus}=169.9-56.5-198\)

= \(111.42 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\).

Example 20. The \(\Delta_{\text {fus }} H^\ominus \text { and } \Delta_{\text {fus }} S^\ominus \text { of } \mathrm{CCl}_4 \text { are } 2.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } 9.99 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively at 298 K. Find the temperature at which solid CCl₄ and its liquid are in equilibrium at 1 atm.

Solution:

⇒ \(\Delta G^{\ominus}=\Delta H^{\ominus}-T \Delta S^{\ominus}\).

But at equilibrium, \(\Delta G^{\ominus}\) = 0.

∴ \(\Delta H^{\ominus}\) = T\(\Delta S^{\ominus}\)

or T = \(\frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\)=\(\frac{2.5 \times 10^3}{9.99}=250.2 \mathrm{~K}\)

Example 21. Consider the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g}).\)

The standard enthalpy change at 298 K is -186.62 kj. The standard molar entropies of H₂(g), Cl₂(g) and HCl(g) are respectively 130.684, 223.07 and 186.91 J K^-1 mol^-1 respectively. Calculate the standard Gibbs energy change at 298 Kfor the reaction.

Solution:

⇒ \(\Delta G^{\ominus}\) = \(\Delta H^{\ominus}\) – T \(\Delta S^{\ominus}\)

Given \(\Delta H^{\ominus}\) = -186.62 k] and T = 298 K

∴ \(\Delta S^{\ominus}\) = \(\Delta S^{\ominus}\)(products) – \(\Delta S^{\ominus}\)(reactants)

= \(2 \times S^{\ominus}(\mathrm{HCl}(\mathrm{g}))-S^{\ominus}\left(\mathrm{H}_2(\mathrm{~g})\right)-S^{\ominus}\left(\mathrm{Cl}_2(\mathrm{~g})\right)\)

= 2 x 186.91- 130.684- 223.07 = 20.07 JK^-1.

Substituting this value of \(\Delta S^{\ominus}\), we get

⇒ \(\Delta G^{\ominus}\) = -186.62 x 10³ – 298 x 20.07

= -186,620-5980.86 = -192,600.86 J.

The standard free energy change for the reaction is -192.6 kj.

Example 22. Predict whether the entropy of the following reactions wll increase or decrease when the reaction occurs.

1. \(\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{CaSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)
2. \(\mathrm{CH}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)
3. \(\mathrm{Cu}(\mathrm{s})+\mathrm{S}(\mathrm{g}) \longrightarrow \mathrm{CuS}(\mathrm{s})\)
4. \(\mathrm{N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})\) .

Solution:

Verify your predictions by calculating \(\Delta S^{\ominus}\) using \(S^{\ominus}\) values.

1. In the reactant side we have only a solid whereas in the product side, we also have a gas apart from a solid. Since gases have large entropy, the entropy of system increases.

∴ \(\Delta S^{\ominus}=S^{\ominus}\left(\mathrm{CaSO}_4(\mathrm{~s})+2 \times S^{\ominus}\left(\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right)-S^{\ominus}\left(\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}\right)\right.\)

= 106.5 + 2 x 188.71-194.0 = 289.92 JK^-1

2. The number of gas molecules (two) is the same on both the reactant and the product side.

Depending on the actual entropies of the substances, \(\Delta S^{\ominus}\) may increase or decrease but only slightly.

∴ \(\Delta S^{\ominus}=S^{\ominus}(\mathrm{C}(\mathrm{s}))+2 \times S^{\ominus}\left(\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right)-S^{\ominus}\left(\mathrm{CH}_4(\mathrm{~g})\right)-S^{\ominus}\left(\mathrm{O}_2(\mathrm{~g})\right)\)

= 5.74 + 2 x188.71- 186.15- 205.03 = -8.02JK^-1

3. One of the reactants is in the gaseous state but the only product is a solid. Hence, the entropy will decrease.

∴ \(\Delta S^{\ominus}=S^{\ominus}(\mathrm{CuS}(\mathrm{s}))-S^{\ominus}(\mathrm{Cu}(\mathrm{s}))-S^{\ominus}(\mathrm{S}(\mathrm{g}))\)

= 66.5- 33.15-167.75 =1344 J K^-1.

4. The total number of gas molecules decreases from three to two as the reaction proceeds from left to right. This results in a decrease in entropy.

∴ \(\Delta S^{\ominus}=2 \times S^{\ominus}(\mathrm{NO}(\mathrm{g}))-S^{\ominus}\left(\mathrm{N}_2(\mathrm{~g})\right)-2 \times S^{\ominus}\left(\mathrm{O}_2(\mathrm{~g})\right)\)

= 2 x 210.65-1915- 2 x 205.03 = -180.26 JK^-1

Thermodynamics Multiple Choice Questions

Question 1. A well-stoppered thermos flask containing some ice cubes is an example of a

1. Closed system
2. Open system
3. Isolated system
4. Nonthermodynamic system

Answer: 3. Isolated system

Question 2. Which of the following is correct in the context of an adiabatic process?

1. pΔV = 0
2. q = +W
3. ΔU=q
4. q = 0

Answer: 4. q = 0

Question 3. The enthalpy change for the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\) is

1. Positive
2. Negative
3. Zero
4. None of these

Answer: 2. Negative

Question 4. The enthalpy of formation of ammonia is – 46.0 kJ mol^-1 What is the enthalpy change for the following reaction?

⇒ \(2 \mathrm{NH}_3(\mathrm{~g}) \longrightarrow \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})\)

1. + 46.0 kj mol^-1
2. – 23.0 kj mol^-1
3. – 92.0 kj mol^-1
4. + 92.0 kj mol^-1

Answer: 4. + 92.0 kj mol^-1

Question 5. The heat released when 0.6 mol of HNO₃ solution is mixed with 0.2 mol of NaOH is

1. 57.0 kj
2. 11.4 kj
3. 28.5kj
4. 34.9 kj

Answer: 2. 11.4 kj

Question 6. When NaCl dissolves in water, the entropy

1. Increases
2. Decreases
3. Remains the same
4. None of these

Answer: 1. Increases

Question 7. The enthalpy change for the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{1/2}\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})\) is called

1. Enthalpy of formation of water
2. Enthalpy of combustion of hydrogen
3. Enthalpy of vaporisation of water
4. None of these

Answer: 1. Enthalpy of formation of water and 2. Enthalpy of combustion of hydrogen

Question 8. The energy required to dissociate 4 g of gaseous hydrogen into free gaseous atoms is 208 kcal at 25°C. The bond energy of the H—H bond is

1. 104 kcal mol^-1
2. 10.4 kcal mol^-1
3. 208 kcal mol^-1
4. 416 kcal mol^-1

Answer: 1. 104 kcal mol^-1

Question 9. In which of the following does entropy decrease?

1. Crystallisation of sugar from a solution
2. Rusting of iron
3. Melting of ice
4. Vaporisation of camphor

Answer: 1. Crystallisation of sugar from a solution

Question 10. For the reaction X→y Y, ΔH and ΔS are negative at a certain temperature. This reaction will be spontaneous at

1. Low temperature
2. High temperature
3. The same temperature
4. Very high temperature

Answer: 1. Low temperature

Question 11. A spontaneous reaction is impossible if

1. Both ΔH and ΔS are positive
2. ΔH and ΔS are negative
3. ΔH is positive and ΔS is negative
4. ΔH is negative and ΔS is positive

Answer: 3. ΔH is positive and ΔS is negative

Question 12. When a solid melts, there is

1. An increase in entropy
2. An increase in enthalpy
3. A decrease in internal energy
4. Decrease in enthalpy

Answer: 1. An increase in entropy and 2. An increase in enthalpy

Question 13. Which of the following conditions are favourable for a spontaneous process?

1. ΔH=-ve TΔS = + ve
2. ΔH = -ve TΔS = -ve TΔS<ΔH
3. ΔH=+ve TΔS = + ve TΔS<ΔH
4. ΔH = +ve TΔS = + ve TΔS>ΔH

Answer: 1. ΔH=-ve TΔS = + ve,

2. ΔH = -ve TΔS = -ve TΔS<ΔH and

4. ΔH = +ve 7ΔS = + ve TΔS>ΔH

Question 14. The enthalpies of elements in their standard states are taken as zero. Thus, the enthalpy of formation of a compound

1. Will always be positive
2. Will always be negative
3. Will always be zero
4. May be negative or positive

Answer: 4. May be negative or positive

Question 15. The enthalpy of combustion of a substance

1. Is always positive
2. Is always negative
3. Is numerically equal to the enthalpy of formation
4. Cannot be predicted

Answer: 2. Is always negative

Question 16. When ammonium chloride is dissolved in water, the solution becomes cold. The change is

1. Exothermic
2. Endothermic
3. Super cooling
4. None of these

Answer: 2. Endothermic