Unit 8 - Chemical Equilibrium

Learning Objectives

After studying this unit, the students will be able to

• describe the meaning of equilibrium
• explain the dynamic nature of equilibrium involved in physical and chemical processes
• state the law of mass action
• deduce the expression for equilibrium constants, \( \mathrm{K_c} \) and \( \mathrm{K_p} \)
• establish the relationship between \( \mathrm{K_p} \) and \( \mathrm{K_c} \)
• predict the extent of reaction using equilibrium constant
• state Le-Chatelier’s principle
• explain the effect of various factors that affect a system at equilibrium
• derive Van’t Hoff equation

8.1 Introduction

In our daily life, we observe several chemical and physical changes. For example, a banana gets ripened after few days, silver gets tarnished in few months and iron gets rusted slowly. These processes proceed in one direction. Now let us consider the transport of oxygen by hemoglobin in our body as an illustration for a reversible change. The hemoglobin combines with oxygen in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing oxygen. In fact, in our lungs all the three species coexist.

Few chemical reactions proceed in only one direction whereas many reactions proceed in both the directions and these reactions are called reversible reactions.

In chemical reactions, the concentration of the reactants decreases and that of the products increases with time. In reversible reactions, initially the reaction proceeds towards the formation of the product. Upon formation of the product, the reverse reaction begins to take place. At a particular stage, the rate of the reverse reaction is equal to that of the forward reaction indicating a state of equilibrium.

It is desirable to know the three crucial aspects of chemical reactions namely the feasibility, the rate of the reaction and the extent of reaction. We know that the feasibility of a reaction is given by thermodynamics. Chemical kinetics will tell about the rate of the reaction. The equilibrium constant tells about the extent of a reaction which we will discuss in this chapter. We will also discuss the types of equilibrium, the significance of equilibrium constant and its relationship to thermodynamic quantities and the response of chemical equilibrium to change in the reaction conditions.

8.2 Physical and Chemical Equilibrium

There are different types of equilibrium. For example, if two persons with same weight sit on opposite sides of a see-saw at equal distance from the fulcrum, then the see-saw will be stationary and straight and it is said to be in equilibrium.

Another example of a state of equilibrium is the game of “tug-of-war.” In this game a rope is pulled taut between two teams. There may be a situation when both the teams are pulling the rope with equal force and the rope is not moving in either direction. This state is said to be in equilibrium.

In reversible processes, the rate of two opposing reactions equals at a particular stage. At this stage the concentration of reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate.

8.2.1 Physical Equilibrium

A system in which the amount of matter constituting different phases does not change with time is said to be in physical equilibrium. This involves no perceptible physical change in the system. To understand the physical equilibrium let us analyse the following phase changes.

Solid-Liquid Equilibrium

Let us consider the melting of ice in a closed container at \( 273 \ \mathrm{K} \). This system will reach a state of physical equilibrium in which the amount of water in the solid phase and liquid phase does not change with time. In the process the total number of water molecules leaving from and returning to the solid phase at any instant are equal.

If some ice-cubes and water are placed in a thermos flask (at \( 273 \ \mathrm{K} \) and 1 atm pressure), then there will be no change in the mass of ice and water.

At equilibrium,

The above equilibrium exists only at a particular temperature and pressure. The temperature at which the solid and liquid phases of a substance are at equilibrium is called the melting point or freezing point of that substance.

Liquid-Vapour Equilibrium

Similarly, there exists an equilibrium between the liquid phase and the vapour phase of a substance. For example, liquid water is in equilibrium with its vapour at \( 373 \ \mathrm{K} \) and 1 atm pressure in a closed vessel.

Here, Rate of evaporation \( = \) Rate of condensation

The temperature at which the liquid and vapour phases of a substance are at equilibrium is called the boiling point and condensation point of that substance.

Solid-Vapour Equilibrium

Consider a closed system in which the solid sublimes to vapour. In this process also, equilibrium can be established between these two phases. When solid iodine is placed in a closed transparent vessel, after sometime, the vessel gets filled up with violet vapour due to sublimation of iodine. Initially, the intensity of the violet colour increases, after sometime it decreases and finally it becomes constant, as the following equilibrium is attained.

More examples:

8.2.2 Equilibrium Involving Dissolution of Solids or Gases in Liquids

Solid in Liquids

When you add sugar to water at a particular temperature, it dissolves to form sugar solution. If you continue to add much sugar, you will reach a stage at which the added sugar remains as solid and the resulting solution is called a saturated solution. Here, as in the previous cases a dynamic equilibrium is established between the solute molecules in the solid phase and in the solution phase.

In this process,

Gas in Liquids

When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.

Example:

In carbonated beverages the following equilibrium exists.

Henry’s law is used to explain such gas-solution equilibrium processes.

8.3 Chemical Equilibrium

Similar to physical processes chemical reactions gradually attain a state of equilibrium after sometime. Let us consider a general reversible reaction.

Initially only A and B are present. Soon, a few molecules of the products C and D are formed by the forward reaction. As the concentration of the products increases, more products collide and react in the backward direction. This leads to an increase in the rate of backward reaction. As the rate of reverse reaction increases, the rate of the forward reaction decreases. Eventually, the rate of both reactions becomes equal.

8.4 Dynamic Nature of Equilibrium

Let us consider a situation in multi storey building, people are moving from first floor to second floor and vice versa. Assume that a certain number of people moves up from first floor to second floor in a specific time, and the same number of people moves down from second floor to the first floor in the same time. Now, the rate of movement of people from first to second floor equals the rate of movement of people from second to first floor, and hence the number of people in each floor will remain the same. Thus the population of people on the two floors is in a dynamic equilibrium. Let us extend this analogy to understand dynamic nature of equilibrium.

Chemical reactions which are reversible do not cease, when equilibrium is attained. At equilibrium the forward and the backward reactions are proceeding at the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.

For example,

8.5 Homogeneous and Heterogeneous Equilibria

8.5.1 Homogeneous Equilibrium

In a homogeneous equilibrium, all the reactants and products are in the same phase.

Example:

In the above equilibrium, \( \mathrm{H_2} \), \( \mathrm{I_2} \) and HI are in the gaseous state.

Similarly, for the following reaction, all the reactants and products are in homogeneous solution phase.

8.5.2 Heterogeneous Equilibrium

If the reactants and products of a reaction in equilibrium are in different phases, then it is called as heterogeneous equilibrium.

Examples:

8.6 Law of Mass Action

In 1864 two Norwegian chemists namely Maximilian Guldberg and Peter Waage formulated the law of mass action, based on the experimental studies of many reversible reactions. The law states that,

“At any instant, the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant”.

where \( x \) is the stoichiometric coefficient of the reactant and the square bracket represents the active mass (concentration) of the reactants.

8.6.1 Equilibrium Constants \( \mathrm{K_p} \) and \( \mathrm{K_c} \)

Let us consider a reversible reaction,

where A and B are the reactants, C and D are the products and \( x, y, l \) and \( m \) are the stoichiometric coefficients of A, B, C and D, respectively.

Applying the law of mass action, the rate of the forward reaction,

Similarly, the rate of the backward reaction,

where \( k_f \) and \( k_b \) are proportionality constants.

At equilibrium,

where \( \mathbf{K_c} \) is the equilibrium constant in terms of concentration (active mass).

At a given temperature, the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants is a constant, known as equilibrium constant.

If the reactants and products of the above reaction are in gas phase, then the equilibrium constant can be written in terms of partial pressures as indicated below,

where \( P_A, P_B, P_C \) and \( P_D \) are the partial pressures of the gases A, B, C and D, respectively.

8.6.2 Relation between \( \mathrm{K_p} \) and \( \mathrm{K_c} \)

Let us consider the general reaction in which all reactants and products are ideal gases.

The equilibrium constant \( \mathbf{K_c} \) is

and \( K_p \) is,

The ideal gas equation is \( PV = nRT \).

Active mass \( = \) molar concentration \( = \frac{n}{V} \)

Since \( \frac{n}{V} = \frac{P}{RT} \), the partial pressure can be expressed as,

On substitution,

where \( \Delta n_g \) is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.

When \( \Delta n_g = 0 \)

Examples:

1. \( \mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)} \)
2. \( \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \)

When \( \Delta n_g = +ve \)

Examples:

1. \( \mathrm{2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)} \)
2. \( \mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)} \)

When \( \Delta n_g = -ve \)

Examples:

1. \( \mathrm{2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)} \)
2. \( \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \)

8.6.3 Equilibrium Constants for Heterogeneous Equilibrium

Consider the following heterogeneous equilibrium.

The equilibrium constant for the above reaction can be written as

A pure solid always has the same concentration at a given temperature, as it does not expand to fill its container. i.e., it has same number of moles per litre of its volume. Therefore, the concentration of a pure solid is a constant. The above expression can be modified as follows

The equilibrium constant for the above reaction depends only on the concentration of carbon dioxide and not on calcium carbonate or calcium oxide. Similarly, the active mass (concentration) of the pure liquid does not change at a given temperature. Consequently, the concentration terms of pure liquids can also be excluded from the expression of the equilibrium constant.

For example,

Since \( \mathrm{H_2O(l)} \) is a pure liquid, the \( K_c \) can be expressed as

Table 8.1 Relation between equilibrium constants for some reversible reactions

Evaluate Yourself

1. Consider the following reaction

A solution is made with initial \( \mathrm{Fe^{3+}} \), SCN concentration of \( 1 \times 10^{-3} \ \mathrm{M} \) and \( 8 \times 10^{-4} \ \mathrm{M} \) respectively. At equilibrium \( [\mathrm{Fe(SCN)}]^{2+} \) concentration is \( 2 \times 10^{-4} \ \mathrm{M} \). Calculate the value of equilibrium constant.

2. The atmospheric oxidation of NO

was studied with initial pressure of 1 atm of NO and 1 atm of \( \mathrm{O_2} \). At equilibrium, partial pressure of oxygen is 0.52 atm. Calculate \( K_p \) of the reaction.

8.7 Application of Equilibrium Constant

The knowledge of equilibrium constant helps us to

1. predict the direction in which the net reaction will take place
2. predict the extent of the reaction
3. calculate the equilibrium concentrations of the reactants and products

It is to be noted that these constants do not provide any information regarding the rates of the forward or reverse reactions.

8.7.1 Predicting the Extent of a Reaction

The value of equilibrium constant, \( \mathrm{K_c} \) tells us the extent of a reaction, i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.

A large value of \( \mathrm{K_c} \) indicates that the reaction reaches equilibrium with high product yield. On the other hand, a low value of \( \mathrm{K_c} \) indicates that the reaction reaches equilibrium with low product yield. In general,

• if \( K_c > 10^3 \), the reaction proceeds nearly to completion
• if \( K_c < 10^{-3} \), the reaction rarely proceeds
• if \( K_c \) is in the range \( 10^{-3} \) to \( 10^3 \), significant amount of both reactants and products are present at equilibrium

Table 8.2 Dependence of extent of reaction on \( \mathbf{K_c} \)

8.7.2 Predicting the Direction of a Reaction

From the knowledge of equilibrium constant, it is possible to predict the direction in which the net reaction is taking place for a given concentration or partial pressure of reactants and products.

Consider a general homogeneous reversible reaction,

For the above reaction under non-equilibrium conditions, reaction quotient ‘Q’ is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants.

Under non-equilibrium conditions, the reaction quotient Q can be calculated using the following expression.

As the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches the equilibrium. At equilibrium Q is equal to \( K_c \) at a particular temperature. Once the equilibrium is attained, there is no change in the Q value. By knowing the Q value, we can predict the direction of the reaction by comparing it with \( K_c \).

• If \( Q = K_c \), the reaction is in equilibrium state.
• If \( Q > K_c \), the reaction will proceed in the reverse direction i.e., formation of reactants.
• If \( Q < K_c \), the reaction will proceed in the forward direction i.e., formation of products.

Example 1

The value of \( K_c \) for the following reaction at 717 K is 48.

At a particular instant, the concentration of \( \mathrm{H_2} \), \( \mathrm{I_2} \) and HI are found to be \( 0.2 \ \mathrm{mol \ L^{-1}} \), \( 0.2 \ \mathrm{mol \ L^{-1}} \) and \( 0.6 \ \mathrm{mol \ L^{-1}} \) respectively. From the above information we can predict the direction of reaction as follows.

Since \( Q < K_c \), the reaction will proceed in the forward direction.

Example 2

The value of \( K_c \) for the reaction

\( K_c = 0.21 \) at \( 373 \ \mathrm{K} \). The concentrations of \( \mathrm{N_2O_4} \) and \( \mathrm{NO_2} \) are found to be \( 0.125 \ \mathrm{mol \ dm^{-3}} \) and \( 0.5 \ \mathrm{mol \ dm^{-3}} \) respectively at a given time. From the above information we can predict the direction of reaction as follows.

The Q value is greater than \( K_c \). Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21.

Evaluate Yourself

3. The following water gas shift reaction is an important industrial process for the production of hydrogen gas.

At a given temperature \( K_p = 2.7 \). If 0.13 mol of CO, 0.56 mol of water, 0.78 mol of \( \mathrm{CO_2} \) and 0.28 mol of \( \mathrm{H_2} \) are introduced into a 2 L flask, find out in which direction must the reaction proceed to reach equilibrium.

8.7.3 Calculation of Concentration of Reactants and Products at Equilibrium

If the equilibrium concentrations of reactants and products are known for a reaction, then the equilibrium constant can be calculated and vice versa.

Let us consider the formation of HI in which \( a \) moles of hydrogen and \( b \) moles of iodine gas are allowed to react in a container of volume V. Let \( x \) moles of each of \( \mathrm{H_2} \) and \( \mathrm{I_2} \) react together to form \( 2x \) moles of HI.

Applying law of mass action,

The equilibrium constant \( K_p \) can also be calculated as follows:

We know the relationship between \( K_c \) and \( K_p \)

Here \( \Delta n_g = n_p - n_r = 2 - 2 = 0 \). Hence \( K_p = K_c \).

Solved Problem

One mole of \( \mathrm{H_2} \) and one mole of \( \mathrm{I_2} \) are allowed to attain equilibrium in 1 litre container. If the equilibrium mixture contains 0.4 mole of HI, calculate the equilibrium constant.

Given data:

At equilibrium, \( [HI] = 0.4 \ \mathrm{mol \ L^{-1}} \). \( K_c = ? \)

Solution:

Dissociation of \( \mathrm{PCl_5} \)

Consider that \( a \) moles of \( \mathrm{PCl_5} \) is taken in a container of volume V. Let \( x \) moles of \( \mathrm{PCl_5} \) be dissociated into \( x \) moles of \( \mathrm{PCl_3} \) and \( x \) moles of \( \mathrm{Cl_2} \).

Applying law of mass action,

The equilibrium constant \( K_p \) can also be calculated as follows:

We know the relationship between \( K_c \) and \( K_p \)

Here \( \Delta n_g = n_p - n_r = 2 - 1 = 1 \). Hence \( K_p = K_c (RT) \).

We know that \( PV = nRT \), so \( RT = \frac{PV}{n} \), where n is the total number of moles at equilibrium.

Synthesis of Ammonia

Let us consider the formation of ammonia in which \( a \) moles of nitrogen and \( b \) moles of hydrogen gas are allowed to react in a container of volume V. Let \( x \) moles of nitrogen react with \( 3x \) moles of hydrogen to give \( 2x \) moles of ammonia.

Applying law of mass action,

The equilibrium constant \( K_p \) can also be calculated as follows:

\( \Delta n_g = n_p - n_r = 2 - 4 = -2 \)

Total number of moles at equilibrium, \( n = a-x + b-3x + 2x = a + b - 2x \)

Solved Problems:

1. The equilibrium concentrations of \( \mathrm{NH_3} \), \( \mathrm{N_2} \) and \( \mathrm{H_2} \) are \( 1.8 \times 10^{-2} \ \mathrm{M} \), \( 1.2 \times 10^{-2} \ \mathrm{M} \) and \( 3 \times 10^{-2} \ \mathrm{M} \) respectively. Calculate the equilibrium constant for the formation of \( \mathrm{NH_3} \) from \( \mathrm{N_2} \) and \( \mathrm{H_2} \). [Hint: M = mol \( \mathrm{L^{-1}} \)]

Solution:

2. The equilibrium constant at 298 K for a reaction is 100.

If the initial concentration of all the four species is 1 M, the equilibrium concentration of D (in mol \( \mathrm{lit^{-1}} \)) will be

Given data:

Solution:

Let \( x \) be the number of moles of reactants reacted.

Evaluate Yourself

4. 1 mol of \( \mathrm{PCl_5} \) kept in a closed container of volume \( 1 \ \mathrm{dm^3} \) and was allowed to attain equilibrium at \( 423 \ \mathrm{K} \). Calculate the equilibrium composition of reaction mixture. (The \( K_c \) value for \( \mathrm{PCl_5} \) dissociation at \( 423 \ \mathrm{K} \) is 2)

8.8 Le-Chatelier’s Principle

Many chemical reactions that have industrial importance such as synthesis of ammonia are reversible in nature. It is important to know the reaction conditions to produce maximum yield. If a system at equilibrium is disturbed by modifying the reaction conditions, then the system adjusts itself to the new conditions to re-establish the equilibrium.

The effect of change in reaction conditions such as temperature, pressure and concentration etc. on a system at equilibrium can be predicted by Le Chatelier-Braun principle.

It states that “If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.”

8.8.1 Effect of Concentration

At equilibrium, the concentration of the reactants and the products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentrations.

According to Le Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance.

Let us consider the reaction

The addition of \( \mathrm{H_2} \) or \( \mathrm{I_2} \) to the equilibrium mixture disturbs the equilibrium. In order to minimize the stress, the system shifts the reaction in a direction where \( \mathrm{H_2} \) and \( \mathrm{I_2} \) are consumed. i.e., the formation of additional HI would balance the effect of added reactant. Hence, the equilibrium shifts to the right (forward direction) i.e., the forward reaction takes place until the equilibrium is re-established. Similarly, removal of HI (product) also favours the forward reaction.

If HI is added to the equilibrium mixture, the concentration of HI is increased, and the system proceeds in the reverse direction to nullify the effect of increase in concentration of HI.

Let us explain the effect of change in concentration by considering the formation of HI from \( \mathrm{H_2} \) and \( \mathrm{I_2} \). At equilibrium, the concentrations of HI, \( \mathrm{H_2} \) and \( \mathrm{I_2} \) are 1 M, 0.2 M and 0.1 M respectively.

The equilibrium was disturbed by adding 0.1 M iodine to the reaction mixture. After sometime, the concentration of HI is found to be 1.092 M. Let us verify whether the system proceeds to re-establish the equilibrium according to Le Chatelier’s principle.

Now,

Concentration of \( \mathrm{HI} = 1 + 2x = 1.092 \ \mathrm{M} \)

Therefore, the concentration of hydrogen and iodine at this stage,

The reaction quotient Q at this stage,

The Q value is equal to K value. So, we can conclude that the system has re-established the equilibrium at this stage and the addition of iodine results in the increase in the HI concentration.

In the large scale production of CaO from \( \mathrm{CaCO_3} \), constant removal of \( \mathrm{CO_2} \) from the kiln drives the reaction to completion. As in the case of manufacture of \( \mathrm{NH_3} \) by the Haber’s process, \( \mathrm{NH_3} \) is liquefied and removed so that the reaction keeps moving in forward direction.

8.8.2 Effect of Pressure

The change in pressure has significant effect only on equilibrium systems with gaseous components. When the pressure on the system is increased, the volume decreases proportionately and the system responds by shifting the equilibrium in a direction that has fewer moles of gaseous molecules.

Let us consider the synthesis of ammonia from nitrogen and hydrogen.

Let the system be allowed to attain equilibrium in a cylinder with a piston. If we press the piston down to increase the pressure, the volume decreases. The system responds to this effect by reducing the number of gas molecules. i.e., it favours the formation of ammonia. If we pull the piston upwards to reduce the pressure, the volume increases. It favours the decomposition of ammonia.

However, when the total number of the moles of the gaseous reactants and the gaseous products are equal, the change in pressure has no effect on system at equilibrium.

Let us consider the following reaction

Here, the number of moles of reactants and products are equal. So, the pressure has no effect on such equilibrium with \( \Delta n_g = 0 \).

8.8.3 Effect of Temperature

If the temperature of a system at equilibrium is changed, the system responds by shifting the equilibrium in the direction that attempts to nullify the effect of temperature.

Let us consider the formation of ammonia.

In this equilibrium, the forward reaction is exothermic i.e., heat is liberated while the reverse reaction is endothermic i.e., heat is absorbed.

If the temperature of the system is increased (by supplying heat energy), the system responds by decomposing some of the ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy. Similarly, the system responds to a drop in the temperature by forming more ammonia molecules from nitrogen and hydrogen, which releases heat energy.

We have already learnt that the change in pressure or concentration causes a change in the equilibrium concentration such that the equilibrium constant remains the same. However, in case of change in temperature, the equilibrium is re-established with a different equilibrium constant.

8.8.4 Effect of a Catalyst

Addition of a catalyst does not affect the state of the equilibrium. The catalyst increases the rate of both the forward and the reverse reactions to the same extent. Hence, it does not change the equilibrium composition of the reaction mixture. However, it speeds up the attainment of equilibrium by providing a new pathway having a lower activation energy.

For example, in the synthesis of \( \mathrm{NH_3} \) by the Haber’s process, iron is used as a catalyst. Similarly, in the contact process of manufacturing \( \mathrm{SO_3} \), platinum or \( \mathrm{V_2O_5} \) is used as a catalyst.

8.8.5 Effect of Inert Gas

When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles of gases present in the container increases, that is, the total pressure of gases increases. The partial pressure of the reactants and the products are unchanged. Hence, at constant volume, addition of inert gas has no effect on equilibrium.

Table 8.3 Effect of concentration, pressure, temperature, catalyst and inert gas on equilibrium

8.9 Van’t Hoff Equation

This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change \( (\Delta G^{\circ}) \) and equilibrium constant is

We know that

Substituting (1) in equation (2)

Rearranging

Differentiating equation (3) with respect to temperature,

Equation (4) is known as differential form of van’t Hoff equation.

On integrating the equation (4) between \( T_1 \) and \( T_2 \) with their respective equilibrium constants \( K_1 \) and \( K_2 \)

Equation (5) is known as integrated form of van’t Hoff equation.

Problem:

For an equilibrium reaction \( K_p = 0.0260 \) at \( 25^{\circ} \mathrm{C} \), \( \Delta H = 32.4 \ \mathrm{kJ \ mol^{-1}} \). Calculate \( K_p \) at \( 37^{\circ} \mathrm{C} \).

Solution:

Evaluate Yourself

5. The equilibrium constant for the following reaction is 0.15 at 298 K and 1 atm pressure.

The reaction conditions are altered as follows.

a) The reaction temperature is altered to \( 100^{\circ} \mathrm{C} \) keeping the pressure at 1 atm. Calculate the equilibrium constant.

Clinical Correlate: Hemoglobin and Oxygen Transport

In a pregnant woman, the oxygen supply for fetus is provided by the maternal blood in the placenta where the blood vessels of both mother and fetus are in close proximity. Both fetal and maternal hemoglobin binds to oxygen reversibly as follows.

In the above two equilibria, the equilibrium constant value for the oxygenation of fetal hemoglobin is higher, which is due to its higher affinity for oxygen compared to the adult hemoglobin. Hence in placenta, the oxygen from the mother’s blood is effectively transferred to the fetal hemoglobin.

Summary

• In our daily life, we observe several chemical and physical changes.
• In some chemical reactions, at a particular stage, the rate of the reverse reaction is equal to that of the forward reaction indicating a state of equilibrium.
• There are different types of equilibrium: Physical equilibrium and Chemical equilibrium.
• Chemical reactions which are reversible do not cease when equilibrium is attained. At equilibrium the forward and the backward reactions are proceeding at the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.
• The rate law states that, “At any instant, the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant”.
• The value of equilibrium constant, \( K_c \) tells us the extent of a reaction, i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.
• Under non-equilibrium conditions, reaction quotient ‘Q’ is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants.
• Le-Chatelier’s Principle: “If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.”
• The effect of change in reaction conditions such as temperature, pressure and concentration etc. on a system at equilibrium can be predicted by Le Chatelier-Braun principle.
• Van’t Hoff equation gives the quantitative temperature dependence of equilibrium constant (K).

Evaluation

I. Choose the best answer.

1. If \( K_b \) and \( K_f \) for a reversible reaction are \( 0.8 \times 10^{-5} \) and \( 1.6 \times 10^{-4} \) respectively, the value of the equilibrium constant is, a) 20 b) \( 0.2 \times 10^{-1} \) c) 0.05 d) none of these

2. At a given temperature and pressure, the equilibrium constant values for the equilibria

   \[ 3\mathrm{A_2} + \mathrm{B_2} + 2\mathrm{C} \xrightarrow{K_1} 2\mathrm{A_3BC} \]
   \[ \mathrm{A_3BC} \xrightarrow{K_2} \mathrm{A_2} + \frac{1}{2}\mathrm{B_2} + \mathrm{C} \]

   The relation between \( K_1 \) and \( K_2 \) is a) \( K_1 = \frac{1}{\sqrt{K_2}} \) b) \( K_2 = K_1^{-1/2} \) c) \( K_1^2 = 2K_2 \) d) \( K_2 = K_2 \)

3. The equilibrium constant for a reaction at room temperature is \( K_1 \) and that at \( 700 \ \mathrm{K} \) is \( K_2 \). If \( K_1 > K_2 \), then a) The forward reaction is exothermic

4. The formation of ammonia from \( \mathrm{N_2(g)} \) and \( \mathrm{H_2(g)} \) is a reversible reaction

   \[ \mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) + Heat} \]

   What is the effect of increase of temperature on this equilibrium reaction? a) equilibrium is unaltered b) formation of ammonia is favoured c) equilibrium is shifted to the left d) reaction rate does not change

5. Solubility of carbon dioxide gas in cold water can be increased by a) increase in pressure b) decrease in pressure c) increase in volume d) none of these

6. Which one of the following is incorrect statement? a) for a system at equilibrium, Q is always less than the equilibrium constant b) equilibrium can be attained from either side of the reaction c) presence of catalyst affects both the forward reaction and reverse reaction to the same extent d) Equilibrium constant varied with temperature

7. \( K_1 \) and \( K_2 \) are the equilibrium constants for the reactions respectively.

   \[ \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \]
   \[ 2\mathrm{NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)} \]

   What is the equilibrium constant for the reaction \( \mathrm{NO_2(g) \rightleftharpoons \frac{1}{2} N_2(g) + O_2(g)} \) a) \( \frac{1}{\sqrt{K_1 K_2}} \) b) \( (K_1 K_2)^{\frac{1}{2}} \) c) \( \frac{1}{2K_1 K_2} \) d) \( \left(\frac{1}{K_1 K_2}\right)^{\frac{1}{2}} \)

8. In the equilibrium,

   \[ 2\mathrm{A(g) \rightleftharpoons 2B(g) + C_2(g)} \]

   the equilibrium concentrations of A, B and \( \mathrm{C_2} \) at \( 400 \ \mathrm{K} \) are \( 1 \times 10^{-4} \ \mathrm{M} \), \( 2.0 \times 10^{-3} \ \mathrm{M} \) and \( 1.5 \times 10^{-4} \ \mathrm{M} \) respectively. The value of \( K_c \) for the equilibrium at \( 400 \ \mathrm{K} \) is a) 0.06 b) 0.09 c) 0.62 d) \( 3 \times 10^{-2} \)

9. An equilibrium constant of \( 3.2 \times 10^{-6} \) for a reaction means, the equilibrium is a) largely towards forward direction b) largely towards reverse direction c) never established d) none of these

10. \( \frac{K_c}{K_p} \) for the reaction,

    \[ \mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \ \text{is} \]

    a) \( \frac{1}{RT} \) b) \( \sqrt{RT} \) c) \( RT \) d) \( (RT)^2 \)

11. For the reaction \( \mathrm{AB(g) \rightleftharpoons A(g) + B(g)} \) at equilibrium, AB is 20% dissociated at a total pressure of P. The equilibrium constant \( K_p \) is related to the total pressure by the expression a) \( P = 24K_p \) b) \( P = 8K_p \) c) \( 24P = K_p \) d) none of these

12. In which of the following equilibrium, \( K_p \) and \( K_c \) are not equal? a) \( 2\mathrm{NO(g)} \rightleftharpoons \mathrm{N_2(g)} + \mathrm{O_2(g)} \) b) \( \mathrm{SO_2(g)} + \mathrm{NO_2(g)} \rightleftharpoons \mathrm{SO_3(g)} + \mathrm{NO(g)} \) c) \( \mathrm{H_2(g)} + \mathrm{I_2(g)} \rightleftharpoons 2\mathrm{HI(g)} \) d) \( \mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)} \)

13. If \( x \) is the fraction of \( \mathrm{PCl_5} \) dissociated at equilibrium in the reaction \( \mathrm{PCl_5} \rightleftharpoons \mathrm{PCl_3} + \mathrm{Cl_2} \), then starting with 0.5 mole of \( \mathrm{PCl_5} \), the total number of moles of reactants and products at equilibrium is a) \( 0.5 - x \) b) \( x + 0.5 \) c) \( 2x + 0.5 \) d) \( x + 1 \)

14. The values of \( K_{p_1} \) and \( K_{p_2} \) for the reactions \( \mathrm{X} \rightleftharpoons \mathrm{Y} + \mathrm{Z} \) and \( \mathrm{A} \rightleftharpoons 2\mathrm{B} \) are in the ratio 9 : 1. If degree of dissociation and initial concentration of X and A be equal, then total pressure at equilibrium \( P_1 \) and \( P_2 \) are in the ratio a) 36 : 1 b) 1 : 1 c) 3 : 1 d) 1 : 9

15. In the reaction, \( \mathrm{Fe(OH)_3(s)} \rightleftharpoons \mathrm{Fe^{3+}(aq)} + 3\mathrm{OH^-(aq)} \), if the concentration of \( \mathrm{OH^-} \) ions is decreased by \( \frac{1}{4} \) times, then the equilibrium concentration of \( \mathrm{Fe^{3+}} \) will a) not changed b) also decreased by \( \frac{1}{4} \) times c) increase by 4 times d) increase by 64 times

16. Consider the reaction where \( K_p = 0.5 \) at a particular temperature

    \[ \mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)} \]

    If the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true? a) more \( \mathrm{PCl_5} \) will be produced b) more \( \mathrm{Cl_2} \) will be produced c) more \( \mathrm{PCl_3} \) will be produced d) none of these

17. Equimolar concentrations of \( \mathrm{H_2} \) and \( \mathrm{I_2} \) are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of \( \mathrm{H_2} \) has reacted at equilibrium if rate constant for both forward and reverse reactions are equal? a) 33% b) 66% c) \( (33)^2 \% \) d) 16.5%

18. In a chemical equilibrium, the rate constant for the forward reaction is \( 2.5 \times 10^2 \) and the equilibrium constant is 50. The rate constant for the reverse reaction is, a) 11.5 b) 5 c) \( 2 \times 10^2 \) d) \( 2 \times 10^{-3} \)

19. Which of the following is not a general characteristic of equilibrium involving physical process? a) Equilibrium is possible only in a closed system at a given temperature b) The opposing processes occur at the same rate and there is a dynamic but stable condition c) All the physical processes stop at equilibrium d) All measurable properties of the system remains constant

20. For the formation of two moles of \( \mathrm{SO_3(g)} \) from \( \mathrm{SO_2} \) and \( \mathrm{O_2} \), the equilibrium constant is \( K_1 \). The equilibrium constant for the dissociation of one mole of \( \mathrm{SO_3} \) into \( \mathrm{SO_2} \) and \( \mathrm{O_2} \) is a) \( \left(\frac{K_1}{2}\right)^{\frac{1}{2}} \) b) \( \left(\frac{1}{2K_1}\right)^{\frac{1}{2}} \) c) \( \frac{1}{2K_1} \) d) \( \left(\frac{1}{K_1}\right)^{\frac{1}{2}} \)

21. Match the equilibria with the corresponding conditions,

    	Equilibrium	Condition
    i	Liquid ⇌ Vapour	1) melting point
    ii	Solid ⇌ Liquid	2) Saturated solution
    iii	Solid ⇌ Vapour	3) Boiling point
    iv	Solute (s) ⇌ Solute (Solution)	4) Sublimation point
    		5) Unsaturated solution

    a) (i)-1, (ii)-2, (iii)-3, (iv)-4 b) (i)-3, (ii)-1, (iii)-4, (iv)-2 c) (i)-2, (ii)-1, (iii)-3, (iv)-4 d) (i)-3, (ii)-2, (iii)-4, (iv)-5

22. Consider the following reversible reaction at equilibrium, \( \mathrm{A + B \rightleftharpoons C} \). If the concentration of the reactants A and B are doubled, then the equilibrium constant will a) be doubled b) become one fourth c) be halved d) remain the same

23. \( \mathrm{[Co(H_2O)_6]^{2+}(aq)(pink) + 4Cl^- (aq) \rightleftharpoons [CoCl_4]^{2-}(aq)(blue) + 6H_2O(l)} \) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one of the following is true? a) \( \Delta H > 0 \) for the forward reaction b) \( \Delta H = 0 \) for the reverse reaction c) \( \Delta H < 0 \) for the forward reaction d) Sign of the \( \Delta H \) cannot be predicted based on this information.

24. The equilibrium constants of the following reactions are:

    \[ \mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}; \quad K_1 \]
    \[ \mathrm{N_2 + O_2 \rightleftharpoons 2NO}; \quad K_2 \]
    \[ \mathrm{H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O}; \quad K_3 \]

    The equilibrium constant (K) for the reaction

    \[ 2\mathrm{NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O}, \ \text{will be} \]

    a) \( \frac{K_2 K_3^3}{K_1} \) b) \( \frac{K_1 K_3^3}{K_2} \) c) \( \frac{K_1 K_2^2}{K_3} \) d) \( \frac{K_2 K_3^2}{K_1} \)

25. A 20 litre container at \( 400 \ \mathrm{K} \) contains \( \mathrm{CO_2(g)} \) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of \( \mathrm{CO_2} \) attains its maximum value will be: Given that: \( \mathrm{SrCO_3(s) \rightleftharpoons SrO(s) + CO_2(g)} \), \( K_p = 1.6 \ \mathrm{atm} \) (NEET 2017) a) 2 litre b) 5 litre c) 10 litre d) 4 litre

II. Write brief answer to the following questions.

26. If there is no change in concentration, why is the equilibrium state considered dynamic?

27. For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.

28. What is the relation between \( K_p \) and \( K_c \)? Give one example for which \( K_p \) is equal to \( K_c \).

29. For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is \( K_c \) larger or smaller than \( K_p \)?

30. When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?

31. For the reaction,

    \[ \mathrm{A_2(g) + B_2(g) \rightleftharpoons 2AB(g)}; \quad \Delta H \text{ is -ve.} \]

    the following molecular scenes represent different reaction mixture (A - green, B - blue) i) Calculate the equilibrium constant \( K_p \) and \( (K_c) \). ii) For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions? iii) What is the effect of increase in pressure for the mixture at equilibrium.

32. State Le-Chatelier principle.

33. Consider the following reactions, a) \( \mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)} \) b) \( \mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)} \) c) \( \mathrm{S(s) + 3F_2(g) \rightleftharpoons SF_6(g)} \) In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.

34. State law of mass action.

35. Explain how will you predict the direction of an equilibrium reaction.

36. Derive a general expression for the equilibrium constant \( K_p \) and \( K_c \) for the reaction

    \[ 3\mathrm{H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g)} \]

37. Write a balanced chemical equation for the equilibrium reaction for which the equilibrium constant is given by expression

    \[ K_c = \frac{[NH_3]^4 [O_2]^5}{[NO]^4 [H_2O]^6} \]

38. What is the effect of added inert gas on the reaction at equilibrium at constant volume?

39. Derive the relation between \( K_p \) and \( K_c \).

40. One mole of \( \mathrm{PCl_5} \) is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.

41. For the reaction

    \[ \mathrm{SrCO_3(s) \rightleftharpoons SrO(s) + CO_2(g)}, \]

    the value of equilibrium constant \( K_p = 2.2 \times 10^{-4} \) at \( 1002 \ \mathrm{K} \). Calculate \( K_c \) for the reaction.

42. To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at \( 500^{\circ} \mathrm{C} \). At equilibrium he found the concentration of HI which is equal to \( 0.05 \ \mathrm{M} \). Calculate \( K_c \) and \( K_p \) for this reaction.

43. 1 mol of \( \mathrm{CH_4} \), 1 mole of \( \mathrm{CS_2} \) and 2 mol of \( \mathrm{H_2S} \) and 2 mol of \( \mathrm{H_2} \) are mixed in a 500 ml flask. The equilibrium constant for the reaction \( K_c = 4 \times 10^{-2} \ \mathrm{mol^2 \ lit^{-2}} \). In which direction will the reaction proceed to reach equilibrium?

44. At particular temperature \( K_c = 4 \times 10^{-2} \) for the reaction

    \[ \mathrm{H_2S(g) \rightleftharpoons H_2(g) + \frac{1}{2} S_2(g)} \]

    Calculate \( K_c \) for each of the following reaction i) \( 2\mathrm{H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)} \) ii) \( 3\mathrm{H_2S(g) \rightleftharpoons 3H_2(g) + \frac{3}{2}S_2(g)} \)

45. 28 g of Nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17 g \( \mathrm{NH_3} \) was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.

46. The equilibrium for the dissociation of \( \mathrm{XY_2} \) is given as,

    \[ 2\mathrm{XY_2(g) \rightleftharpoons 2XY(g) + Y_2(g)} \]

    if the degree of dissociation \( x \) is so small compared to one. Show that \( 2K_p = P x^3 \) where P is the total pressure and \( K_p \) is the dissociation equilibrium constant of \( \mathrm{XY_2} \).

47. A sealed container was filled with 1 mol of \( \mathrm{A_2(g)} \), 1 mol \( \mathrm{B_2(g)} \) at \( 800 \ \mathrm{K} \) and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that \( K = 1 \) for the reaction

    \[ \mathrm{A_2(g) + B_2(g) \rightleftharpoons 2AB(g)} \]

48. Deduce the Van’t Hoff equation.

49. The equilibrium constant \( K_p \) for the reaction

50. The partial pressure of carbon dioxide in the reaction

    \[ \mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)} \]

    is \( 1.017 \times 10^{-3} \) atm at \( 500^{\circ} \mathrm{C} \). Calculate \( K_p \) at \( 600^{\circ} \mathrm{C} \) for the reaction. \( \Delta H \) for the reaction is \( 181 \ \mathrm{kJ \ mol^{-1}} \) and does not change in the given range of temperature.

ICT Corner

Effect of temperature and pressure in an equilibrium process

By using this tool, we can determine the effect of pressure and temperature in the equilibrium concentration of the components in ammonia synthesis (Haber process).

Please go to the URL http://www.freezeray.com/flashFiles/ammoniaConditions.htm (or) Scan the QR code.

Steps:

• Open the Browser and type the URL given (or) Scan the QR Code.
• The website will show the equilibrium reaction involved in ammonia synthesis and the relative concentration of the components. The visual representation and the actual concentration values are given in the box 1.
• Now change the pressure or temperature using the corresponding slider indicated in the box 2.
• As you move the slider you will be able to see the change in the equilibrium concentration of reactants and products.
• Now you can understand that if a stress is applied on the system at equilibrium, the system will adjust itself to nullify the effect of the stress.