Application of equilibrium constant

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 and

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.

1. Predicting the extent of a reaction

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.

A large value of K_c indicates that the reaction reaches equilibrium with high product yield. On the other hand, a low value of K_c indicates that the reaction reaches equilibrium with low product formed. In general, if the K_c is greater than the 10³, the reaction proceeds nearly to completion. If it is less than 10^-3, the reaction rarely proceeds. If the K_c is in the range 10^-3 to10³, significant amount of both reactants and products are present at equilibrium.

Table8.2 Dependence of extent of reaction on Kc

Example

Consider the following equilibrium reactions and relate their equilibrium, constants

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,

xA + yB ⇌ lC + mD

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 Kc.

·           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.

H₂(g) + I₂(g) ⇌ 2HI(g)

At a particular instant, the concentration of H₂, I₂and HI are found to be 0.2 mol L^-1, 0.2 mol L^-1 and 0.6 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 Kc for the reaction

N₂O₄ (g) ⇌ 2NO₂(g)

Kc = 0.21 at 373 K. The concentrations N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 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 Kc. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21

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 H₂ and I₂ react together to form 2x moles of HI.

H₂ (g)+ I_{2 (g)} ⇌ 2HI₂(g)

Applying law of mass action,

The equilibrium constant K_p can also be calculated as follows:

We know the relationship between the K_c and K_p

Solved Problem

One mole of H₂ and one mole of I₂ are allowed to attain equilibrium. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.

Given data:

[H₂] = 1 mole [I₂] = 1 mole

At equilibrium, [HI] = 0.4 mole Kc= ?

Solution:

Dissociation of PCl₅:

Consider that ‘a’ moles of PCl₅ is taken in a container of volume V. Let ‘x’ moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Applying law of mass action,

The equilibrium constant K_p can also be calculated as follows:

We know the relationship between the Kc and Kp

Where n is the total number of moles at equilibrium.

n = (a-x) + x + x = (a+x)

Synthesis of ammonia:

Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and ‘b’ moles 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.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Applying law of mass action,

The equilibrium constant K_p can also be calculated as follows:

Total number of moles at equilibrium,

n= a-x + b-3x+ 2x = a+b-2x

Solved Problems:

1. The equilibrium concentrations of NH₃, N₂ and H₂ are 1.8 × 10^-2 M, 1.2 × 10^-2 M and 3 × 10^-2 M respectively. Calculate the equilibrium constant for the formation of NH₃ from N₂ and H₂. [Hint: M= mol lit^-1]

Given data:

[NH₃] = 1.8 × 10^-2 M

[N₂] = 1.2 × 10^-2 M

[H₂] = 3 × 10^-2 M

K_c = ?

Solution:

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

A + B ⇌ C + D

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

Given data:

[A] = [B] = [C]= [D] =1 M

K_c = 100

[D]_eq = ?

Solution:

Let x be the no moles of reactants reacted