Chapter 12, Problem 14QAP

Consider the following reaction at 1000 C:

$\text{NO}\left(g\right)+\frac{1}{2}{\text{Cl}}_2\left(g\right)\rightleftharpoons \text{NOCl}\left(g\right)$

(a) Write an equilibrium constant expression for the reaction and call it K'.

(b) Write an equilibrium constant expression for the decomposition of NOCl to produce one mole of chlorine gas. Call the constant K".

(c) Relate K' and K".

Interpretation Introduction

(a)

Interpretation:

For the given equilibrium reaction, the expression for the equilibrium constant needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 14QAP

$K^{\text{'}}=\frac{\left({\text{P}}_{\text{NOCl}}\right)}{\left(P_{\text{NO}}\right){\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{1/2}}$

Explanation of Solution

The reaction at $100{}^{\circ }\text{C}$ is as follows:

$\text{NO}\left(g\right)+\frac{1}{2}{\text{Cl}}_{\text{2}}\left(g\right)\rightleftharpoons \text{NOCl}\left(g\right)$

The expression for the equilibrium constant of the reaction represented by symbol $K^{\text{'}}$ will be:

$K^{\text{'}}=\frac{\left({\text{P}}_{\text{NOCl}}\right)}{\left(P_{\text{NO}}\right){\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{1/2}}$

Interpretation Introduction

(b)

Interpretation:

The expression for the decomposition of 1 mol of NOCl gas needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 14QAP

$K^{\text{'}\text{'}}=\frac{{\left(P_{\text{NO}}\right)}^2\left(P_{{\text{Cl}}_{\text{2}}}\right)}{{\left({\text{P}}_{\text{NOCl}}\right)}^2}$

Explanation of Solution

The reaction for the decomposition of NOCl that produces 1 mol of the chlorine gas is represented as follows:

$2\text{NOCl}\left(g\right)\rightleftharpoons \text{2NO}\left(g\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

Thus, the expression for the equilibrium constant of the reaction represented by symbol $K^{\text{'}\text{'}}$ will be:

$K^{\text{'}\text{'}}=\frac{{\left(P_{\text{NO}}\right)}^2\left(P_{{\text{Cl}}_{\text{2}}}\right)}{{\left({\text{P}}_{\text{NOCl}}\right)}^2}$

Interpretation Introduction

(c)

Interpretation:

The relation between equilibrium constant $K\text{'}$ and $K"$ needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 14QAP

$K^{\text{'}\text{'}}=\frac{1}{{\left(K^{\text{'}}\right)}^2}$

Explanation of Solution

From the part (a) and (b) expression for equilibrium constant $K^{\text{'}}$ and $K^{\text{'}\text{'}}$ is as follows:

$K^{\text{'}}=\frac{\left({\text{P}}_{\text{NOCl}}\right)}{\left(P_{\text{NO}}\right){\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{1/2}}$.. ....... (1)

$K^{\text{'}\text{'}}=\frac{{\left(P_{\text{NO}}\right)}^2\left(P_{{\text{Cl}}_{\text{2}}}\right)}{{\left({\text{P}}_{\text{NOCl}}\right)}^2}$.. ........ (2)

Dividing (1) and (2)

$\frac{K^{\text{'}}}{K^{\text{'}\text{'}}}=\frac{\frac{\left({\text{P}}_{\text{NOCl}}\right)}{\left(P_{\text{NO}}\right){\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{1/2}}}{\frac{{\left(P_{\text{NO}}\right)}^2\left(P_{{\text{Cl}}_{\text{2}}}\right)}{{\left({\text{P}}_{\text{NOCl}}\right)}^2}}=\frac{\left({\text{P}}_{\text{NOCl}}\right){\left({\text{P}}_{\text{NOCl}}\right)}^2}{\left(P_{\text{NO}}\right){\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{1/2}{\left(P_{\text{NO}}\right)}^2\left(P_{{\text{Cl}}_{\text{2}}}\right)}=\frac{{\left({\text{P}}_{\text{NOCl}}\right)}^3}{{\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{3/2}{\left(P_{\text{NO}}\right)}^3}$

Or,

$\frac{K^{\text{'}}}{K^{\text{'}\text{'}}}=\frac{{\left({\text{P}}_{\text{NOCl}}\right)}^3}{{\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{3/2}{\left(P_{\text{NO}}\right)}^3}$.. ....... (1)

Since,

$K^{\text{'}}=\frac{\left({\text{P}}_{\text{NOCl}}\right)}{\left(P_{\text{NO}}\right){\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{1/2}}$

Thus,

$K^{\text{'}}{}^3=\frac{{\left({\text{P}}_{\text{NOCl}}\right)}^3}{{\left(P_{\text{NO}}\right)}^3{\left(P_{{\text{Cl}}_{\text{2}}}\right)}^{3/2}}$

Putting the value in equation (1)

$\begin{array}{l}\frac{K^{\text{'}}}{K^{\text{'}\text{'}}}={\left(K^{\text{'}}\right)}^3\\ K^{\text{'}\text{'}}=\frac{1}{{\left(K^{\text{'}}\right)}^2}\end{array}$

Therefore, the relation between $K^{\text{'}}$ and $K^{\text{'}\text{'}}$ will be:

$K^{\text{'}\text{'}}=\frac{1}{{\left(K^{\text{'}}\right)}^2}$