Chapter 12, Problem 23E

(a)

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K\right)$ for each of the given gas-phase reactions is to be stated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$.

To determine: The equilibrium constant for the given reaction.

Answer to Problem 23E

Answer

The equilibrium constant for the given reaction is $\underset{\_}{0.11}$.

Explanation of Solution

Given

The stated reaction is,

${\text{N}}_2{}_{\left(g\right)}+3{\text{H}}_2{}_{\left(g\right)}\rightleftharpoons 2{\text{NH}}_3{}_{\left(g\right)}$

The equilibrium constant $\left(K\right)$ for the stated reaction is $1.3\times {10}^{-2}$.

At equilibrium, the equilibrium constant expression is stated,

$K=\frac{{\left[{\text{NH}}_3\right]}^2}{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}$

Where,

• $K$ is the equilibrium constant.

The stated reaction is,

$\frac{1}{2}{\text{N}}_2{}_{\left(g\right)}+\frac{3}{2}{\text{H}}_2{}_{\left(g\right)}\rightleftharpoons {\text{NH}}_3{}_{\left(g\right)}$

At equilibrium, the equilibrium constant expression is stated,

$K_1=\frac{\left[{\text{NH}}_3\right]}{{\left[{\text{N}}_2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[{\text{H}}_2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Where,

• $K_1$ is the equilibrium constant for this reaction.

Simplify the above expression.

$\begin{array}{c}{K}_1={\left[{\left(\frac{\left[{\text{NH}}_3\right]}{{\left[{\text{N}}_2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[{\text{H}}_2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[\frac{{\left[{\text{NH}}_3\right]}^2}{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[K\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Substitute the given value of $K$ in the above expression.

$\begin{array}{c}{K}_1={\left[K\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[1.3\times {10}^{-2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\underset{\_}{0.11}\end{array}$

Conclusion

The equilibrium constant for the given reaction is $\underset{\_}{0.11}$.

(b)

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K\right)$ for each of the given gas-phase reactions is to be stated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$.

To determine: The equilibrium constant for the given reaction.

Answer to Problem 23E

Answer

The equilibrium constant for the given reaction is $\underset{\_}{77}$.

Explanation of Solution

Given

The stated reaction is,

${\text{2NH}}_3{}_{\left(g\right)}\rightleftharpoons {\text{N}}_2{}_{\left(g\right)}+3{\text{H}}_2{}_{\left(g\right)}$

At equilibrium, the equilibrium constant expression is stated,

$K_2=\frac{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}{{\left[{\text{NH}}_3\right]}^2}$

Where,

• $K_2$ is the equilibrium constant for this reaction.

Simplify the above expression.

$\begin{array}{c}{K}_2=\left[\frac{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}{{\left[{\text{NH}}_3\right]}^2}\right]\\ =\left[\frac{1}{K}\right]\end{array}$

Substitute the given value of $K$ in the above expression.

$\begin{array}{c}{K}_2=\left[\frac{1}{K}\right]\\ =\left[\frac{1}{1.3\times {10}^{-2}}\right]\\ =\underset{\_}{77}\end{array}$

Conclusion

The equilibrium constant for the given reaction is $\underset{\_}{77}$.

(c)

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K\right)$ for each of the given gas-phase reactions is to be stated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$.

To determine: The equilibrium constant for the given reaction.

Answer to Problem 23E

Answer

The equilibrium constant for the given reaction is $\underset{\_}{8.8}$.

Explanation of Solution

Given

The stated reaction is,

${\text{2NH}}_3{}_{\left(g\right)}\rightleftharpoons \frac{1}{2}{\text{N}}_2{}_{\left(g\right)}+\frac{3}{2}{\text{H}}_2{}_{\left(g\right)}$

At equilibrium, the equilibrium constant expression is stated,

$K_3=\frac{{\left[{\text{N}}_2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[{\text{H}}_2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left[{\text{NH}}_3\right]}$

Where,

• $K_3$ is the equilibrium constant for this reaction.

Simplify the above expression.

$\begin{array}{c}{K}_3={\left[{\left(\frac{{\left[{\text{N}}_2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[{\text{H}}_2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left[{\text{NH}}_3\right]}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[\frac{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}{{\left[{\text{NH}}_3\right]}^2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[\frac{1}{K}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Substitute the given value of $K$ in the above expression.

$\begin{array}{c}{K}_3={\left[\frac{1}{K}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[\frac{1}{1.3\times {10}^{-2}}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\underset{\_}{8.8}\end{array}$

Conclusion

The equilibrium constant for the given reaction is $\underset{\_}{8.8}$.

(d)

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K\right)$ for each of the given gas-phase reactions is to be stated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented $K$.

To determine: The equilibrium constant for the given reaction.

Answer to Problem 23E

Answer

The equilibrium constant for the given reaction is $\underset{\_}{1.7\times 10^{-4}}$.

Explanation of Solution

Given

The stated reaction is,

$2{\text{N}}_2\left(g\right)+6{\text{H}}_2\left(g\right)\rightleftharpoons {\text{4NH}}_3\left(g\right)$

At equilibrium, the equilibrium constant expression is stated,

$\begin{array}{c}{K}_4=\frac{{\left[{\text{NH}}_3\right]}^4}{{\left[{\text{N}}_2\right]}^2{\left[{\text{H}}_2\right]}^6}\\ ={\left[\frac{{\left[{\text{NH}}_3\right]}^2}{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}\right]}^2\\ ={\left[K\right]}^2\end{array}$

Where,

• $K_4$ is the equilibrium constant for this reaction.

Substitute the given value of $K$ in the above expression.

$\begin{array}{c}{K}_4={\left[K\right]}^2\\ ={\left[1.3\times {10}^{-2}\right]}^2\\ =\underset{\_}{1.7\times 10^{-4}}\end{array}$

Conclusion

The equilibrium constant for the given reaction is $\underset{\_}{1.7\times 10^{-4}}$