Chapter U6.120, Problem 4E

(a)

Interpretation Introduction

Interpretation: The equilibrium constant for the following reaction needs to be calculated and the reaction favors products or not is to be determined.

  $\text{2CO}(g)\text{+}{\text{O}}_2(g)\to 2{\text{CO}}_{\text{2}}(g)$

Concept Introduction : The equilibrium constant K is a measure of whether starting substances or products are favored in an equilibrium mixture. A large value of $K$ indicates a large amount of product in the equilibrium mixture. A small value of K indicates very little product in the equilibrium mixture.

Answer to Problem 4E

Value of $K$ is $1.024\times {10}^5$ . So, the equilibrium favors products.

Explanation of Solution

The given reaction is as follows:

  $\text{2CO}(g)\text{+}{\text{O}}_2(g)\to 2{\text{CO}}_{\text{2}}(g)$

  $\begin{array}{c}K=\frac{{{\text{[CO}}_{\text{2}}]}^2}{[{\text{O}}_{\text{2}}]{[\text{CO}]}^2}\\ =\frac{{[3.2\times {10}^{-2}]}^2}{[1.6\times {10}^{-3}]{[2.5\times {10}^{-3}]}^2}\\ =\frac{10.24\times {10}^{-4}}{1.6\times {10}^{-3}\times 6.25\times {10}^{-6}}\\ =\frac{10.24\times {10}^{-4}}{10\times {10}^{-9}}\\ =1.024\times {10}^5\end{array}$

As value of $K$ is large, the equilibrium favors products.

(b)

Interpretation Introduction

Interpretation: The equilibrium constant for the following reaction needs to be calculated and the reaction favors products or not is to be determined.

  ${\text{N}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}(g)\to 2{\text{NH}}_{\text{3}}(g)$

Concept Introduction : The equilibrium constant K is a measure of whether starting substances or products are favored in an equilibrium mixture. A large value of $K$ indicates a large amount of product in the equilibrium mixture. A small value of K indicates very little product in the equilibrium mixture.

Answer to Problem 4E

Value of K is $1.521\times {10}^{-1}$ and the equilibrium favors reactants.

Explanation of Solution

Given concentrations are as follows:

  $\begin{array}{c}\left[{\text{H}}_{\text{2}}\right]=1.2\times {10}^{-1}\\ \left[{\text{NH}}_3\right]=3.4\times {10}^{-3}\\ \left[{\text{N}}_{\text{2}}\right]=4.4\times {10}^{-2}\end{array}$

Now calculate K for ${\text{N}}_{\text{2}}(g)+3{\text{H}}_{\text{2}}(g)\to 2{\text{NH}}_{\text{3}}(g)$ .

  $\begin{array}{c}K=\frac{{[{\text{NH}}_{\text{3}}]}^2}{[{\text{N}}_{\text{2}}]{[{\text{H}}_{\text{2}}]}^3}\\ =\frac{{\left(3.4\times {10}^{-3}\right)}^2}{4.4\times {10}^{-2}\times {\left(1.2\times {10}^{-1}\right)}^3}\\ =\frac{11.56\times {10}^{-6}}{4.4\times {10}^{-2}\times 1.728\times {10}^{-3}}\\ =\frac{11.56\times {10}^{-6}}{7.60\times {10}^{-5}}\\ =1.521\times {10}^{-1}\end{array}$

As the $K$ value is smaller, the equilibrium favors reactants.