Chapter 15, Problem 15.97QP

Interpretation Introduction

Interpretation: The equilibrium concentration of all four gases at $700\text{K}$ is to be calculated for the given reaction.

Concept introduction: The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

$K_{\text{c}}=\frac{{\left[\text{Product}\right]}^y}{{\left[\text{Reactant}\right]}^x}$

To determine: The equilibrium concentration of all four gases at $700\text{K}$.

Answer to Problem 15.97QP

Solution

The equilibrium concentration of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.0307atm}$, the equilibrium concentration of $\text{CO}$ is $\underset{\_}{0.0307atm}$, the equilibrium concentration of ${\text{H}}_2$ is $\underset{\_}{0.0693atm}$ and the equilibrium concentration of ${\text{CO}}_2$ is $\underset{\_}{0.0693atm}$.

Explanation of Solution

Explanation

Given

The balanced chemical equation is,

$\text{CO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\rightleftharpoons {\text{CO}}_2\left(g\right)+{\text{H}}_2\left(g\right)$

The equilibrium constant at $700\text{K}$ is $5.1$.

The initial concentration of $\text{CO}$ is $0.050\text{M}$.

The initial concentration of ${\text{H}}_2\text{O}$ is $0.050\text{M}$.

The initial concentration of ${\text{CO}}_2$ is $0.050\text{M}$.

The initial concentration of ${\text{H}}_2$ is $0.050\text{M}$.

The equilibrium constant is calculated by the formula,

$K_{\text{c}}=\frac{\left[{\text{CO}}_2\right]\left[{\text{H}}_2\right]}{\left[\text{CO}\right]\left[{\text{H}}_2\text{O}\right]}$ (1)

Where,

• $\left[{\text{CO}}_2\right]$ is the equilibrium concentration of ${\text{CO}}_2$.
• $\left[{\text{H}}_2\right]$ is the equilibrium concentration of ${\text{H}}_2$.
• $\left[\text{CO}\right]$ is the equilibrium concentration of $\text{CO}$.
• $\left[{\text{H}}_2\text{O}\right]$ is the equilibrium concentration of ${\text{H}}_2\text{O}$.

The table showing the values of concentration at initial and at equilibrium stage,

$\begin{array}{cccccccc}& \text{CO}\left(g\right)& +& {\text{H}}_2\text{O}\left(g\right)& \rightleftharpoons & {\text{CO}}_2\left(g\right)& +& {\text{H}}_2\left(g\right)\\ \text{Initial}& 0.050& & 0.050& & 0.050& & 0.050\\ \text{Equilibrium}& 0.050-x& & 0.050-x& & 0.050+x& & 0.050+x\end{array}$ (2)

Where,

• $x$ is the change in concentration occurred during the reaction.

Substitute the values of concentration of ${\text{CO}}_2$, ${\text{H}}_2$, $\text{CO}$, ${\text{H}}_2\text{O}$ and equilibrium constant in equation (1).

$\begin{array}{c}5.1=\frac{\left[0.050+x\right]\left[0.050+x\right]}{\left[0.050-x\right]\left[0.050-x\right]}\\ 5.1=\frac{{\left[0.050+x\right]}^2}{{\left[0.050-x\right]}^2}\\ \sqrt{5.1}=\frac{\left[0.050+x\right]}{\left[0.050-x\right]}\\ 2.258=\frac{\left[0.050+x\right]}{\left[0.050-x\right]}\end{array}$

Cross multiply the terms,

$\begin{array}{c}2.258\left(0.050-x\right)=0.050+x\\ 0.1129-0.050=2.258+x\\ 0.0629=3.258x\\ x=0.0193\end{array}$

The value if $x$ is $0.0193$.

Substitute the value of $x$  in equation (2).

The equilibrium concentration of ${\text{H}}_2\text{O}$ is $\begin{array}{l}=0.050-x\\ =0.050-0.0193\\ =\underset{\_}{0.0307atm}\end{array}$

The equilibrium concentration of $\text{CO}$ is $\begin{array}{l}=0.050-x\\ =0.050-0.0193\\ =\underset{\_}{0.0307atm}\end{array}$

The equilibrium concentration of ${\text{H}}_2$ is $\begin{array}{l}=0.050+x\\ =0.050+0.0193\\ =\underset{\_}{0.0693atm}\end{array}$

The equilibrium concentration of ${\text{CO}}_2$ is $\begin{array}{l}=0.050+x\\ =0.050+0.0193\\ =\underset{\_}{0.0693atm}\end{array}$

Therefore, the equilibrium concentration of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.0307atm}$, the equilibrium concentration of $\text{CO}$ is $\underset{\_}{0.0307atm}$, the equilibrium concentration of ${\text{H}}_2$ is $\underset{\_}{0.0693atm}$ and the equilibrium concentration of ${\text{CO}}_2$ is $\underset{\_}{0.0693atm}$.

Conclusion

The equilibrium concentration of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.0307atm}$, the equilibrium concentration of $\text{CO}$ is $\underset{\_}{0.0307atm}$, the equilibrium concentration of ${\text{H}}_2$ is $\underset{\_}{0.0693atm}$ and the equilibrium concentration of ${\text{CO}}_2$ is $\underset{\_}{0.0693atm}$.