Chapter 5, Problem 5I.24E

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations for all substances have to be calculated.

Explanation of Solution

The required reaction is given below.

  $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$

Given that, $1.00mol{H}_{2}Oand1.00molCO$ are placed in a reaction vessel of volume $10.00L.$  Their concentrations can be calculated as given below.

  $\begin{array}{l}\left[CO\right]=\frac{moles}{volume}=\frac{1mol}{10.00L}=0.1M\\ \\ \left[H_{2}O\right]=\frac{moles}{volume}=\frac{1mol}{10.00L}=0.1M\end{array}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition(M)& CO& H_{2}O& C{O}_2& H_2\\ Initialconcentration& 0.1& 0.1& 0& 0\\ Changeinconcentration& -x& -x& +x& +x\\ Equilibriumconcentration& 0.1-x& 0.1-x& x& x\end{array}$

It is given that, at equilibrium the number of moles of carbon dioxide gas is $0.665mol.$  Its concentration is given below.

  $\begin{array}{l}Atequilibrium\left[C{O}_2\right]=\frac{moles}{volume}=\frac{0.665mol}{10.00L}=0.0665M\\ So,x=0.0665M\\ \\ \left[CO\right]=0.1-x=0.0335M\\ \\ \left[H_{2}O\right]=0.1-x=0.0335M\\ \\ \left[C{O}_2\right]=x=0.0665M\\ \\ \left[H_2\right]=x=0.0665M\end{array}$

Therefore, the equilibrium concentrations of $CO,H_{2}O,C{O}_2,and{H}_2$ are $0.0335M,0.0335M,0.0665M,and0.0665M$ respectively.

(b)

Interpretation Introduction

Interpretation:

The value of $K_c$ for the reaction $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$ has to be calculated.

Explanation of Solution

The required reaction is given below.

  $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$

The expression for equilibrium constant for the above reaction can be written as shown below,

  $K_c=\frac{\left[C{O}_2\right]\left[H_2\right]}{\left[CO\right]\left[H_{2}O\right]}$

Given that, $1.00mol{H}_{2}Oand1.00molCO$ are placed in a reaction vessel of volume $10.00L.$  Their concentrations can be calculated as given below.

  $\begin{array}{l}\left[CO\right]=\frac{moles}{volume}=\frac{1mol}{10.00L}=0.1M\\ \\ \left[H_{2}O\right]=\frac{moles}{volume}=\frac{1mol}{10.00L}=0.1M\end{array}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition(M)& CO& H_{2}O& C{O}_2& H_2\\ Initialconcentration& 0.1& 0.1& 0& 0\\ Changeinconcentration& -x& -x& +x& +x\\ Equilibriumconcentration& 0.1-x& 0.1-x& x& x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_c=\frac{\left(x\right)\left(x\right)}{\left(0.1-x\right)\left(0.1-x\right)}$

It is given that, at equilibrium the number of moles of carbon dioxide gas is $0.665mol.$  Its concentration is given below.

  $\begin{array}{l}Atequilibrium\left[C{O}_2\right]=\frac{moles}{volume}=\frac{0.665mol}{10.00L}=0.0665M\\ \\ So,x=0.0665M\end{array}$

By plugging all data in the equilibrium constant expression, the value of $K_c$ can be obtained.

  $\begin{array}{c}{K}_c=\frac{\left(x\right)\left(x\right)}{\left(0.1-x\right)\left(0.1-x\right)}\\ \\ =\frac{{\left(0.0665\right)}^2}{{\left(0.1-0.0665\right)}^2}\\ \\ K_c=3.94\end{array}$

Therefore, the value of $K_c$ for the reaction $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$ is $3.94.$