Chapter 18.3, Problem 22SP

Interpretation Introduction

Interpretation: The equilibrium constant value for the given below reaction needs to be determined.

  ${\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{g}\right)+\text{CO}\left(\text{g}\right)$

Concept Introduction: The reaction in which the rate of the forward reaction is equal to the rate of the backward reaction is called an equilibrium reaction.

Answer to Problem 22SP

The equilibrium constant value for the reaction is 0.786.

Explanation of Solution

Given;

Initial moles of ${\text{H}}_{\text{2}}$ are 0.10 mol.

Initial moles of ${\text{CO}}_{\text{2}}$ are 0.10 mol.

Moles of ${\text{H}}_{\text{2}}\text{O}$ at equilibrium are 0.047 mol.

Moles of $\text{CO}$ at equilibrium are 0.047 mol.

Volume is $1\text{L}$ .

The chemical equation is given as;

  ${\text{H}}_{\text{2}}\left(\text{g}\right)+{\text{CO}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{g}\right)+\text{CO}\left(\text{g}\right)$

Now, write the expression for the equilibrium constant for the above reaction as;

  $K_{eq}=\frac{\left[{\text{H}}_2\text{O}\right]\left[\text{CO}\right]}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{CO}}_2\right]}$

The moles of hydrogen and carbon dioxide at equilibrium will be calculated as;

  $\begin{array}{c}\text{Moles}\text{of}{\text{H}}_{\text{2}}=\left(0.10-0.047\right)\text{mol}\\ =0.053\text{mol}\\ \text{Moles}\text{of}{\text{CO}}_{\text{2}}=\left(0.10-0.047\right)\text{mol}\\ =0.053\text{mol}\end{array}$

Calculating the value of the equilibrium constant as;

  $\begin{array}{c}{K}_{eq}=\frac{\left[\frac{0.047}{1}\right]\left[\frac{0.047}{1}\right]}{\left[\frac{0.053}{1}\right]\left[\frac{0.053}{1}\right]}\\ K_{eq}=0.786\end{array}$