Chapter 18.3, Problem 24SP

Interpretation Introduction

Interpretation: The equilibrium concentration of hydrogen and hydrogen iodide needs to be determined.

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 24SP

The equilibrium concentration of hydrogen and hydrogen iodide is 0.50 mol/L and 3.5 mol/L respectively.

Explanation of Solution

The chemical equation is given as;

  $\text{2HI}\left(g\right)\rightleftharpoons {\text{H}}_2\left(g\right){\text{+I}}_2\left(g\right)$

It is given that the volume is 1 L, moles of hydrogen are 0.50 moles and the equilibrium constant is 0.020.

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

  $K_{eq}=\frac{\left[{\text{H}}_2\right]\left[{\text{I}}_2\right]}{{\left[\text{HI}\right]}^2}----\left(eq1\right)$

At equilibrium concentration of hydrogen will be equal to the concentration of iodine. Thus, substituting the given values in equation 1 as;

  $\begin{array}{c}0.020=\frac{\left[\frac{\text{0}\text{.50}\text{mol}}{1\text{L}}\right]\left[\frac{\text{0}\text{.50}\text{mol}}{1\text{L}}\right]}{{\left[\text{HI}\right]}^2}\\ {\left[\text{HI}\right]}^2=\frac{\left[0.25{\text{M}}^2\right]}{0.020}\\ \left[\text{HI}\right]=\sqrt{\frac{\left[0.25\right]}{0.020}}\text{M}\\ \left[\text{HI}\right]=3.5\text{M}\end{array}$

Therefore, the equilibrium concentration of hydrogen and hydrogen iodide is 0.5 mol/L and 3.5 mol/L respectively.