Chapter 18.3, Problem 31LC

Interpretation Introduction

Interpretation: The number of moles of ${\text{I}}_{\text{2}}$ and $\text{HI}$ present at equilibrium needs to be determined.

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

Answer to Problem 31LC

The number of moles of ${\text{I}}_{\text{2}}$ and $\text{HI}$ present at equilibrium are 0.050 mol and 0.371 mol.

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 moles of hydrogen are 0.050 moles and the equilibrium constant is 0.018.

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

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

The concentration of iodine will be equal to 0.050 and assuming the concentration of hydrogen iodide be 2x.

Now, equation 1 is given as;

  $0.018=\frac{\left[0.050\right]\left[0.050\right]}{{\left(\text{2x}\right)}^2}$

Now, solving for x as;

  $\begin{array}{c}2x^2=\frac{\left(0.050\right)\left(0.050\right)}{0.018}\\ x=\sqrt{\frac{\left(0.050\right)\left(0.050\right)}{0.018}}\\ x=0.371\end{array}$

Therefore, the moles of iodide and hydrogen iodide are 0.050 mol and 0.371 mol.