Chapter 18.3, Problem 19SP

Interpretation Introduction

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

  ${\text{N}}_2\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NH}}_{\text{3}}\left(\text{g}\right)$

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 19SP

The equilibrium constant value for the reaction is 11.84.

Explanation of Solution

Given;

Moles of ${\text{N}}_{\text{2}}$ are 0.25 mol.

Moles of ${\text{H}}_{\text{2}}$ are 0.15 mol.

Moles of ${\text{NH}}_3$ are 0.10 mol.

Volume is $1\text{L}$ .

The chemical equation is given as;

  ${\text{N}}_2\left(\text{g}\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NH}}_{\text{3}}\left(\text{g}\right)$

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

  $K_{eq}=\frac{{\left[{\text{NH}}_3\right]}^2}{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\right]}^3}$

Substituting the given values in the above expression as;

  $\begin{array}{c}{K}_{eq}=\frac{{\left[\frac{0.10\text{mol}}{1\text{L}}\right]}^2}{\left[\frac{0.25\text{mol}}{1\text{L}}\right]{\left[\frac{0.15\text{mol}}{1\text{L}}\right]}^3}\\ K_{eq}=11.84\end{array}$