Chapter 18.3, Problem 23SP

Interpretation Introduction

Interpretation: The equilibrium concentration of ${\text{NO}}_2$ 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 23SP

The equilibrium concentration of ${\text{NO}}_2$ is 0.34 M.

Explanation of Solution

The chemical equation is given as;

  $\text{2NO}\left(g\right)\rightleftharpoons {\text{N}}_2{\text{O}}_4\left(g\right)$

It is given that the moles of dinitrogen tetroxide are 0.66 moles and the equilibrium constant is 5.6.

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

  $K_{eq}=\frac{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right]}{{\left[\text{NO}\right]}^2}----\left(eq1\right)$

Substituting the given values in equation 1 as;

  $\begin{array}{c}5.6=\frac{\left[\frac{\text{0}\text{.66}}{1}\right]}{{\left[\text{NO}\right]}^2}\\ {\left[\text{NO}\right]}^2=\frac{\left[\frac{\text{0}\text{.66}}{1}\right]}{5.6}\\ \left[\text{NO}\right]=\sqrt{\frac{\left[0.66\right]}{5.6}}\\ \left[\text{NO}\right]=0.34\text{M}\end{array}$

Therefore, the equilibrium concentration of nitrogen dioxide is 0.34 M.