Chapter 18, Problem 1STP

Interpretation Introduction

Interpretation: The reaction which gives the given below equilibrium constant expression is to be determined.

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{CO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}{{\left[{\text{CO}}_{\text{2}}\right]}^2}$

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 1STP

The reaction depicting the given equilibrium constant expression is;

  ${\text{2CO}}_2\rightleftharpoons {\text{O}}_2+2\text{CO}$

Explanation of Solution

The ratio of the concentration of the products to the concentration of the reactants raised to some power is termed an equilibrium constant.

The given expression is shown as;

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{CO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}{{\left[{\text{CO}}_{\text{2}}\right]}^2}$

As per the expression given above for the equilibrium constant, the power of carbon monoxide, oxygen, and carbon dioxide is 2, 1, and 2 respectively. That means the stoichiometric coefficient of $\text{CO,}{\text{O}}_{\text{2}}\text{,}\text{and}{\text{CO}}_{\text{2}}$ is 2, 1, and 2. Therefore, the reaction depicting the given equilibrium constant expression is;

  ${\text{2CO}}_2\rightleftharpoons {\text{O}}_2+2\text{CO}$