Chapter 6, Problem 49E

Interpretation Introduction

Interpretation:Equilibrium concentrations of ${\text{CO}}_{\text{2}}$ , $\text{CO}$ , and ${\text{O}}_{\text{2}}$ is to be calculated.

Concept introduction:Law of mass action states that rate of any chemical reaction is proportional to product of molar concentrations of the reactants, each raised with power of its coefficient in a balanced chemical reaction.

Consider a hypothetical reversible reaction as follows:

  $\text{aA}+\text{bB}\rightleftharpoons \text{cC}+\text{dD}$

The expression used to calculate $K_{\text{eq}}$ is as follows:

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{C}\right]}_{\text{eq}}^{\text{c}}{\left[\text{D}\right]}_{\text{eq}}^{\text{d}}}{{\left[\text{A}\right]}_{\text{eq}}^{\text{a}}{\left[\text{B}\right]}_{\text{eq}}^{\text{b}}}$

Where,

• ${\text{K}}_{\text{f}}$is the rate constant of forward reaction.
• ${\text{K}}_{\text{b}}$is the rate constant of backward reaction.
• ${\text{K}}_{\text{eq}}$is the equilibrium constant.
• ${\left[\text{A}\right]}_{\text{eq}}$ and ${\left[\text{B}\right]}_{\text{eq}}$ are the concentration of reactant ‘A’ and ‘B’ respectively at equilibrium.
• ${\left[\text{C}\right]}_{\text{eq}}$ and ${\left[\text{D}\right]}_{\text{eq}}$ are the concentration of product ‘C’ and ‘D’ respectively at equilibrium.
• ‘a’ and ‘b’ are the stoichiometric coefficient of reactant ‘A’ and ‘B’ respectively.
• c’ and ‘d’ are the stoichiometric coefficient of product ‘C’ and ‘D’ respectively.

Equilibrium constant is only function of temperature. So at fixed temperature equilibrium constant value for a particular reaction is constant.