Chapter 13, Problem 13.61SP

Interpretation Introduction

Interpretation:

The rate constant of reverse reaction in given equilibriumhas to be calculated.

Concept introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

$\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.