Chapter 17, Problem 66A

(a)

Interpretation Introduction

Interpretation:

The equilibrium expression of the given reaction has to be written.

Concept Introduction:

At chemical equilibrium, the rate of forward reaction and that of backward reaction becomes equal. The chemical reaction follows the law of mass action. The law of mass action states that the rate of a chemical reaction is directly proportional to the product of active masses of the reactants.

  $aA+bB\rightleftharpoons cC+dD$

  $Rateoftheforwardreaction{r}_f=k_f{[A]}^a{[B]}^b$

  $Rateofthebackwardreaction{r}_b=k_b{[C]}^c{[D]}^d$

  $k_{f}and{k}_{b}aretherateconstantsofforwardandbackwardreactionrespectively$

  $Atequilibriumtherateoftheforwardreaction=therateofthebackwardreaction$

  $k_f{[A]}^a{[B]}^b=k_b{[C]}^c{[D]}^d$

  $\frac{k_f}{k_b}=K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

  $K_{eq}istheequilibriumconstantofthereaction.$

Answer to Problem 66A

  $K_{eq}=\frac{{[HBr]}^2}{[{\mathrm{H}}_2][B{r}_2]}$

Explanation of Solution

The given chemical equation is-

  ${\mathrm{H}}_2(s)+B{r}_2(g)\rightleftharpoons 2HBr(g)$

  $\begin{array}{l}{\mathrm{H}}_2(s)+B{r}_2(g)\rightleftharpoons 2HBr(g)\\ EquilibriumConstant{K}_{eq}=\frac{{[HBr]}^2}{[{\mathrm{H}}_2][B{r}_2]}\end{array}$