Chapter 17, Problem 68A

Interpretation Introduction

Interpretation:

The value for K for the backward reaction $2N{H}_3(g)\to N_2(g)+3H_2(g)$ have to be calculated.

Concept Introduction:

Chemical equilibrium: Chemical equilibrium is a state of a chemical reaction when the rate of forward reaction is equal to the rate of backward reaction. 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.

  $aA+bB\rightleftharpoons cC+dD$

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

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

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

  $Atequilibriumtherateoftheforwardreaction=therateofthebackwardreaction$

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

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

  $K_{eq}istheequilibriumconstantofthereaction.$

  $\begin{array}{l}{N}_2(g)+3H_2(g)\to 2N{H}_3(g)\\ K_1=\frac{{[N{H}_3]}^2}{[N_2]{[H_2]}^3}\\ 2N{H}_3(g)\to N_2(g)+3H_2(g)\\ K_2=\frac{[N_2]{[H_2]}^3}{{[N{H}_3]}^2}=\frac{1}{K_1}\end{array}$

Answer to Problem 68A

The value for K for the backward reaction $2N{H}_3(g)\to N_2(g)+3H_2(g)$ is 8.33 × 10⁴.

Explanation of Solution

Data given: K₁ = 1.2 × 10-5

  $\begin{array}{l}{N}_2(g)+3H_2(g)\to 2N{H}_3(g)\\ K_1=\frac{{[N{H}_3]}^2}{[N_2]{[H_2]}^3}\\ 2N{H}_3(g)\to N_2(g)+3H_2(g)\\ K_2=\frac{[N_2]{[H_2]}^3}{{[N{H}_3]}^2}=\frac{1}{K_1}\end{array}$

  $\begin{array}{l}{K}_2=\frac{1}{K_1}\\ =\frac{1}{1.2\times 10^{-5}}\\ =8.33\times 10^4\end{array}$