Question

Consider the reaction:

$N{O}_2\rightleftharpoons \frac{1}{2}{N}_2+O_2,K_1$

$N_2{O}_4\rightleftharpoons 2N{O}_2,K_2$

Give the equilibrium constant for the formation of $N_2{O}_4$ from $N_2$ and $O_2$.

• A. $\frac{1}{K_1^2}\frac{1}{K_2}$
• B. $\frac{1}{K_1{K}_2}$
• C. $\sqrt{\frac{1}{K_1{K}_2}}$
• D. $\frac{K_2}{K_1}$

Answer 1

Answer: (A)

$\frac{1}{K_1^2}\frac{1}{K_2}$

Given,
$N{O}_2\rightleftharpoons \frac{1}{2}{N}_2+O_2,\dots K_1$ ....(i)

$N_2{O}_4\rightleftharpoons 2N{O}_2,\dots K_2$ ....(ii)

Required reaction,
$N_2+2O_2\rightleftharpoons N_2{O}_4,\dots K$ ...(iii)

By squaring equation (i), we get-
$2N{O}_2\rightleftharpoons N_2+2O_2,\dots K_1^2$ ...(iv)

By adding equations (iii) and (iv):

$N_2+2O_2\rightleftharpoons 2N{O}_2,\frac{1}{K_1^2}$
$2N{O}_2\rightleftharpoons N_2{O}_4,\frac{1}{K_2}$
______________________
$N_2+2O_2\rightleftharpoons 2N_2{O}_4$

$\Rightarrow K=\frac{1}{K_1^2}\frac{1}{K_2}$

Hence, the correct answer is option $\text{A}$.