Question

$K_1,K_2$ and $K_3$ are the equilibrium constants of the following reactions (I), (II) and (III) respectively:
(I) $N_2+2O_2\rightleftharpoons 2N{O}_2$
(II) $2N{O}_2\rightleftharpoons N_2+2O_2$
(III) $N{O}_2\rightleftharpoons \frac{1}{2}{N}_2+O_2$
The correct relation from the following is

• A. $K_1=\frac{1}{K_2}=\frac{1}{K_3}$
• B. $K_1=\frac{1}{K_2}=\frac{1}{{\left(K_3\right)}^2}$
• C. $K_1=\sqrt{K_2}=K_3$
• D. $K_1=\frac{1}{K_2}=K_3$

Answer 1

Answer: (B)

(I) $N_2+2O_2\stackrel{K_1}{\rightleftharpoons }2N{O}_2$
$K_1=\frac{{\left[N{O}_2\right]}^2}{\left[N_2\right]{\left[O_2\right]}^2}\dots (i)$
(II) $2N{O}_2\stackrel{K_2}{\rightleftharpoons }{N}_2+2O_2$
$K_2=\frac{\left[N_2\right]{\left[O_2\right]}^2}{{\left[N{O}_2\right]}^2}\dots (ii)$
(III) $N{O}_2\stackrel{A_3}{\rightleftharpoons }\frac{1}{2}{N}_2+O_2$
$K_3=\frac{{\left[N_2\right]}^{1/2}{O}_2}{\left[N{O}_2\right]}$
$\therefore {\left(K_3\right)}^2=\frac{\left[N_2\right]{\left[O_2\right]}^2}{{\left[N{O}_2\right]}^2}...\left(iii\right)$
$\therefore$ from equation (i), (ii) and (iii)
$K_1=\frac{1}{K_2}=\frac{1}{{\left(K_3\right)}^2}$