Q. Equilibrium constant $K_{1}$ and $K_{2}$ for the following equilibria $NO(g) + \frac{1}{2}O_{2}NO_{2}(g)$ and, $2NO_{2}(g)\ 2NO(g) + O_{2}(g)$ are related as
A) $K_{1} = \frac{1}{K_{2}}$
B) $K_{2} = \frac{1}{K_{1}}$
C) $K_{2} = \frac{1}{K_{1}^{2}}$
D) $K_{1} = \frac{1}{K_{2}^{2}}$
Explanation
(c)
$K_{1} = \frac{\left\lbrack NO_{2} \right\rbrack}{\lbrack NO\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}}$ …(i)
$K_{2} = \frac{\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack}{\left\lbrack NO_{2} \right\rbrack^{2}}$ …(ii)
$\therefore\frac{1}{K_{2}} = \frac{\left\lbrack NO_{2} \right\rbrack^{2}}{\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack}$
or, $\frac{1}{\sqrt{K_{2}}} = \frac{\lbrack NO_{2}\rbrack}{\lbrack NO\rbrack\left\lbrack O_{2} \right\rbrack^{1/2}}$ …(iii)
From Eqs.(i) and (iii),
$K_{1} = \frac{1}{\sqrt{K_{2}}}$
$\therefore K_{2} = \frac{1}{K_{1}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.