Q. The equilibrium constant for the reaction, $2NO_{2}(g) \rightleftharpoons 2NO(g) + O_{2}(g)\ is\ 2 \times 10^{- 6}\ at\ 185℃$. Then the equilibrium constant for the reaction, $4NO(g) + 2O_{2}(g) \rightleftharpoons 4NO_{2}(g)$ at the same temperature would be
A) $2.5 \times 10^{- 5}$
B) $4 \times 10^{- 12}$
C) $2.5 \times 10^{11}$
D) $2 \times 10^{6}$
Explanation
(c)
$2NO_{2}(g) \rightleftharpoons 2NO(g) + O_{2}(g)$
$K = \frac{\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack}{\left\lbrack NO_{2} \right\rbrack^{2}} = 2 \times 10^{- 6}\$
$4NO(g) + O_{2}(g) \rightleftharpoons 4NO_{2}(g)$
$K^{'} = \frac{\left\lbrack NO_{2} \right\rbrack^{4}}{\lbrack NO\rbrack^{4}\left\lbrack O_{2} \right\rbrack^{2}}$
$= \frac{1}{(K)^{2}} = \frac{1}{\left( 2 \times 10^{- 6} \right)^{2}}$
Equilibrium constant $K^{'} = 0.25 \times 10^{12}$
$= 2.5 \times 10^{11}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.