Equilibrium Ques 59
59. For the reaction $N_2(g)+O_2(g) \rightarrow 2 NO(g)$, the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for the reaction
[2011]
$ 2 NO(g)+O_2(g) \rightarrow 2 NO_2(g) $
What is $K$ for the reaction
$ NO_2(g) \rightarrow \frac{1}{2} N_2(g)+O_2(g) ? $
(a) $1 /(2 K_1 K_2)$
(b) $1 /(4 K_1 K_2)$
(c) $[1 / K_1 K_2]^{1 / 2}$
(d) $1 /(K_1 K_2)$
Show Answer
Solution:
- (c) For the reaction
$ \begin{aligned} & N_2(g)+O_2(g) \rightarrow 2 NO(g) \\ & K_1=\frac{[NO]^{2}}{[N_2][O_2]} \end{aligned} $
For the reaction
$ \begin{aligned} & 2 NO(g)+O_2(g) \rightarrow 2 NO_2(g) \\ & K_2=\frac{[NO_2]^{2}}{[NO^{2}[O_2].} \end{aligned} $
For the reaction
$ \begin{aligned} & NO_2(g) \rightarrow \frac{1}{2} N_2(g)+O_2(g) \\ & K=\frac{[N_2]^{\frac{1}{2}}[O_2]}{[NO_2]} \end{aligned} $
Hence, $K=\sqrt{\frac{1}{K_1} \times \frac{1}{K_2}}$
$K=\frac{[N_2]^{\frac{1}{2}}[O_2]^{\frac{1}{2}}}{[NO]} \times \frac{[NO][O_2]^{\frac{1}{2}}}{[NO_2]}=\frac{[N_2]^{\frac{1}{2}}[O_2]}{[NO_2]}$
BETA
