Equilibrium - Result Question 15
####16. $K_1$ and $K_2$ are equilibrium constant for reactions (i) and (ii)
$N_2(g)+O_2(g) \rightarrow 2 NO(g)$
$NO(g) \rightarrow \frac{1}{2} N_2(g)+\frac{1}{2} O_2(g)$
Then,
[1989, 94, 2005]
(a) $K_1=(\frac{1}{K_2})^{2}$
(b) $K_1=K_2^{2}$
(c) $K_1=\frac{1}{K_2}$
(d) $K_1=(K_2)^{0}$
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Solution:
- (a) For reaction (i)
$ K_P=\frac{P _{CO_2}}{P _{CH_4} \times[P _{O_2}]^{2}} $
$ K_1 \frac{[NO]^{2}}{[N_2][O_2]} $
and for reaction (ii)
$ K_2 \quad \frac{[N_2]^{1 / 2}[O_2]^{1 / 2}}{[NO]} \text{ therefore } K_1=\frac{1}{K_2^{2}} $
