Question

Equilibrium constants $ {{K}_{1}} $ and $ {{K}_{2}} $
for the following equilibria
$ NO(g)+\frac{1}{2}{{O}_{2}} \rightleftharpoons N{{O}_{2}}(g) $
and $ 2N{{O}_{2}}(g) \rightleftharpoons 2NO(g)+{{O}_{2}}(g) $
are related as

• A. $ {{K}_{2}}=\frac{1}{{{K}_{1}}} $
• B. $ {{K}_{2}}={{K}_{1}}^{2} $
• C. $ {{K}_{2}}=\frac{{{K}_{1}}}{2} $
• D. $ {{K}_{2}}=\frac{1}{={{K}_{1}}^{2}} $

Answer 1

Answer: (B)

(i) $ NO(g)+\frac{1}{2}{{O}_{2}}(g)N{{O}_{2}}(g) $
So, $ {{K}_{1}}=\frac{[N{{O}_{2}}]}{[NO]{{[{{O}_{2}}]}^{1/2}}} $ ..(i) (ii)
$ 2N{{O}_{2}}(g)2NO+{{O}_{2}}(g) $
So, $ {{K}_{2}}=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}} $ ..(ii)
From Eq. (i) $ {{K}_{1}}^{2}=\frac{{{[N{{O}_{2}}]}^{2}}}{{{[NO]}^{2}}[{{O}_{2}}]} $
or $ \frac{1}{K_{1}^{2}}=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}} $ ..(iii)
From Eqs. (i) and (iii)
$ {{K}_{2}}=\frac{1}{K_{1}^{2}} $