K₁, K₂and K₃ are the equilibrium constants of the following reactions (I), (II) and (III) respectively:
$(I) N_{2} + 2 O_{2} ⇌ 2 {NO}_{2} (II) 2 {NO}_{2} ⇌ N_{2} + 2 O_{2} (III) {NO}_{2} ⇌ \frac{1}{2} N_{2} + O_{2}$
The correct relation from the following is

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$K_{1} = \frac{1}{K_{2}} = \frac{1}{{(K_{3})}^{2}}$

$K_{1} = \sqrt{K_{2}} = K_{3}$

$K_{1} = \frac{1}{K_{2}} = K_{3}$

$K_{1} = \frac{1}{K_{2}} = \frac{1}{K_{3}}$

answer is B.

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Detailed Solution

(I) $N_{2} + 2 O_{2} \overset{K_{1}}{⇌} 2 {NO}_{2}$

$K_{1} = \frac{{[{NO}_{2}]}^{2}}{[N_{2}] {[O_{2}]}^{2}} . . . . (i)$

(II) $2 {NO}_{2} \overset{K_{2}}{⇌} N_{2} + 2 O_{2}$

$K_{2} = \frac{[N_{2}] {[O_{2}]}^{2}}{{[{NO}_{2}]}^{2}} . . . (ii)$

(III) ${NO}_{2} \underset{K_{3}}{⇌} \frac{1}{2} N_{2} + O_{2}$

$K_{3} = \frac{{[N_{2}]}^{1 / 2} [O_{2}]}{[{NO}_{2}]}$

$∴ {(K_{3})}^{2} = \frac{[N_{2}] {[O_{2}]}^{2}}{{[{NO}_{2}]}^{2}}$

$∴$ from equation (i), (ii) and (iii)

$K_{1} = \frac{1}{K_{2}} = \frac{1}{{(K_{3})}^{2}}$

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