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For the reaction : $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$
Equilibrium constant $K_{c} = \frac{{[N H_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
Some reactions are written below in column-I and their equilibrium constants in terms of $K_{c}$ are written in column-II. Match the following reactions with the corresponding equilibrium constant
Column-I (Reaction) Column-II (Equilibrium constant)
A $2 N_{2} (g) + 6 H_{2} (g) ⇌ 4 N H_{3} (g)$ P $2 K_{c}$
B $2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ Q $K_{c}^{\frac{1}{2}}$
C $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ N H_{3} (g)$ R $\frac{1}{K_{c}}$
D $N H_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$ S $K_{c}^{2}$
T $\frac{1}{K_{c}^{\frac{1}{2}}}$

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$A - T, B - R, C - S, D - R$

$A - S, B - R, C - Q, D - T$

$A - P, B - T, C - R, D - Q$

$A - P, B - S, C - R, D - T$

answer is D.

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Detailed Solution

$N_{2} + 3 H_{2} ⇌ 2 N H_{3}; K_{c} A) 2 N_{2} + 6 H_{2} ⇌ 4 N H_{3} = K_{c}^{2} B) 2 N H_{3} ⇌ N_{3} + 3 H_{2} = \frac{1}{K_{c}} C) \frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3} = K_{c}^{1 / 2} D) N H_{3} ⇌ \frac{1}{2} N_{2} + \frac{3}{2} H_{2} = K_{c}^{- 1 / 2}$

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