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Q.

If $K_{C} is$ the equilibrium constant for the formation of ${NH}_{3},$
the dissociation constant of ammonia under the same temperature will be

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a

$K_{C}$

b

$\sqrt{K_{C}}$

c

$K_{c}^{2}$

d

$\frac{1}{K_{C}}$

answer is D.

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Detailed Solution

$\begin{array}{l} {NH}_{3} ⇌ \frac{1}{2} N_{2} + \frac{3}{2} H_{2} \\ K_{C_{1}} = \frac{{[N_{2}]}^{\frac{1}{2}} {[H_{2}]}^{\frac{3}{2}}}{{NH}_{3}} {.. For the decomposition of NH}_{3} \\ \frac{1}{2} N_{2} {+3H}_{2} ⇌ {NH}_{3} \\ K_{C_{2}} = \frac{[{NH}_{3}]}{{[N_{2}]}^{\frac{1}{2}} {[H_{2}]}^{\frac{3}{2}}} . . . For the formation of {NH}_{3} \\ K_{C_{2}} = \frac{1}{K_{C_{1}}} \end{array}$

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