Question

The equilibrium constant for the reversible reaction, $ N_{2}+3H_{2} \rightleftharpoons 2NH_{3} $ is $K$ and for the reaction $ \frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightleftharpoons NH_{3}, $ the equilibrium constant is $ K' $ . $K$ and $ K' $ will be related as

• A. $ K=K' $
• B. $ K' =\sqrt{K} $
• C. $ K=\sqrt{K'} $
• D. $ K\times K'=1 $

Answer 1

Answer: (B)

$ N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g) $
From law of mass action,
$K=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}} \ldots$ (i)
Similarly, for reaction
$ \frac{1}{2}N_{2}+\frac{3}{2}H_{2} \rightleftharpoons NH_{3} $
$K' =\frac{\left[N H_{3}\right]}{\left[N_{2}\right]^{1 / 2}\left[H_{2}\right]^{3 / 2}} ...$(ii)
So, from Eqs.(i) and (ii)
$ K'=\sqrt{K} $