Question

The equilibrium constant for the reaction,
$N_2+3H_2\rightleftharpoons 2N{H}_3$
at $400K$ is $41$. The equilibrium constant for the reaction, $\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3$
at the same temperature will be closest to

• A. 41
• B. 20.5
• C. 6.4
• D. 1681

Answer 1

Answer: (C)

$N_2+3H_2\rightleftharpoons 2N{H}_3{K}_C=\mathrm{4.1...}(i)$
$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3....(ii)$
We divide equation $(i)$ by $2$ to get equation $(ii)$
$\therefore$ The new equilibrium constant, $K_C^{\prime }$
for Eq. (ii) will be the square root of $K_C^{\prime }$
Eq. (i), i.e.
$K_C^{\prime }=\sqrt{K_C}$
$K_C^{\prime }=\sqrt{41}=6.4$