Question

For the reaction : $N_{2(g)}+3H_{2(g)}\rightleftharpoons 2N{H}_{3(g)}$
equilibrium constant, $K_c=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$
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)
i	$2{\mathrm{N}}_{2(g)}+6{\mathrm{H}}_{2(g)}\rightleftharpoons 4{\mathrm{NH}}_{3(g)}$	(a)	$2K_c$
ii	$2{\mathrm{NH}}_{3(g)}\rightleftharpoons {\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)}$	(b)	$K_c^{1/2}$
iii	$\frac{1}{2}{\mathrm{N}}_{2(g)}+\frac{3}{2}{\mathrm{H}}_{2(g)}\rightleftharpoons {\mathrm{NH}}_{3(g)}$	(c)	$\frac{1}{K_c}$
		(d)	$K_c^2$

• A. $(i)\to (b),(ii)\to (a),(iii)\to (c)$
• B. $(i)\to (c),(ii)\to (d),(iii)\to (a)$
• C. $(i)\to (d),(ii)\to (c),(iii)\to (b)$
• D. $(i)\to (c),(ii)\to (a),(iii)\to (d)$

Answer 1

Answer: (B)

$(\mathrm{i})\to (\mathrm{d}),(\mathrm{ii})\to (\mathrm{c}),(\mathrm{iii})\to (\mathrm{b})$
For ${\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)}\rightleftharpoons 2{\mathrm{NH}}_{3(g)}$
$K_c=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$
(i) $2{\mathrm{N}}_{2(g)}+6{\mathrm{H}}_{2(\mathrm{g}}\rightleftharpoons 4{\mathrm{NH}}_{3(g)};{\mathrm{K}}_c^{\prime }=K_c^2$
(ii) $2{\mathrm{NH}}_{3(g)}\rightleftharpoons {\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)};K_c^{\prime }=\frac{1}{K_c}$
(iii) $\frac{1}{2}{\mathrm{N}}_{2(g)}+\frac{3}{2}{\mathrm{H}}_{2(g)}\rightleftharpoons {\mathrm{NH}}_{3(g)};K_c^{\prime }=K_c^{1/2}$