39. For the reaction, ${\mathrm{N}}_{2}\left(\mathrm{g}\right)+3{\mathrm{H}}_{2}\left(\mathrm{g}\right)⇌2{\mathrm{NH}}_{3}\left(\mathrm{g}\right)$

Equilibrium constant, ${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{NH}}_{3}\right]}^{2}}{\left[{\mathrm{N}}_{2}\right]{\left[{\mathrm{H}}_{2}\right]}^{3}}$

Some reaction are written below in Column I and their equilibrium constants in terms of Kc are written in Column II. Match the following reactions with the corresponding equilibrium constant.

 Column I (Reaction) Column II (Equilibrium constant) A. $2{\mathrm{N}}_{2}\left(\mathrm{g}\right)+6{\mathrm{H}}_{2}\left(\mathrm{g}\right)⇌4{\mathrm{NH}}_{3}\left(\mathrm{g}\right)$ 1. $2{\mathrm{K}}_{\mathrm{c}}$ B. $2{\mathrm{NH}}_{3}\left(\mathrm{g}\right)⇌2{\mathrm{N}}_{2}\left(\mathrm{g}\right)+3{\mathrm{H}}_{2}\left(\mathrm{g}\right)$ 2. ${\mathrm{K}}_{\mathrm{c}}^{1/2}$ C. $\frac{1}{2}{\mathrm{N}}_{2}\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)⇌{\mathrm{NH}}_{3}\left(\mathrm{g}\right)$ 3. $\frac{1}{{\mathrm{K}}_{\mathrm{c}}}$ 4. ${\mathrm{K}}_{\mathrm{c}}^{2}$
A.$\to$(4) B.$\to$(3) C.$\to$(2)
For the reaction, ${\mathrm{N}}_{2}\left(\mathrm{g}\right)+3{\mathrm{H}}_{2}\left(\mathrm{g}\right)⇌2{\mathrm{NH}}_{3}\left(\mathrm{g}\right)$
Equilibrium constant ${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{NH}}_{3}\right]}^{2}}{\left[{\mathrm{N}}_{2}\right]{\left[{\mathrm{H}}_{2}\right]}^{3}}$
A. The given reaction $2{\mathrm{N}}_{2}\left(\mathrm{g}\right)+6{\mathrm{H}}_{2}\left(\mathrm{g}\right)⇌4{\mathrm{NH}}_{3}\left(\mathrm{g}\right)$ is twice the above reaction. Hence, $\mathrm{K}={\mathrm{K}}_{\mathrm{c}}^{2}$
B. The reaction $2{\mathrm{NH}}_{3}\left(\mathrm{g}\right)⇌2{\mathrm{N}}_{2}\left(\mathrm{g}\right)+3{\mathrm{H}}_{2}\left(\mathrm{g}\right)$ is reverse of the above reaction. Hence, $\mathrm{K}=\frac{1}{{\mathrm{K}}_{\mathrm{c}}}$'
C. The reaction $\frac{1}{2}{\mathrm{N}}_{2}\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)⇌{\mathrm{NH}}_{3}\left(\mathrm{g}\right)$ is half of the above reaction. Hence, $\mathrm{K}=\sqrt{{\mathrm{K}}_{\mathrm{c}}}={\mathrm{K}}_{\mathrm{c}}^{1/2}$