For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_p = 41$ at $400 \ K$. Calculate $K_c$ for the following reactions at $400 \ K$:
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$

(A) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,$\Delta n_g = 2 - (1 + 3) = -2$.
Given $K_p = 41$,$T = 400 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
$41 = K_c(0.0821 \times 400)^{-2} \implies K_c = 41 \times (32.84)^2 \approx 44216.7$.
$(a)$ For $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$,the reaction is multiplied by $2$. Thus,$K_c' = (K_c)^2 = (44216.7)^2 \approx 1.95 \times 10^9$.
$(b)$ For $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$,the reaction is reversed and multiplied by $2$. Thus,$K_c'' = (1/K_c)^2 = (1/44216.7)^2 \approx 5.11 \times 10^{-10}$.
$(c)$ For $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$,the reaction is multiplied by $1/2$. Thus,$K_c''' = (K_c)^{1/2} = \sqrt{44216.7} \approx 210.28$.