For the reaction $2 \mathrm{NO}_{2(g)} \rightleftharpoons \mathrm{N}_2 \mathrm{O}_{4(g)}, K_p / K_c$

Question

For the reaction $2 \mathrm{NO}_{2(g)} \rightleftharpoons \mathrm{N}_2 \mathrm{O}_{4(g)}, K_p / K_c$ is $x$ and for reaction, $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}, K_p / K_c$ is $y$ then $\frac{x}{y}$ is

Professor PC Thomas Classes · Answer

(c): For reaction, $2 \mathrm{NO}_{2(g)} ightleftharpoons \mathrm{N}_2 \mathrm{O}_{4(g)}$ 
$ \Delta n_g=1-2=-1 ; \K_p=K_c(R T)^{\Delta n_g} \Rightarrow \frac{K_p}{K_c}=(R T)^{-1}=\frac{1}{R T}=x $
For reaction, $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} ightleftharpoons 2 \mathrm{NH}_{3(g)}$ 
$ \Delta n_g=2-4=-2 $
 $ \begin{aligned} & K_p=K_c(R T) \Rightarrow \frac{K_p}{K_c}=(R T)^{-2} \ & \Rightarrow y=\frac{1}{(R T)^2} \ & \frac{x}{y}=\frac{1}{R T} 	imes(R T)^2=R T \end{aligned} $