At $350 \mathrm{~K}, K_p$ for the reaction given below is $3.0 imes 10^{10}$ bar $^{-1}$ at equili

Question

At $350 \mathrm{~K}, K_p$ for the reaction given below is $3.0 	imes 10^{10}$ bar $^{-1}$ at equilibrium. What will be the value of $K_c$ at this temperature? 
$ 2 \mathrm{~N}_{2(g)}+\mathrm{O}_{2(g)} ightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{(g)} $

Professor PC Thomas Classes · Answer

(d): $K_p=K_c(R T)^{\Delta n_g} ; \Delta n_g=2-3=-1$ $ \begin{aligned} T & =350 \mathrm{~K}, R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \ K_c & =\frac{K_p}{(R T)^{\Delta n_g}} ; K_c=\frac{3 	imes 10^{10} \mathrm{bar}^{-1}}{\left(0.083 \mathrm{~L} \mathrm{bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} 	imes 350 \mathrm{~K}ight)^{-1}} \ K_c & =3 	imes 10^{10} \mathrm{bar}^{-1} 	imes 0.083 \mathrm{~L} \mathrm{bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} 	imes 350 \mathrm{~K} \ & =8.715 	imes 10^{11} \mathrm{~L} \mathrm{~mol} \mathrm{~mol}^{-1} \end{aligned} $