One mole of nitrogen is mixed with three moles of hydrogen in a $4 ext{ litre}$ container. At equi

Question

One mole of nitrogen is mixed with three moles of hydrogen in a $4 	ext{ litre}$ container. At equilibrium, it is observed that $0.25$ percent of the nitrogen is converted to ammonia by the reaction $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$. Let the equilibrium constant for this reaction be $K_c$. What will be the value of the equilibrium constant $K_c'$ for the reaction $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} ightleftharpoons NH_{3(g)}$ at the same temperature?

Professor PC Thomas Classes · Answer

From the previous calculation, the equilibrium constant for the reaction $N_2 + 3H_2 ightleftharpoons 2NH_3$ is $K_c \approx 1.496 	imes 10^{-5}$. The new reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 ightleftharpoons NH_3$ is obtained by multiplying the original reaction by a factor of $1/2$. Therefore, its equilibrium constant $K_c'$ is the square root of the original equilibrium constant: $K_c' = (K_c)^{1/2} = \sqrt{1.496 	imes 10^{-5}} = \sqrt{14.96 	imes 10^{-6}} \approx 3.867 	imes 10^{-3}$. This corresponds to option A.