Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
Medium$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are taken in a $4 \ L$ closed vessel at a constant temperature. The reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. If $0.25\%$ of $N_2$ is converted into ammonia,calculate $K_c$ for this reaction and the $K_c'$ for the reaction $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$.
(N/A) Initial moles: $[N_2] = 1 \ mol$,$[H_2] = 3 \ mol$,$V = 4 \ L$.
$0.25\%$ of $N_2$ reacts: $n(N_2)_{reacted} = 1 \times 0.0025 = 0.0025 \ mol$.
Reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
At equilibrium: $[N_2] = (1 - 0.0025)/4 = 0.249375 \ M$,$[H_2] = (3 - 3 \times 0.0025)/4 = 0.748125 \ M$,$[NH_3] = (2 \times 0.0025)/4 = 0.00125 \ M$.
$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.00125)^2}{(0.249375)(0.748125)^3} \approx 1.49 \times 10^{-5} \ (mol \ L^{-1})^{-2}$.
For the second reaction,$K_c' = \sqrt{K_c} = \sqrt{1.49 \times 10^{-5}} \approx 3.86 \times 10^{-3} \ (mol \ L^{-1})^{-1}$.
$0.25\%$ of $N_2$ reacts: $n(N_2)_{reacted} = 1 \times 0.0025 = 0.0025 \ mol$.
Reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
At equilibrium: $[N_2] = (1 - 0.0025)/4 = 0.249375 \ M$,$[H_2] = (3 - 3 \times 0.0025)/4 = 0.748125 \ M$,$[NH_3] = (2 \times 0.0025)/4 = 0.00125 \ M$.
$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.00125)^2}{(0.249375)(0.748125)^3} \approx 1.49 \times 10^{-5} \ (mol \ L^{-1})^{-2}$.
For the second reaction,$K_c' = \sqrt{K_c} = \sqrt{1.49 \times 10^{-5}} \approx 3.86 \times 10^{-3} \ (mol \ L^{-1})^{-1}$.
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