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(B) Given reaction: $N_2 + 3H_2 \rightleftharpoons 2NH_3$Initial moles: $n(N_2) = 1, n(H_2) = 1, n(NH_3) = 1$Let $x$ be the moles of $N_2$ reacted at

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$3657.14$

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(B) Given reaction: $N_2 + 3H_2 ightleftharpoons 2NH_3$Initial moles: $n(N_2) = 1, n(H_2) = 1, n(NH_3) = 1$Let $x$ be the moles of $N_2$ reacted at equilibrium.Equilibrium moles of $N_2 = 1 - x = 0.7 \Rightarrow x = 0.3 \ mol$Equilibrium moles of $H_2 = 1 - 3x = 1 - 3(0.3) = 0.1 \ mol$Equilibrium moles of $NH_3 = 1 + 2x = 1 + 2(0.3) = 1.6 \ mol$Since volume $V = 1 \ L$,concentrations are equal to the number of moles.$[N_2] = 0.7 \ M, [H_2] = 0.1 \ M, [NH_3] = 1.6 \ M$Equilibrium constant $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$K_c = \frac{(1.6)^2}{(0.7)(0.1)^3} = \frac{2.56}{0.7 	imes 0.001} = \frac{2.56}{0.0007} = 3657.14$

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