The following equilibria are given:N_2 + 3H_2 | vedclass

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(C) Given equilibria:$(I)$ $N_2 + 3H_2 \rightleftharpoons 2NH_3 ; K_1$$(II)$ $N_2 + O_2 \rightleftharpoons 2NO ; K_2$$(III)$ $H_2 + \frac{1}{2}O_2 \ri

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$\frac{K_2 K_3^3}{K_1}$

Vedclass · Answer

(C) Given equilibria:$(I)$ $N_2 + 3H_2 ightleftharpoons 2NH_3 ; K_1$$(II)$ $N_2 + O_2 ightleftharpoons 2NO ; K_2$$(III)$ $H_2 + \frac{1}{2}O_2 ightleftharpoons H_2O ; K_3$To obtain the target reaction $2NH_3 + \frac{5}{2}O_2 ightleftharpoons 2NO + 3H_2O$,we perform the following operations:Reverse reaction $(I)$: $2NH_3 ightleftharpoons N_2 + 3H_2$ with constant $K' = \frac{1}{K_1}$Add reaction $(II)$: $N_2 + O_2 ightleftharpoons 2NO$ with constant $K_2$Add $3 	imes$ reaction $(III)$: $3H_2 + \frac{3}{2}O_2 ightleftharpoons 3H_2O$ with constant $K_3^3$Summing these reactions:$(2NH_3 + N_2 + O_2 + 3H_2 + \frac{3}{2}O_2) ightleftharpoons (N_2 + 3H_2 + 2NO + 3H_2O)$Canceling common terms gives: $2NH_3 + \frac{5}{2}O_2 ightleftharpoons 2NO + 3H_2O$The equilibrium constant $K_c = \frac{1}{K_1} 	imes K_2 	imes K_3^3 = \frac{K_2 K_3^3}{K_1}$.

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