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

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(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 $K_c = \frac{1}{K_1} 	imes K_2 	imes K_3^3 = \frac{K_2 K_3^3}{K_1}$.

Similar Questions

$A$ reaction is $A + B \rightleftharpoons C + D$. Initially,we start with equal concentrations of $A$ and $B$. At equilibrium,the number of moles of $C$ is two times that of $A$. What is the equilibrium constant $(K_c)$ of the reaction?

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For the reaction $A + 2B \rightleftharpoons C$,the expression for equilibrium constant is

For the reactions $A \rightleftharpoons B; K_c = 2$,$B \rightleftharpoons C; K_c = 4$,and $C \rightleftharpoons D; K_c = 6$,the value of $K_c$ for the reaction $A \rightleftharpoons D$ is:

The reaction quotient $Q$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is given by $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$. The reaction will proceed from right to left when:

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