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(A) To obtain the target reaction $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$,we manipulate the given reactions:$1$. Reverse the first rea

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$\frac{k_2 k_3^3}{k_1}$

Vedclass · Answer

(A) To obtain the target reaction $2NH_3 + \frac{5}{2}O_2 ightleftharpoons 2NO + 3H_2O$,we manipulate the given reactions:$1$. Reverse the first reaction: $2NH_3 ightleftharpoons N_2 + 3H_2$,$K_a = \frac{1}{k_1}$$2$. Keep the second reaction as is: $N_2 + O_2 ightleftharpoons 2NO$,$K_b = k_2$$3$. Multiply the third reaction by $3$: $3H_2 + \frac{3}{2}O_2 ightleftharpoons 3H_2O$,$K_c = k_3^3$Adding these reactions gives: $2NH_3 + N_2 + O_2 + 3H_2 + \frac{3}{2}O_2 ightleftharpoons N_2 + 3H_2 + 2NO + 3H_2O$Simplifying,we get: $2NH_3 + \frac{5}{2}O_2 ightleftharpoons 2NO + 3H_2O$The equilibrium constant $K = K_a 	imes K_b 	imes K_c = \frac{1}{k_1} 	imes k_2 	imes k_3^3 = \frac{k_2 k_3^3}{k_1}$.

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