Equilibrium - Result Question 1
####1. The equilibrium constant of the following are :
$ \begin{matrix} N_2+3 H_2 \rightarrow 2 NH_3 & K_1 \\ N_2+O_2 \rightarrow 2 NO & K_2 \\ H_2+\frac{1}{2} O_2 \to H_2 O & K_3 \end{matrix} $
The equilibrium constant $(K)$ of the reaction :
$2 NH_3+\frac{5}{2} O_2 \stackrel{K}{\rightarrow} 2 NO+3 H_2 O$, will be;
(a) $K_2 K_3^{3} / K_1$
(b) $K_2 K_3 / K_1$
(c) $K_2^{3} K_3 / K_1$
(d) $K_1 K_3^{3} / K_2$
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Solution:
- (a)
(i) $N_2+3 H_2 \rightarrow 2 NH_3 ; K_1=\frac{[NH_3]^{2}}{[N_2][H_2]^{3}}$
(ii) $N_2+O_2 \rightarrow 2 NO ; K_2=\frac{[NO]^{2}}{[N_2][O_2]}$
(iii) $H_2+\frac{1}{2} O_2 \longrightarrow H_2 O ; K_3=\frac{[H_2 O]}{[H_2][O_2]^{1 / 2}}$
Applying (II $+3 \times III-I)$ we will get
$2 NH_3+\frac{5}{2} O_2 \stackrel{K}{\rightarrow} 2 NO+3 H_2 O$;
$ K=\frac{[NO]^{2}}{[N_2][O_2]} \times \frac{[H_2 O]^{3}}{[H_2]^{3} \times[O_2]^{3 / 2}} / \frac{[NH_3]^{2}}{[N_2][H_2]^{3}} $
$\therefore \quad K=K_2 \times K_3^{3} / K_1$
