Equilibrium - Result Question 13
####14. The following equilibrium constants are given:
$N_2+3 H_2 \rightarrow 2 NH_3 ; K_1$
[2003, 2007]
$N_2+O_2 \rightarrow 2 NO ; K_2$
$H_2+\frac{1}{2} O_2 \rightarrow H_2 O ; K_3$
The equilibrium constant for the oxidation of 2 moles $NH_3$ by oxygen to give $NO$ is
(a) $\frac{K_2 K_3^{2}}{K_1}$
(b) $\frac{K_2^{2} K_3}{K_1}$
(c) $\frac{K_1 K_2}{K_3}$
(d) $\frac{K_2 K_3^{3}}{K_1}$
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Solution:
- (d) Given,
$N_2+3 H_2 \rightarrow 2 NH_3 ; K_1$
$N_2+O_2 \rightarrow 2 NO ; K_2$
$H_2+\frac{1}{2} O_2 \rightarrow H_2 O ; K_3$
We have to calculate
$4 NH_3+5 O_2 \longrightarrow 4 NO+6 H_2 O ; K=$ ?
or $2 NH_3+\frac{5}{2} O_2 \longrightarrow 2 NO+3 H_2 O$
For this equation, $K=\frac{[NO]^{2}[H_2 O]^{3}}{[NH_3]^{2}[O_2]^{5 / 2}}$
but $K_1=\frac{[NH_3]^{2}}{[N_2][H_2]^{3}}, K_2=\frac{[NO]^{2}}{[N_2][O_2]}$
$& K_3=\frac{[H_2 O]}{[H_2][O_2]^{1 / 2}}$ or $K_3^{3}=\frac{[H_2 O]^{3}}{[H_2]^{3}[O_2]^{3 / 2}}$
Now operate, $\frac{K_2 \cdot K_3{ }^{3}}{K_1}$
$=\frac{[NO]^{2}}{[N_2][O_2]} \times \frac{[H_2 O]^{3}}{[H_2]^{3}[O_2]^{3 / 2}} \cdot \frac{[N_2][H_2]^{3}}{[NH_3]^{2}}$
$=\frac{[NO^{2}[H_2 O]^{3}.}{[NH_3]^{2}[O_2]^{5 / 2}}=K$
$\therefore K=\frac{K_2 \cdot K_3{ }^{3}}{K_1}$
