The following equilibrium constant are given: N_2+3 H_2 leftharpoons 2 NH_3 ; K_1 ~N_2+O_2 leftharpoons 2 NO ; K_2 H_2+1 / 2 O_2 leftharpoons H_2 O ; K_3 The equilibrium constant for the oxidation of NH_3 by oxygen to give NO is:

Correct answer: (D) (K_2 K_3^3)/K_1. Given, N_2+3 H_2 leftharpoons 2 NH_3 ; K_1 ~N_2+O_2 leftharpoons 2 NO ; K_2 H_2+1/2 O_2 leftharpoons 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+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 =([NH_3 ]^2)/([~N_2 ] [H_2 ]^3) ~K_2 =([NO]^2)/([~N_2 ] [O_2 ]) K_3 =frac [H_2 O ] [H_2 ] [O_2 ]^1 / 2 ~K_3 =frac [H_2 O^3 . [H_2 ]^3 [O_2 ]^3 / 2 Now fracK_2 · K_3^3~K_1 =([NO]^2)/([~N_2 ] [O_2 ]) × frac [H_2 O ]^3 [H_2 ]^3 [O_2 ]^3 / 2 × ([N_2 ] [H_2 ]^3)/([NH_3 ]^2) =frac[NO]^2 [H_2 O ]^3 [NH_3 ]^2 [O_2 ]^5 / 2=K ∴ K=fracK_2 · K_3^3~K_1

Q58 / 200 · NEET 2007 · Chemistry

The following equilibrium constant are given: $\begin{aligned} \mathrm{N}_2+3 \mathrm{H}_2 & \rightleftharpoons 2 \mathrm{NH}_3 ; \mathrm{K}_1 \\ \mathrm{~N}_2+\mathrm{O}_2 & \rightleftharpoons 2 \mathrm{NO} ; \mathrm{K}_2 \\ \mathrm{H}_2+1 / 2 \mathrm{O}_2 & \rightleftharpoons \mathrm{H}_2 \mathrm{O} ; \mathrm{K}_3 \end{aligned}$ The equilibrium constant for the oxidation of $\mathrm{NH}_3$ by oxygen to give $\mathrm{NO}$ is:

A
$\frac{\mathrm{K}_2 \mathrm{~K}_3^2}{\mathrm{~K}_1}$
B
$\frac{\mathrm{K}_2^2 \mathrm{~K}_3}{\mathrm{~K}_1}$
C
$\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}$
Correct answer
D
$\frac{K_2 K_3^3}{K_1}$

Solution

Given, $\begin{aligned} & \mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 ; \mathrm{K}_1 \\ & \mathrm{~N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} ; \mathrm{K}_2 \\ & \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{H}_2 \mathrm{O} ; \mathrm{K}_3 \end{aligned}$ We have to calculate $\begin{aligned} & 4 \mathrm{NH}_3+5 \mathrm{O}_2 \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} ; \mathrm{K}=\text { ? } \\ & \text { or } 2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \longrightarrow 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O} \end{aligned}$ For this equation $\mathrm{K}=\frac{[\mathrm{NO}]^2\left[\mathrm{H}_2 \mathrm{O}\right]^3}{\left[\mathrm{NH}_3\right]^2\left[\mathrm{O}_2\right]^{5 / 2}}$ but $\begin{aligned} \mathrm{K}_1 & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \\ \mathrm{~K}_2 & =\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \\ \mathrm{K}_3 & =\frac{\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{H}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}} \\ \mathrm{~K}_3 & =\frac{\left[\mathrm{H}_2 \mathrm{O}^3\right.}{\left[\mathrm{H}_2\right]^3\left[\mathrm{O}_2\right]^{3 / 2}} \end{aligned}$ Now $\begin{aligned} & \frac{\mathrm{K}_2 \cdot \mathrm{K}_3{ }^3}{\mathrm{~K}_1} =\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \times \frac{\left[\mathrm{H}_2 \mathrm{O}\right]^3}{\left[\mathrm{H}_2\right]^3\left[\mathrm{O}_2\right]^{3 / 2}} \times \frac{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2} \\ & =\frac{[\mathrm{NO}]^2\left[\mathrm{H}_2 \mathrm{O}\right]^3}{\left[\mathrm{NH}_3\right]^2\left[\mathrm{O}_2\right]^{5 / 2}}=\mathrm{K} \\ & \therefore \mathrm{K}=\frac{\mathrm{K}_2 \cdot \mathrm{K}_3{ }^3}{\mathrm{~K}_1} \end{aligned}$

Hence, the correct answer is option: 'D'.

Solution

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