At 25°C, the equilibrium constants K1,K2 and K3 of three reactions are:N2+3H2⇌2NH3;K1N2+O2⇌2NO;K2H2+12O2⇌H2O;K3The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:

K1=NH32N2H23K2=[NO]2N2O2K3=H2OH2O21/2∴For NH3+52O2⟶2NO+3H2OK=[NO]2H2O3NH32O25/2=K2K33K1

At 25°C, the equilibrium constants K1,K2 and K3 of three reactions are:N2+3H2⇌2NH3;K1N2+O2⇌2NO;K2H2+12O2⇌H2O;K3The equilibrium constants for the oxidation of NH3 by oxygen to give NO is:

K1=NH32N2H23K2=[NO]2N2O2K3=H2OH2O21/2∴For NH3+52O2⟶2NO+3H2OK=[NO]2H2O3NH32O25/2=K2K33K1

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At 25°C, the equilibrium constants K₁,K₂ and K₃ of three reactions are:
$\begin{array}{l} N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}; K_{1} \\ N_{2} + O_{2} ⇌ 2 NO; K_{2} \\ H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O; K_{3} \end{array}$
The equilibrium constants for the oxidation of NH₃ by oxygen to give NO is:

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$K_{1} K_{2} / K_{3}$

$K_{2}^{2} K_{3} / K_{1}$

$K_{2} K_{3}^{3} / K_{1}$

$K_{3}^{2} / K_{1}$

answer is D.

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Detailed Solution

$\begin{array}{l} K_{1} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} \\ K_{2} = \frac{[NO]^{2}}{[N_{2}] [O_{2}]} \end{array}$
$K_{3} = \frac{[H_{2} O]}{[H_{2}] {[O_{2}]}^{1 / 2}}$
$∴$ For ${NH}_{3} + \frac{5}{2} O_{2} ⟶ 2 NO + 3 H_{2} O$
$K = \frac{[NO]^{2} {[H_{2} O]}^{3}}{{[{NH}_{3}]}^{2} {[O_{2}]}^{5 / 2}} = \frac{K_{2} K_{3}^{3}}{K_{1}}$