The equilibrium constant of the following are : The equilibrium constant $(\mathrm{K})$ of the reaction : $2…

The equilibrium constant of the following are :


The equilibrium constant $(\mathrm{K})$ of the reaction :
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2} \stackrel{\mathrm{K}}{ightleftharpoons} 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$, will be
  1. $\mathrm{K}_{2} \mathrm{~K}_{3}^{3} / \mathrm{K}_{1}$
  2. $\mathrm{K}_{2} \mathrm{~K}_{3} / \mathrm{K}_{1}$
  3. $\mathrm{K}_{2}^{3} \mathrm{~K}_{3} / \mathrm{K}_{1}$
  4. $\mathrm{K}_{1} \mathrm{~K}_{3}^{3} / \mathrm{K}_{2}$

Solution

$\mathrm{N}_{2}+3 \mathrm{H}_{2} ightleftharpoons 2 \mathrm{NH}_{3} ; \mathrm{K}_{1}=\frac{\left[\mathrm{NH}_{3}ight]^{2}}{\left[\mathrm{~N}_{2}ight]\left[\mathrm{H}_{2}ight]^{3}}$
(ii) $\mathrm{N}_{2}+\mathrm{O}_{2} ightleftharpoons 2 \mathrm{NO} ; \mathrm{K}_{2}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{~N}_{2}ight]\left[\mathrm{O}_{2}ight]}$
(iii) $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} ; \mathrm{K}_{3}=\frac{\left[\mathrm{H}_{2} \mathrm{O}ight]}{\left[\mathrm{H}_{2}ight]\left[\mathrm{O}_{2}ight]^{1 / 2}}$
Applying (II $+3 \times$ III $-$ I) we will get $2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2} \stackrel{\mathrm{K}}{ightleftarrows} 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{K}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{~N}_{2}ight]\left[\mathrm{O}_{2}ight]} \times \frac{\left[\mathrm{H}_{2} \mathrm{O}ight]^{3}}{\left[\mathrm{H}_{2}ight]^{3} \times\left[\mathrm{O}_{2}ight]^{3 / 2}} / \frac{\left[\mathrm{NH}_{3}ight]^{2}}{\left[\mathrm{~N}_{2}ight]\left[\mathrm{H}_{2}ight]^{3}}$
$\therefore \mathrm{K}=\mathrm{K}_{2} \times \mathrm{K}_{3}^{3} / \mathrm{K}_{1}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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