$2 {NO}_{2} (g) + 7 H_{2} (g) ⇌ 2 {NH}_{3} (g) + 4 H_{2} O (l)$
What is the correct equilibrium expression for this reaction?

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$K_{C} = \frac{{[{NH}_{3}]}^{2}}{{[{NO}_{2}]}^{2} {[H_{2}]}^{7}}$

$K_{C} = \frac{{[{NO}_{2}]}^{2} {[H_{2}]}^{7}}{{[{NH}_{3}]}^{2}}$

$K_{C} = \frac{{[{NH}_{3}]}^{2} {[H_{2} O]}^{4}}{{[{NO}_{2}]}^{2}}$

$K_{C} = \frac{{[{NH}_{3}]}^{2} [H_{2} O^{4}}{{[{NO}_{2}]}^{2} {[H_{2}]}^{7}}$

answer is A.

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Detailed Solution

Consider the following reversible reaction:

$c C + d D \Rightarrow a A + b B$

The reaction's equilibrium constant, measured in moles per liter of concentration:

$\Rightarrow K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}$

For the given reaction,

$2 {NO}_{2} (g) + 7 H_{2} (g) ⇌ 2 {NH}_{3} (g) + 4 H_{2} O (l)$

$K_{C} = \frac{{[{NH}_{3}]}^{2}}{{[{NO}_{2}]}^{2} {[H_{2}]}^{7}}$

Where $K_{c}$ is the equilibrium constants of gaseous mixtures

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