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Effective Equilibrium Constant

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Question

A container containing ammonia is allowed to decompose and the following equilibrium is established.
$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ .
Then, the equilibrium constant, $K_{p}$ is expressed as:
(Where $p_{N_{2}}$ = partial pressure of $N_{2}$ and $p_{N H_{3}}$ is the partial pressure of $N H_{3}$ )

A

K_{p} = \frac{3^{3} \times p_{N_{2}}^{2}}{p_{N H_{3}}}

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B

K_{p} = \frac{3^{3} \times p_{N_{2}} p_{H_{2}}^{3}}{p_{N H_{3}}^{2}}

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C

K_{p} = \frac{3^{3} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}

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D

K_{p} = \frac{3^{(3 / 2)} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}

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Solution

The correct option is C $K_{p} = \frac{3^{3} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}$
Given, equilibrium is,
$2 N H_{3} ⇌ N_{2} + 3 H_{2}$
$∴ K_{p} = \frac{p_{N_{2}} . p_{H_{2}}^{3}}{p_{N H_{3}}^{2}}$
Where, $p_{N_{2}}, p_{H_{2}} and p_{N H_{3}} are the partial pressure of N_{2}, H_{2} and N H_{3} respectively$

Since initially only $N H_{3}$ is present hence, $3 p_{N_{2}} = p_{H_{2}}$
$∴ K_{p} = \frac{p_{N_{2}} \times (3 p_{N_{2}})^{3}}{p_{N H_{3}}^{2}}$

$= \frac{p_{N_{2}} \times 3^{3} . p_{N_{2}}^{3}}{p_{N H_{3}}^{2}} = \frac{3^{3} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}} = \frac{3^{3} \times p_{N_{2}}^{4}}{p_{N H_{3}}^{2}}$

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