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Introduction

If the equili...

Question

If the equilibrium constant for the reaction:
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
at $750 K$ is $49$ , then the equilibrium constant for the reaction,
${N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$
at the same temperature is:

1 / 49

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49

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7

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49^{2}

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1 / 7

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Solution

The correct option is E $1 / 7$
If the equilibrium constant for the reaction:
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ .....(1)
at $750 K$ is $49$ , then the equilibrium constant for the reaction,
${N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$ .....(2)
at the same temperature is $1 / 7$ .
When reaction (1) is reversed and divided by 2, the reaction (2) is obtained.
$K_{2} = \sqrt{\frac{1}{K_{1}}}$
$K_{2} = \sqrt{\frac{1}{49}}$
$K_{2} = \frac{1}{7}$

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