The equilibrium constant $K_{c}$ for the reaction $N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ 2 $N H_{3}$ (g) is:

$\frac{[2 N H_{3}]}{[N_{2}] [3 H_{2}]}$

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$\frac{[N H_{3}]}{[N_{2}] {[3 H_{2}]}^{3}}$

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$\frac{{[N H_{3}]}^{2}}{[N_{2}] [3 H_{2}]}$

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$\frac{{[N H_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

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Solution

The correct option is D
$\frac{{[N H_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

$N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ 2 $N H_{3}$ (g)

We can write Rate of forward reaction, $r_{f}$ = $k_{f}$ ${[N_{2}]}^{1}$ ${[H_{2}]}^{3}$

And the rate of backward reaction, $r_{b}$ = $k_{b}$ ${[N H_{3}]}^{2}$

When $r_{f}$ = $r_{b}$
$\frac{{[N H_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

Suggest Corrections

Similar questions

3H₂(g) + N₂ (g) ⇌ 2NH₃ (g)

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2 N H_{3} ⇌ N_{2} + 3 H_{2}

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