$N_{2} (g) + {3 H}_{2} (g) ⇌ 2 {N H}_{3} (g) + h e a t$

For the reaction above, what is the equilibrium expression?

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

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

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

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\frac{{[N_{2}]}^{2} {[H_{2}]}^{6}}{{[{N H}_{3}]}^{6}}

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Solution

The correct option is C $\frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
The given reaction is :

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g) + h e a t$

$\Rightarrow K_{C} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$ .

Hence, the correct option is $C$

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Similar questions

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

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

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

K_{C} = L^{2} {m o l}^{- 2}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

K_{C} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}

N H_{3}

N_{2}

H_{2}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

[N_{2}] = 1.5 \times 10^{- 2} M

[H_{2}] = 3.0 \times 10^{- 2} M

[N H_{3}] = 1.2 \times 10^{- 2} M

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

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

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