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Molality

The equilibri...

Question

The equilibrium constant for the reaction: $N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 {N H}_{3 (g)}$ at $715 K$ , is $6.0 \times 10^{- 2}$ . If, in a particular reaction, there are $0.25 m o l L^{- 1}$ of $H_{2}$ and $0.06 m o l L^{- 1}$ of ${N H}_{3}$ present, calculate the concentration of $N_{2}$ at equilibrium.

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Solution

Given: Concentration of $[H_{2}] = 0.25 m o l L^{- 1}$
Concentration of $[{N H}_{3}] = 0.06 m o l L^{- 1}$
now, for the reaction $N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 {N H}_{3 (g)}$
$K = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
or
$[N_{2}] = \frac{{[{N H}_{3}]}^{2}}{K \times {[H_{2}]}^{3}}$
$[N_{2}] = \frac{{[{N H}_{3}]}^{2}}{K \times {[H_{2}]}^{3}}$
$= \frac{{(0.06)}^{2}}{6.0 \times 10^{- 2} \times {(0.25)}^{3}} m o l L^{- 1} = 3.84 m o l L^{- 1}$
Thus, the concentration of $[N_{2}]$ at equilibrium is $3.84 m o l L^{- 1}$ .

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N H_{3}

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

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1.2 \times 10^{- 2}, 3.0 \times 10^{- 2}

1.5 \times 10^{- 2} M

N H_{3}

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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} (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

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

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