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Gay Lussac's Law, Avagadro's Law

67.2 liters o...

Question

67.2 liters of $H_{2}$ combines with 44.8 liters of $N_{2}$ to form $N H_{3}$ as per the reaction.
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
The vol. of $N H_{3}$ produced is :

11.2 L

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22.4 L

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44.8 L

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5.56 L

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Solution

The correct option is D 44.8 L
$N_{2} + H_{2} ⟶ 2 N H_{3}$
According to mole concept, 1 volume of nitrogen will react with 3 volumes of hydrogen to give 2 volumes of ammonia.
22.4 L of nitrogen will react with 67.2 L of hydrogen to give 44.8 L of ammonia.

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2

N_{2}

6

H_{2}

1

50 %

N_{2}

{N H}_{3}

K_{c}

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

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}}

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

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

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

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