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Introduction to Alcohols

N2g and H2g...

Question

$N_{2} (g)$ and $H_{2} (g)$ are taken in a vessel in mass ratio of $7 : 1$ . The only reaction, $N_{2} + 2 H_{2} ⇌ N_{2} H_{4} (g)$ occurs. Pressure due to $N_{2} H_{4}$ at equilibrium is $0.2$ times of total pressure 'P'. Then at equilibrium.

A

Partial pressure of

N_{2} = 2 P / 15

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B

Partial pressure of

H_{2} = 8 P / 25

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C

2 p_{N_{2}} = p_{H_{2}}

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D

P_{N_{2}} = 2 p_{H_{2}}

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Solution

The correct option is C $2 p_{N_{2}} = p_{H_{2}}$
The gases are taken in mass ratio of $7 : 1$ .
$N_{2} + 2 H_{2} ⇋ N_{2} H_{4}$
$t = 0$ $P$ $2 P$ $0$
$t = e q u i l i b r i u m$ $P - α$ $(2 P - 2 α)$ $α$
Here, $α = 0.2 P$

where,

P is Total Pressure
$P_{N_{2}} = 0.8 P$
$P_{H_{2}} = 1.6 P$
$P_{H_{2}} = 2 P_{N_{2}}$
Hence, option (C) is correct.

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Similar questions

N_{2} (g)

H_{2} (g)

are taken in a vessel in mass ratio of

7 : 1

N_{2} + 2 H_{2} ⇌ N_{2} H_{4} (g)

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N_{2} H_{4}

at equilibrium is

0.2

times of total pressure '

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H_{2}

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molar ratio in a closed vessel to attain the following equilibrium

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

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P_{N_{2}}

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H_{2}

are taken in 1 : 3 molar ratio in a closed vessel to attained the following equilibrium

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

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Q. In a vessel, two equilibrium are simultaneously established at the same temperature as follows :

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