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Equilibrium Constant and Standard Free Energy Change

N2g and H2g...

Question

$N_{2} (g)$ and $H_{2} (g)$ are taken in a vessel in mass ratio of $7 : 1$ . The 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:

partial pressure of

N_{2} = 2 P / 15

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partial pressure of

H_{2} = 8 P / 25

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

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

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Solution

The correct option is C $2 p_{N_{2}} = p_{H_{2}}$
$N_{2} + 2 H_{2} ⇌ N_{2} H_{4} (g)$
Initially P 2P 0
Equilibrium P-0.2P 2(P-0.2P) 0.2P
At equilibrium 9 pressure of $N_{2} = P_{N_{2}} = P - 0.2 P = 0.8 P$
pressure of $H_{2} = P_{H_{2}} = 2 P - 0.4 P = 0.16 P$
$T h u s, 2 P_{N_{2}} = P_{H_{2}}$

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