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Standard XII

Chemistry

Nitrogen

When 1 mole o...

Question

When 1 mole of $N_{2}$ and 1 mole of $H_{2}$ is enclosed in 3L vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is $^{'} x^{'}$ mole of $H_{2}$ . The number of moles oh $N H_{3}$ would be:

\frac{2 x}{3}

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\frac{2 (1 + x)}{3}

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\frac{2 (1 - x)}{3}

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\frac{1 - x}{3}

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Solution

The correct option is B $\frac{2 (1 - x)}{3}$
Solution:- (C) $\frac{2 (1 - x)}{3}$
$Initially At equilibrium N_{2} 1 1 - y + 3 H_{2} 1 (1 - 3 y) ⇌ 2 N H_{3} 0 2 y$
Given that at equilibrium, $x$ moles of $H_{2}$ are found, i.e.,
$1 - 3 y = x$
$\Rightarrow y = \frac{1 - x}{3}$
Therefore,
No. of moles of $N H_{3}$ formed $= 2 y = \frac{2 (1 - x)}{3}$

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Similar questions

Q. When 1 mole of

N_{2}

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H_{2}

is enclosed in

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The number of moles of

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N_{2} + 3 H_{2} ⇌ 2 N H_{3}

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Q. At a certain temperature (T), the equilibrium constant

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When 1 mole of N2 and 1 mole of H2 is enclosed in 3L vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is ′x′ mole of H2. The number of moles oh NH3 would be:

The correct option is B 2(1−x)3Solution:- (C) 2(1−x)3InitiallyAt equilibriumN211−y+3H21(1−3y)⇌2NH302yGiven that at equilibrium, x moles of H2 are found, i.e.,1−3y=x⇒y=1−x3Therefore,No. of moles of NH3 formed =2y=2(1−x)3

When 1 mole of $N_{2}$ and 1 mole of $H_{2}$ is enclosed in 3L vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is $^{'} x^{'}$ mole of $H_{2}$ . The number of moles oh $N H_{3}$ would be: