16 moles of H2 and 4 moles of N2 are sealed in one litre vessel. The vessel is kept at constant temperature until the equilibrium N2(g)+2H2(g)⇌2NH3(g) is established; It is found that the pressure in the vessel has fallen to 910th of its original value (P atm.)Now ‘x’ moles of an inert gas is added to the equilibrium mixture at constant temperature till the original pressure (P atm.) is attained. The value of ‘x’ is

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16 moles of $H_{2}$ and 4 moles of $N_{2}$ are sealed in one litre vessel. The vessel is kept at constant temperature until the equilibrium $N_{2} (g) + 2 H_{2} (g) ⇌ 2 N H_{3} (g)$ is established; It is found that the pressure in the vessel has fallen to $\frac{9}{10} t h$ of its original value (P atm.)Now ‘x’ moles of an inert gas is added to the equilibrium mixture at constant temperature till the original pressure (P atm.) is attained. The value of ‘x’ is

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answer is 2.

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Detailed Solution

Ans: 2
Sol $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$t = 046$
$P \times 1 = 20 \times R \times T$
$\frac{9}{10} P \times 1 = (20 - 2 a) \times R \times T \Rightarrow a = 1$
On adding ‘x’ moles of inert gas at constant volume, equilibrium composition does not change