Consider the reaction: $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$ (exothermic process)
How many of given changes will result in decrease in number of moles of $N_{2}$ gas initially present at equilibrium in a closed vessel?
Increase in temperature
Decrease in volume of container
Increase in total pressure
Addition of $He$ gas at constant pressure
Addition of $He$ gas at constant volume
Addition of small amount of $H_{2} O$ (l) in vessel
Addition of $H_{2}$ gas in vessel at constant volume
Removing some moles of ${NH}_{3}$
Addition of $N_{2}$ gas

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answer is 5.

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Detailed Solution

The given reaction is,

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

This is an endothermic reaction.

The decrease in number of moles of $N_{2}$ gas initially present at equilibrium in a closed vessel sees following changes;

(2) Decrease in volume of container

(3) Increase in total pressure

(6) Addition of small amount of $H_{2} O$ (l) in vessel

(7) Addition of $H_{2}$ gas in vessel at constant volume

(8) Removing some moles of ${NH}_{3}$

$[(2), (3), (6), (7), (8)]$

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