Question

In a vessel ${{N}_{2}}$, ${{H}_{2}}$, and $N{{H}_{3}}$ are at equilibrium. Some helium gas is introduced into the vessel so that total pressure increases while temperature and volume remain constant. According to the Le Chatelier's principle, the dissociation of $N{{H}_{3}}$:
(a)- Increases
(b)- Decreases
(c)- Remains unchanged
(d)- Equilibrium is disturbed

Answer 1

Hint: In the equilibrium, the reaction will be $2N{{H}_{3}}\rightleftharpoons {{N}_{2}}+3{{H}_{2}}$. When the number of moles in the reaction increases then the pressure must be decreased and when the number of moles in the reaction decreases then the pressure must be increased, but if there is no change in the number of moles then the pressure doesn't have any effect.

Complete step by step solution:
In the equilibrium the dissociation of ammonia to form hydrogen and nitrogen will be:
$2N{{H}_{3}}\rightleftharpoons {{N}_{2}}+3{{H}_{2}}$
There are many factors that decide the completion of the reaction like pressure, volume, temperature, etc.
- When the number of moles in the reaction increases then the pressure must be decreased and when the number of moles in the reaction decreases then the pressure must be increased, but if there is no change in the number of moles then the pressure doesn't have any effect.
- So, in the dissociation of ammonia, helium gas introduced in the reaction which increases the total pressure of the reaction but then there is no change in the volume and number of moles of the reaction, therefore will be no change in the equilibrium of the reaction and the dissociation remains unchanged.
So, the correct answer is “Option C”.

Note: When the volume of the reactant decreases then the reaction will move backward and when the volume of the product decreases then the reaction will move forward. The exothermic reactions require low temperature and the endothermic reactions require high temperature.