Question

Consider the following reaction:

${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right);∆\mathrm{H}°=+58\mathrm{kJ}$

For each of the following cases (a, b), the direction in which the equilibrium shifts is:

(a) Temperature is decreased.

(b) Pressure is increased by adding ${\mathrm{N}}_2$ at constant $T$.

• A. (a) towards reactant, (b) towards product
• B. (a) towards reactant, (b) no change
• C. (a) towards product, (b) towards reactant
• D. (a) towards product, (b) no change

Answer 1

The correct option is B

(a) towards reactant, (b) no change

Explanation for correct option:

B. (a) towards reactant, (b) no change

1. Equilibrium of a reaction shifts according to Le Chatelier's principle which says that: if an equilibrium of a reaction is subjected to a change in temperature, pressure, or volume then the equilibrium shifts in the direction where the change can be neutralized or undone.
2. As the given reaction is endothermic if the temperature of the reaction has decreased the equilibrium of the reaction will shift towards reactants.
3. This happens so because to complete the reaction and form product temperature should be increased.
4. Now, if pressure is increased by adding ${\mathrm{N}}_2$ there will be no effect on the equilibrium of the reaction because nitrogen is an inert gas and it will not react with reactants or products of the reaction.

Therefore, if the temperature of the reaction has decreased the equilibrium of the reaction shifts towards reactants, and on increasing the pressure of the reaction by adding ${\mathrm{N}}_2$ there will be no effect on the equilibrium.

Hence option B is correct.