Question

The volume of a closed reaction vessel in which the equilibrium reaction \[2S{O_2}({\text{g}}) + {O_2}({\text{g}}) \rightleftharpoons 2S{O_3}({\text{g}})\] sets is halved now.
A.The rates of forward and backward reaction will remain the same.
B.The equilibrium will not shift.
C.The equilibrium will shift to the right.
D.The rate of the forward reaction will become double that of the reverse reaction and the equilibrium will shift to the right.

Answer 1

Hint: To solve this question, one needs to know about the Le Chatelier’s Principle for the equilibrium of reversible reactions. It states that when a system in dynamic equilibrium is subjected to any kind of change in temperature, pressure, volume, or concentration, then the system adjusts to a new state of equilibrium to counteract the changes.

Complete step by step answer:
-According to Le Chatelier’s principle, an increase in the concentration of the reactants should increase the rate of the forward reaction to restore the equilibrium. Similarly, a decrease in the concentration of the reactants should increase the backward reaction.
- In the given equation, there are 3 moles of reactants (2 moles of $S{O_2}$ and 1 mole of ${O_2}$ ) and 2 moles of the product $S{O_3}$ . Therefore we can say that there is a decrease in the concentration in the forward reaction. According to Le Chatelier’s principle, any system would try to restore or counteract any changes in the equilibrium. So, when the volume of the vessel is halved, the equilibrium should shift to the right and the reaction should progress in the forward direction and should be double that of the backward reaction, so more amount of $S{O_3}$ will form.

Hence (D) is the correct answer.

Note:
-If the pressure on a system is increased, then its volume will decrease and the equilibrium will adjust in the direction that involves fewer moles of the gas.
-The effect of temperature is influenced by the heat of reaction. If in any reaction, heat energy is supplied, that is for an endothermic reaction, the reaction should proceed in the forward direction. So increasing the temperature would shift the equilibrium to the right and a decrease in temperature would shift it to the left. The reason is that here heat is considered to be a reactant and hence an increase in it increases the forward reaction. -Likewise, for exothermic reaction, heat is a product and hence an increase in temperature would increase the forward reaction while a decrease in the temperature would shift the equilibrium to the left and increase the backward reaction.