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In the reaction, \[2S{{O}_{2\text{ }\left( g \right)}}+{{O}_{2\text{ }\left( g \right)}}\rightleftharpoons 2S{{O}_{3\text{ }\left( g \right)}}\] (Exothermic), will shift in forward direction by:
(A) Adding $S{{O}_{3}}$ at constant volume
(B) Increasing volume of container
(C) Adding $S{{O}_{2}}$ at constant volume
(D) Adding inert gas at constant volume

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Hint: This reaction is governed by Le Chatelier’s principle of equilibrium. In simple words, we describe Le Chatelier’s principle that as we add some component to a side of the equation, then equilibrium will try to oppose that increase in concentration and will shift to the other side.

Complete answer:
Let us first try and understand what Le Chatelier’s principle really is before going on to analyse this reaction in particular.
- Le Chatelier’s principle can be written as: If we give stress to the equilibrium, then the reaction shifts to reduce the stress.
- There are several ways to stress the equilibrium. We can add or remove a product or a reactant in a chemical reaction at equilibrium to alter its equilibrium. When more reactant is added externally, the equilibrium shifts to reduce this stress: it makes more products. When more product molecules are added externally, the equilibrium shifts to reactants to reduce the stress. If a reactant or product is removed, the equilibrium shifts in a way that makes it more reactant or product, respectively, to make up for the loss.
Now, with these concepts in mind, let us try to analyse the given reaction.
- We can now easily observe that the addition of more reactants at constant volume will move the direction in forward direction according to Le Chatelier’s principle.
- If we add sulphur trioxide at constant volume to the reaction vessel, then equilibrium will shift towards the left side and forward reaction will not occur.
- Adding inert gas at constant volume will not make any change in reaction equilibrium.
- When we increase the volume of the reaction vessel, the decrease in concentration of gases will be more at the left side because there are three molecules and so, the reaction will shift towards the left hand side.

Therefore, we can safely conclude that the answer to this question is (C).

Note: It is worth noting that when reactants or products are added or removed, the value of the ${{K}_{eq}}$ does not change. The chemical reaction simply shifts, in a predictable fashion, to re-establish concentrations so that the ${{K}_{eq}}$ expression reverts to the correct value.


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