Question

Chemical equation of the industrial preparation of $SO_3$ is given:
$2S{{O}_{2}}+{{O}_{2}}\rightleftharpoons 2S{{O}_{3}}+\Delta$
What is the influence of the following factors in this system at equilibrium?
(a) Increasing the amount of $O_2$
(b) $SO_2$ is removed from the system
(c) Decreasing the temperature

Answer 1

The reaction given above is part of the contact process, by which $SO_3$ is produced industrially.$SO_2$ and $O_2$ react in the presence of a catalyst to produce $SO_3$. So, the $SO_3$ produced is passed through a tube so that it doesn’t form $SO_2$ again, as the reaction is reversible. So, to answer the above question, we will use the Le-Chatelier principle.

Complete step-by-step answer: The chemical equation of conversion of $SO_2$ to $SO_3$ is given. Also heat (∆) is released in the product side, so it is an exothermic reaction.
For answering the question, we will use Le-Chatelier’s principle. The principle states that when a system at equilibrium is disturbed the position of the equilibrium shifts in the direction so that the effect of the change gets nullified.
(a) Le-Chatelier principle states that when the reactant of the reaction is increased then the reaction proceeds in the forward direction and formation of product increases. Here, oxygen is the reactant so when the amount of $O_2$ is increased, the production of sulfur dioxide which is the product increases.
(b) Le-Chatelier principle states that when the concentration of the reactant is decreased the reaction favors the backward reaction and more of the reactants is produced and product is reduced, such that the system attains equilibrium again. If $SO_2$ is removed from the system, then the reaction favors the backward direction and more of the sulfur-dioxide is produced to nullify the disbalance.
(c) Le-Chatelier principle states that when the temperature of the exothermic reaction is decreased then the reaction proceeds in the forward direction. As can be seen from the above reaction, the process is exothermic. Thus, decreasing the temperature will favor the conversion and will shift the equilibrium right (forward reaction).

Note: Whenever the equilibrium disturbes, it is restored when the Le-Chatelier's principle comes to play. The $SO_3$ produced in this step is later used to produce sulfuric acid. The $SO_3$ is not directly dissolved in water, but first in $H_2SO_4$ to form oleum or fuming sulfuric acid and then with water to form $H_2SO_4$.