Question

The oxidation of $S{{O}_{2}}\text{ by }{{\text{O}}_{2}}$ to $S{{O}_{3}}$ is an exothermic reaction. The yield of $S{{O}_{3}}$ will be maximum if:
(A) Temperature is increased and the pressure is kept constant.
(B) Temperature is reduced and the pressure is increased
(C) Both temperature and pressure are increased
(D) Both temperature and pressure are reduced

Answer 1

Hint: Exothermic reaction is a reaction in which heat is released, it lets out the heat when the temperature of the surrounding object will increase. Exothermic reactions generally take place in different day to day activities.

Complete step by step solution:
To solve this question first we have to write the reaction involved: as mentioned in the question sulphur dioxide reacts with oxygen and forms sulphur trioxide.
\[S{{O}_{2}}(g)+{{O}_{2}}(g)\to S{{O}_{3}}(g)\]
Balanced chemical reaction:
\[S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to S{{O}_{3}}(g)\]
Given that the reaction is exothermic it means that the value of $\Delta H$ for this reaction is negative.
According to le chatelier's principle the exothermic reactions are favored only when the temperature is low. If we compare the number of moles in the given reaction we notice that there is a decrease in number of moles from reactant to product side.
And according to le chatelier's principles this type of reactions favored at high pressure.
Hence the reaction will give maximum yield when the temperature is low and the pressure is high.

The correct answer is option (B) i.e. temperature is reduced and the pressure is increased.

Additional information:
Le chatelier's principle predicts the favorable condition for a reaction. It states that when there is a disturbance in the system either by change in concentration, change in pressure or change in temperature then the system will respond to this change it means that it tries to restore a new equilibrium state.

Note: Le chatliers principle depends on the number of moles hence it is important to balance the chemical reaction. The principle states that the pressure increase shifts an equilibrium to the side of the reaction which has fewer number of moles and vice versa.