$ ext{For the reaction } 2SO_2 + O_2 ightleftharpoons 2SO_3; \Delta H = -45 ext{ kcal, which one of the following statement is correct?}$

Q: $\text{For the reaction } 2SO_2 + O_2 \rightleftharpoons 2SO_3; \Delta H = -45 \text{ kcal, which one of the following statement is correct?}$

$\text{For the reaction } 2SO_2 + O_2 \rightleftharpoons 2SO_3; \Delta H = -45 \text{ kcal, which one of the following statement is correct?}$

Question

$	ext{For the reaction } 2SO_2 + O_2 ightleftharpoons 2SO_3; \Delta H = -45 	ext{ kcal, which one of the following statement is correct?}$

Accepted Answer

The reaction in question is $2SO_2 + O_2 \rightleftharpoons 2SO_3$ with $\Delta H = -45 \text{ kcal}$, indicating that it is an exothermic reaction. Let's analyze why Option 3 is the correct choice and why the other options are incorrect.

### Explanation for Option 3:
**The reaction is favoured by the increase of pressure.**

- **Le Chatelier's Principle:** According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in pressure, the system will adjust to counteract that change. For gaseous reactions, an increase in pressure will favor the side with fewer moles of gas.
- **Mole Analysis:** In the given reaction, the reactants $2SO_2 + O_2$ have a total of 3 moles of gas, while the product $2SO_3$ has 2 moles of gas. Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side ($2SO_3$).
- **Conclusion:** Therefore, increasing the pressure favors the formation of $SO_3$, making Option 3 the correct choice.

### Explanation for why other options are incorrect:

**Option 1: The reaction is favoured by the addition of inert gas at constant pressure.**

- **Effect of Inert Gas:** Adding an inert gas at constant pressure does not change the partial pressures of the reactants or products. Therefore, it does not affect the position of equilibrium for the reaction.
- **Conclusion:** The addition of an inert gas at constant pressure does not favor the reaction in either direction, making Option 1 incorrect.

**Option 2: The reaction is favoured by the increase in temperature.**

- **Exothermic Reaction:** Since the reaction is exothermic ($\Delta H = -45 \text{ kcal}$), increasing the temperature will shift the equilibrium towards the reactants to absorb the excess heat, according to Le Chatelier's Principle.
- **Conclusion:** Increasing the temperature actually disfavors the formation of $SO_3$, making Option 2 incorrect.

**Option 4: The reaction is favoured by the condition 1, 2 and 3.**

- **Combination of Incorrect Options:** Since Options 1 and 2 are incorrect, the combination of all three conditions cannot be correct.
- **Conclusion:** Therefore, Option 4 is incorrect as it includes conditions that do not favor the reaction.

In summary, the correct answer is Option 3, as increasing the pressure favors the formation of $SO_3$ by shifting the equilibrium towards the side with fewer moles of gas.