2SO2(g) + O2(g) $\iff$2SO3(g) $\bigtriangleup H = -395.7kJmol^{-1}$ In the reaction above, the concentration of 2SO3(g) can be in creased by

Analysis

The reaction given is: 2SO₂(g) + O₂(g) $\rightleftharpoons$ 2SO₃(g) $\Delta H = -395.7 \text{ kJ mol}^{-1}$

This is an exothermic reaction (indicated by the negative $\Delta H$ value). We are looking for the condition that will shift the equilibrium to the right (the product side) to increase the concentration of SO₃.

Concept: Le Chatelier’s Principle

Le Chatelier’s Principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract that change.

1. Temperature: For an exothermic reaction ($\Delta H < 0$), heat is treated as a product. According to Le Chatelier’s Principle, decreasing the temperature will cause the system to shift in the direction that produces heat (the forward reaction) to compensate for the loss. Therefore, lowering the temperature increases the yield of SO₃.
2. Pressure: The reaction has 3 moles of gas on the left (2 + 1) and 2 moles of gas on the right. Increasing the pressure would shift the equilibrium to the side with fewer moles (the right), increasing the yield of SO₃. Decreasing pressure would do the opposite.
3. Catalyst: A catalyst speeds up both the forward and backward reactions equally. It helps the system reach equilibrium faster, but it does not change the position of the equilibrium or the concentration of the products.

Step-by-Step Solution

• Step 1: Identify the nature of the reaction. The negative sign in $\Delta H$ confirms it is exothermic.
• Step 2: Evaluate the options based on Le Chatelier’s Principle. • Option A: Decreasing pressure favors the side with more moles (the reactants), so SO₃ concentration decreases. • Option B: Decreasing temperature favors the exothermic direction (the forward reaction), which produces more SO₃. • Option C: A catalyst has no effect on the equilibrium position. • Option D: Increasing temperature favors the endothermic direction (the backward reaction), which decreases the concentration of SO₃.

Conclusion

By decreasing the temperature, the system shifts to the right to produce more heat, thereby increasing the concentration of SO₃.

Exam Tip: Always remember the "Exo-Cold, Endo-Hot" rule. For exothermic reactions, lower temperatures favor the products. For endothermic reactions, higher temperatures favor the products.