Q.17
Consider the following equilibrium:
SO2 (g) + ½ O2 (g) ⇌ SO3 (g), ΔH = −23.5 kcal mol−1
The formation of SO3 is favoured by
Options:
(A) compression and decreasing the temperature
(B) compression and increasing the temperature
(C) expansion and increasing the temperature
(D) expansion and decreasing the temperature
Formation of SO3 in Chemical Equilibrium
This question is based on chemical equilibrium and
Le Chatelier’s principle.
We analyze how changes in pressure and
temperature affect the formation of
sulfur trioxide (SO3).
Given Equilibrium Reaction
SO2 (g) + 1/2 O2 (g)
⇌
SO3 (g)
ΔH = −23.5 kcal mol−1
Key Observations
- The reaction is exothermic (ΔH < 0).
- Total gaseous moles decrease:
Reactants = 1 + 1/2 = 1.5 moles
Products = 1 mole
Application of Le Chatelier’s Principle
Effect of Pressure
Increasing pressure (compression) favors the side with
fewer moles of gas.
Hence, compression favors the formation of SO3.
Effect of Temperature
For an exothermic reaction, decreasing temperature shifts
equilibrium towards products.
Hence, lowering temperature favors SO3 formation.
Condition Favoring SO3 Formation
Compression and decreasing the temperature
Correct Answer
Option (A)
Summary
| Factor | Effect on SO3 Formation |
|---|---|
| Compression | Increases |
| Expansion | Decreases |
| Decreasing Temperature | Increases |
| Increasing Temperature | Decreases |
Conclusion
According to Le Chatelier’s principle, the formation of
SO3 is favored by increasing pressure and lowering
temperature.
Therefore, the correct answer is Option (A).
