Le Chatelier analysis of solubility: A solute dissolves with an increase in solution volume and liberation of heat (exothermic dissolution). Under which conditions is its solubility favored?

Difficulty: Easy

Correct Answer: Low pressure and low temperature

Explanation:


Introduction / Context:
Le Chatelier's principle predicts how equilibrium responds to external changes. For dissolution equilibria, the signs of enthalpy change and volume change determine how pressure and temperature affect solubility.


Given Data / Assumptions:

  • Dissolution is exothermic (releases heat).
  • Dissolution increases total volume (positive ΔV).
  • We consider qualitative trends at equilibrium.


Concept / Approach:
For exothermic processes, lowering temperature favors products (greater solubility). If dissolution increases volume, reducing external pressure favors the side with larger volume, also increasing solubility. Combine these to choose the condition that simultaneously lowers T and P.


Step-by-Step Solution:
Temperature effect: exothermic → lower T increases solubility.Pressure effect: ΔV > 0 → lower P increases solubility.Best combination: low pressure and low temperature.Therefore, select option (b).


Verification / Alternative check:
Qualitative thermodynamic derivatives (∂ln K/∂T) ∝ −ΔH/(R T^2) and (∂ln K/∂P) ∝ −ΔV/(R T) support the same trends for ΔH < 0 and ΔV > 0.


Why Other Options Are Wrong:
High temperature opposes exothermic dissolution; high pressure opposes processes with positive ΔV.


Common Pitfalls:

  • Assuming all solubilities increase with temperature; not true for exothermic cases.
  • Neglecting pressure effects for liquids; though often small, the sign logic still applies.


Final Answer:
Low pressure and low temperature

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