The solubility of the solids that dissolves in a given solvent with the liberation of heat will

Analysis

This question tests your understanding of Le Chatelier’s Principle as it applies to the solubility of solids in liquids. Specifically, it focuses on how temperature changes affect a system at equilibrium when the dissolution process is exothermic (liberates heat).

Concept Explanation

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

The dissolution process can be represented as: Solid + Solvent ⇌ Solution + Heat

• Exothermic process: Heat is released (liberated) during dissolution.
• Increasing the temperature: You are adding more heat to the system. To counteract this, the system will shift the equilibrium to the left (the side that absorbs heat) to consume the excess heat.
• Result: Shifting to the left means the solid remains undissolved, effectively decreasing the solubility.

Conversely, if the process were endothermic (absorbing heat), increasing the temperature would shift the equilibrium to the right, increasing solubility.

Step-by-Step Solution

1. Identify the process: The question states the solid dissolves with the "liberation of heat." This means the reaction is exothermic ($\Delta H < 0$).
2. Apply Le Chatelier’s Principle: Increasing the temperature adds heat to the system.
3. Predict the shift: The system tries to remove the added heat by favoring the reverse reaction (the endothermic direction).
4. Conclusion: Because the reverse reaction is favored, the amount of solid that can remain dissolved decreases as the temperature rises.

Identifying the Answer

Based on the principle explained above, increasing the temperature causes the solubility of a substance that dissolves exothermically to decrease.

Exam Tip: Remember the "Heat Rule":

• Exothermic (Heat out): Increase Temp = Solubility Decreases.
• Endothermic (Heat in): Increase Temp = Solubility Increases. Think of heat as a "product" in exothermic reactions; adding more product pushes the reaction backward!