Chemistry

Problem

For solid A,

\Delta H^{\circ}_{\text{diss}} = -10\,\text{kJ mol}^{-1}

and

\Delta S^{\circ}_{\text{diss}} = -40\,\text{J mol}^{-1}\,\text{K}^{-1}

. For solid B,

\Delta H^{\circ}_{\text{diss}} = +10\,\text{kJ mol}^{-1}

and

\Delta S^{\circ}_{\text{diss}} = +40\,\text{J mol}^{-1}\,\text{K}^{-1}

. Both solids are sparingly soluble at

0^{\circ}\text{C}

. When the temperature is raised to

60^{\circ}\text{C}

, which solid is expected to show the larger increase in solubility? Briefly explain your choice.

Intuitive explanation

Solubility is governed by the Gibbs free energy of dissolution: $\Delta G° = \Delta H° - T\Delta S°$ . As temperature increases, the $-T\Delta S°$ term grows in magnitude. For solid A, the negative entropy change works against dissolution at all temperatures, making it less soluble when heated. For solid B, the positive entropy change becomes increasingly favorable at higher temperatures, driving dissolution forward. The key insight is that entropy-driven dissolution (solid B) improves dramatically with heat, while enthalpy-driven dissolution (solid A) worsens—so B shows the larger solubility increase.

Intuitive explanation

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