Chemistry

Problem

Explain whether heating a salt with

\Delta H^{\circ}_{\text{diss}} > 0

and

\Delta S^{\circ}_{\text{diss}} > 0

will increase, decrease, or have no effect on its spontaneity of dissolution. Justify in terms of the signs of

\Delta H^{\circ}

and

T\Delta S^{\circ}

Intuitive explanation

Spontaneity depends on the competition between enthalpy and entropy in the Gibbs free energy equation $\Delta G° = \Delta H° - T\Delta S°$ . When both $\Delta H°$ and $\Delta S°$ are positive, heating increases the $T\Delta S°$ term, which grows faster than the positive $\Delta H°$ term stays constant. Eventually $T\Delta S°$ dominates, making $\Delta G°$ more negative and the dissolution more spontaneous—heat tips the balance toward entropy's favor.

Intuitive explanation

Browse the problem library