KEY2CHEM

Thermodynamics of Dissolution

Dissolving a salt requires consideration of changes in both enthalpy ($$\Delta H$$) and entropy ($$\Delta S$$). Changes in enthalpy are associated with energy that must be absorbed to disrupt ionic bonds holding the salt together, as well as energy released when new ion-dipole attractions  form as the ions are solvated by water molecules. Changes in entropy occur with increased freedom of motion the particles have after they are now dissolved. The relative magnitudes of both $$\Delta H$$ and $$\Delta S$$ are important in determining how soluble a salt will be.

Example 1.

Sodium chloride ($$\require{mhchem}\ce{NaCl}$$) is a very soluble ionic compound ($$K_{sp} = 36$$). However, the $$\Delta H_{dissolving}$$ is endothermic for $$\require{mhchem}\ce{NaCl}$$. Which statement about dissolving sodium chloride is true?

A. $$\Delta S_{dissolving} > 0$$

B. $$\Delta H_{dissolving} < 0$$

C. $$\Delta S_{dissolving} < 0$$

Solution

A. $$\Delta S_{dissolving} > 0$$

Since dissolving sodium chloride is thermodynamically favored and $$\Delta H_{dissolving} > 0$$ (endothermic), the $$\Delta S_{dissolving} > 0$$ to compensate for thermodynamically unfavorable enthalpy term.

Example 2.

Which diagram best represents the $$\Delta S$$ of the dissolving of $$\require{mhchem}\ce{NH4NO3}$$ in water? The filled circles represent the ionic compound and the open circles represent water molecules.

A. B. C. Solution

A. Dissolving an ionic solid has $$\Delta S > 0$$, since there are more options for how the ionic particles and water molecules can arrange themselves when mixed.

Example 3.

When solid sodium hydroxide ($$\require{mhchem}\ce{NaOH}$$) is dissolved in water, the resulting solution feels warm. What are the signs of $$\Delta H$$ and $$\Delta S$$ for the dissolving?

A.$$\Delta H > 0, \Delta S > 0$$

B. $$\Delta H < 0, \Delta S < 0$$

C. $$\Delta H < 0, \Delta S > 0$$

Solution

C. $$\Delta H < 0, \Delta S > 0$$

Since the surrounding solution feels warm, $$\Delta H < 0$$ for the system (releases heat to the surroundings). $$\Delta S > 0$$, since the dissolved ions have more freedom of motion compared to being in the isolated solid.