---
title: "The dissolution of a solid ionic compound, \\(\\text{MX}(s)\\), in water is represented by the equation below.  \\[ \\text{MX}(s) \\rightleftharpoons \\text{M}^+(aq) + \\text{X}^-(aq) \\]  During the dissolution, the energy required to separate the ions from the crystal lattice exceeds the energy released by the hydration of the ions, resulting in an endothermic process (\\(\\Delta H^\\circ_{\\text{soln}} > 0\\)). If \\(\\text{MX}(s)\\) dissolves spontaneously in water at \\(298\\text{ K}\\), which of the following statements best explains why the dissolution is thermodynamically favorable?"
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url: "https://nerd-notes.com/ubq/119914/"
date_modified: "2026-08-21T08:17:22+00:00"
---

# The dissolution of a solid ionic compound, \(\text{MX}(s)\), in water is represented by the equation below.

\[ \text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq) \]

During the dissolution, the energy required to separate the ions from the crystal lattice exceeds the energy released by the hydration of the ions, resulting in an endothermic process (\(\Delta H^\circ_{\text{soln}} > 0\)). If \(\text{MX}(s)\) dissolves spontaneously in water at \(298\text{ K}\), which of the following statements best explains why the dissolution is thermodynamically favorable?

The dissolution of a solid ionic compound, \(\text{MX}(s)\), in water is represented by the equation below.

\[ \text{MX}(s) \rightleftharpoons \text{M}^+(aq) + \text{X}^-(aq) \]

During the dissolution, the energy required to separate the ions from the crystal lattice exceeds the energy released by the hydration of the ions, resulting in an endothermic process (\(\Delta H^\circ_{\text{soln}} > 0\)). If \(\text{MX}(s)\) dissolves spontaneously in water at \(298\text{ K}\), which of the following statements best explains why the dissolution is thermodynamically favorable?

- **A.** The dissolution is thermodynamically favorable because the increase in entropy (\(\Delta S^\circ_{\text{soln}} > 0\)) from the dispersal of ions into the solution yields a \(-T\Delta S^\circ_{\text{soln}}\) term that is larger in magnitude than \(\Delta H^\circ_{\text{soln}}\), resulting in \(\Delta G^\circ_{\text{soln}} < 0\).
- **B.** The dissolution is thermodynamically favorable because the exothermic hydration of the ions releases more energy than the lattice energy of the solid, resulting in \(\Delta H^\circ_{\text{soln}} < 0\) and \(\Delta G^\circ_{\text{soln}} < 0\).
- **C.** The dissolution is thermodynamically favorable because the breaking of ionic bonds in the crystal lattice releases energy to the surroundings, driving the process despite a negative entropy change (\(\Delta S^\circ_{\text{soln}} < 0\)).
- **D.** The dissolution is thermodynamically favorable because the strong ion-dipole attractions between the ions and water decrease the total entropy of the system (\(\Delta S^\circ_{\text{soln}} < 0\)), making the \(-T\Delta S^\circ_{\text{soln}}\) term negative and \(\Delta G^\circ_{\text{soln}} < 0\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119914/*