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title: "A student dissolves a sample of solid potassium nitrate, \\(\\text{KNO}_3(s)\\), in water at \\(298\\text{ K}\\) and observes that the solid dissolves completely while the temperature of the solution decreases. The lattice energy required to separate \\(\\text{KNO}_3(s)\\) into ions is greater in magnitude than the energy released by the hydration of \\(\\text{K}^+\\) and \\(\\text{NO}_3^-\\) ions. Which of the following statements best explains why the dissolution of \\(\\text{KNO}_3(s)\\) is thermodynamically favorable at \\(298\\text{ K}\\)?"
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url: "https://nerd-notes.com/ubq/120338/"
date_modified: "2026-08-23T04:23:13+00:00"
---

# A student dissolves a sample of solid potassium nitrate, \(\text{KNO}_3(s)\), in water at \(298\text{ K}\) and observes that the solid dissolves completely while the temperature of the solution decreases. The lattice energy required to separate \(\text{KNO}_3(s)\) into ions is greater in magnitude than the energy released by the hydration of \(\text{K}^+\) and \(\text{NO}_3^-\) ions. Which of the following statements best explains why the dissolution of \(\text{KNO}_3(s)\) is thermodynamically favorable at \(298\text{ K}\)?

A student dissolves a sample of solid potassium nitrate, \(\text{KNO}_3(s)\), in water at \(298\text{ K}\) and observes that the solid dissolves completely while the temperature of the solution decreases. The lattice energy required to separate \(\text{KNO}_3(s)\) into ions is greater in magnitude than the energy released by the hydration of \(\text{K}^+\) and \(\text{NO}_3^-\) ions. Which of the following statements best explains why the dissolution of \(\text{KNO}_3(s)\) is thermodynamically favorable at \(298\text{ K}\)?

- **A.** The dissolution is thermodynamically favorable because \(\Delta S^\circ_{\text{soln}} > 0\) due to the increased dispersal of matter as ions enter solution, and the \(T\Delta S^\circ_{\text{soln}}\) term is greater in magnitude than \(\Delta H^\circ_{\text{soln}}\), which makes \(\Delta G^\circ_{\text{soln}} < 0\).
- **B.** The dissolution is thermodynamically favorable because \(\Delta H^\circ_{\text{soln}} < 0\) due to ion-dipole attractions releasing more energy than the lattice consumes, which makes \(\Delta G^\circ_{\text{soln}} < 0\).
- **C.** The dissolution is not thermodynamically favorable because \(\Delta H^\circ_{\text{soln}} > 0\) due to lattice disruption requiring more energy than hydration releases, which makes \(\Delta G^\circ_{\text{soln}} > 0\).
- **D.** The dissolution is thermodynamically favorable because \(\Delta S^\circ_{\text{soln}} < 0\) due to water molecules forming structured hydration shells around the ions, which makes 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/120338/*