---
title: "A student dissolves a sample of solid ammonium chloride, \\(\\text{NH}_4\\text{Cl}(s)\\), in distilled water at \\(298\\text{ K}\\), as represented by the equation below.  \\[ \\text{NH}_4\\text{Cl}(s) \\rightarrow \\text{NH}_4^+(aq) + \\text{Cl}^-(aq) \\]  Which of the following correctly predicts and justifies the sign of \\(\\Delta S^\\circ\\) for this dissolution process?"
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url: "https://nerd-notes.com/ubq/119984/"
date_modified: "2026-08-21T08:31:37+00:00"
---

# A student dissolves a sample of solid ammonium chloride, \(\text{NH}_4\text{Cl}(s)\), in distilled water at \(298\text{ K}\), as represented by the equation below.

\[ \text{NH}_4\text{Cl}(s) \rightarrow \text{NH}_4^+(aq) + \text{Cl}^-(aq) \]

Which of the following correctly predicts and justifies the sign of \(\Delta S^\circ\) for this dissolution process?

A student dissolves a sample of solid ammonium chloride, \(\text{NH}_4\text{Cl}(s)\), in distilled water at \(298\text{ K}\), as represented by the equation below.

\[ \text{NH}_4\text{Cl}(s) \rightarrow \text{NH}_4^+(aq) + \text{Cl}^-(aq) \]

Which of the following correctly predicts and justifies the sign of \(\Delta S^\circ\) for this dissolution process?

- **A.** \(\Delta S^\circ > 0\), because the ions held in the rigid crystalline lattice become dispersed throughout the solution, increasing the number of accessible microstates.
- **B.** \(\Delta S^\circ > 0\), because thermal energy is absorbed from the surroundings to overcome the electrostatic attractions within the lattice.
- **C.** \(\Delta S^\circ < 0\), because water molecules form organized hydration shells around the dissolved ions, decreasing the disorder of the system.
- **D.** \(\Delta S^\circ < 0\), because the two product ions occupy a smaller total volume than the original solid crystal.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119984/*