---
title: "A student heats a sample of pure calcium carbonate, \\(\\text{CaCO}_3(s)\\), in an open crucible, causing it to decompose according to the balanced equation below.  \\[ \\text{CaCO}_3(s) \\rightarrow \\text{CaO}(s) + \\text{CO}_2(g) \\]  Which of the following correctly predicts and explains the sign of the standard entropy change, \\(\\Delta S^\\circ\\), for this reaction?"
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url: "https://nerd-notes.com/ubq/119731/"
date_modified: "2026-08-21T08:12:10+00:00"
---

# A student heats a sample of pure calcium carbonate, \(\text{CaCO}_3(s)\), in an open crucible, causing it to decompose according to the balanced equation below.

\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]

Which of the following correctly predicts and explains the sign of the standard entropy change, \(\Delta S^\circ\), for this reaction?

A student heats a sample of pure calcium carbonate, \(\text{CaCO}_3(s)\), in an open crucible, causing it to decompose according to the balanced equation below.

\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]

Which of the following correctly predicts and explains the sign of the standard entropy change, \(\Delta S^\circ\), for this reaction?

- **A.** \(\Delta S^\circ < 0\), because one reactant decomposes into multiple product species.
- **B.** \(\Delta S^\circ < 0\), because gas particles occupy a larger volume with fewer accessible microstates than an ordered crystal lattice.
- **C.** \(\Delta S^\circ > 0\), because a gas is produced from a solid reactant, resulting in greater dispersal of matter and more accessible microstates.
- **D.** \(\Delta S^\circ > 0\), because the total mass of the products is greater than the initial mass of the reactant.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119731/*