---
title: "A sample of solid carbon dioxide sublimes at room temperature inside a sealed container according to the following process:  \\[ \\text{CO}_2\\text{(s)} \\rightarrow \\text{CO}_2\\text{(g)} \\]  Which of the following claims about the standard entropy change, \\(\\Delta S^\\circ\\), for this process is correct, and what is the particulate justification?"
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url: "https://nerd-notes.com/ubq/120326/"
date_modified: "2026-08-23T04:23:07+00:00"
---

# A sample of solid carbon dioxide sublimes at room temperature inside a sealed container according to the following process:

\[ \text{CO}_2\text{(s)} \rightarrow \text{CO}_2\text{(g)} \]

Which of the following claims about the standard entropy change, \(\Delta S^\circ\), for this process is correct, and what is the particulate justification?

A sample of solid carbon dioxide sublimes at room temperature inside a sealed container according to the following process:

\[ \text{CO}_2\text{(s)} \rightarrow \text{CO}_2\text{(g)} \]

Which of the following claims about the standard entropy change, \(\Delta S^\circ\), for this process is correct, and what is the particulate justification?

- **A.** \(\Delta S^\circ > 0\), because the molecules in the gas phase have greater freedom of motion and spatial dispersal, leading to a much larger number of accessible microstates than in the solid crystal.
- **B.** \(\Delta S^\circ > 0\), because covalent bonds within the \(\text{CO}_2\) molecules are broken during sublimation, resulting in a larger number of independent particles.
- **C.** \(\Delta S^\circ < 0\), because the gaseous molecules spread out across a larger volume, which lowers their density and reduces the number of accessible microstates.
- **D.** \(\Delta S^\circ < 0\), because sublimation is an endothermic process that absorbs thermal energy, which decreases the entropy of the system.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120326/*