---
title: "A materials scientist considers the conversion of diamond to graphite at standard conditions.  \\(\\text{C(s, diamond)} \\rightarrow \\text{C(s, graphite)}\\)  | Substance | \\(\\Delta_{\\mathrm f}G^\\circ\\) at \\(298\\text{ K}\\) | |—|—:| | \\(\\text{C(s, diamond)}\\) | \\(+2.9\\text{ kJ/mol}\\) | | \\(\\text{C(s, graphite)}\\) | \\(0.0\\text{ kJ/mol}\\) |  After several years at \\(298\\text{ K}\\), no measurable graphite is detected in a sample of diamond. Which statement best reconciles the thermodynamic data with the observation?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/119497/"
date_modified: "2026-08-19T12:40:47+00:00"
---

# A materials scientist considers the conversion of diamond to graphite at standard conditions.

\(\text{C(s, diamond)} \rightarrow \text{C(s, graphite)}\)

| Substance | \(\Delta_{\mathrm f}G^\circ\) at \(298\text{ K}\) |
|—|—:|
| \(\text{C(s, diamond)}\) | \(+2.9\text{ kJ/mol}\) |
| \(\text{C(s, graphite)}\) | \(0.0\text{ kJ/mol}\) |

After several years at \(298\text{ K}\), no measurable graphite is detected in a sample of diamond. Which statement best reconciles the thermodynamic data with the observation?

A materials scientist considers the conversion of diamond to graphite at standard conditions.

\(\text{C(s, diamond)} \rightarrow \text{C(s, graphite)}\)

| Substance | \(\Delta_{\mathrm f}G^\circ\) at \(298\text{ K}\) |
|---|---:|
| \(\text{C(s, diamond)}\) | \(+2.9\text{ kJ/mol}\) |
| \(\text{C(s, graphite)}\) | \(0.0\text{ kJ/mol}\) |

After several years at \(298\text{ K}\), no measurable graphite is detected in a sample of diamond. Which statement best reconciles the thermodynamic data with the observation?

- **A.** The conversion is thermodynamically favorable but effectively unobservable, because the product-favored equilibrium constant makes the forward reaction rate extremely small.
- **B.** The conversion is thermodynamically favorable but effectively unobservable, because graphite has a lower standard Gibbs free energy, whereas the extensive bond rearrangement requires crossing a large activation-energy barrier.
- **C.** The conversion is thermodynamically unfavorable and effectively unobservable, because the large activation energy raises \(\Delta G^\circ\) above zero.
- **D.** The conversion is thermodynamically unfavorable and effectively unobservable, because subtracting the product value from the reactant value gives \(\Delta G^\circ=+2.9\text{ kJ/mol}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119497/*