---
title: "Consider the conversion of diamond to graphite represented by the following equation:  \\[ \\text{C}(\\text{s, diamond}) \\rightarrow \\text{C}(\\text{s, graphite}) \\quad \\Delta G^\\circ_{298} = -2.9\\text{ kJ/mol}_{\\text{rxn}} \\]  Under standard conditions at \\(298\\text{ K}\\) and \\(1\\text{ atm}\\), diamond does not visibly convert to graphite and persists indefinitely. Which of the following statements best explains this observation?"
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url: "https://nerd-notes.com/ubq/120324/"
date_modified: "2026-08-23T04:23:07+00:00"
---

# Consider the conversion of diamond to graphite represented by the following equation:

\[ \text{C}(\text{s, diamond}) \rightarrow \text{C}(\text{s, graphite}) \quad \Delta G^\circ_{298} = -2.9\text{ kJ/mol}_{\text{rxn}} \]

Under standard conditions at \(298\text{ K}\) and \(1\text{ atm}\), diamond does not visibly convert to graphite and persists indefinitely. Which of the following statements best explains this observation?

Consider the conversion of diamond to graphite represented by the following equation:

\[ \text{C}(\text{s, diamond}) \rightarrow \text{C}(\text{s, graphite}) \quad \Delta G^\circ_{298} = -2.9\text{ kJ/mol}_{\text{rxn}} \]

Under standard conditions at \(298\text{ K}\) and \(1\text{ atm}\), diamond does not visibly convert to graphite and persists indefinitely. Which of the following statements best explains this observation?

- **A.** The conversion is under thermodynamic control because the negative value of \(\Delta G^\circ\) indicates that diamond is more stable than graphite at \(298\text{ K}\).
- **B.** The conversion is thermodynamically unfavorable because a large amount of energy is required to break the strong \(\text{C}-\text{C}\) covalent bonds in diamond.
- **C.** The conversion is under kinetic control because the negative value of \(\Delta G^\circ\) causes the rate of the reverse reaction to be much faster than that of the forward reaction.
- **D.** The conversion is under kinetic control because the reaction has a very high activation energy resulting from the need to break strong \(\text{C}-\text{C}\) covalent bonds in the diamond network.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120324/*