---
title: "The catalytic decomposition of hydrogen peroxide, \\(\\text{H}_2\\text{O}_2\\text{(aq)}\\), is represented by the following balanced equation:  \\[ 2\\text{ H}_2\\text{O}_2\\text{(aq)} \\rightarrow 2\\text{ H}_2\\text{O(l)} + \\text{O}_2\\text{(g)} \\]  For this reaction, the standard enthalpy change is negative (\\(\\Delta H^\\circ  0\\)). Which of the following statements best predicts the temperature conditions under which the reaction is thermodynamically favored, and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/120347/"
date_modified: "2026-08-23T04:23:15+00:00"
---

# The catalytic decomposition of hydrogen peroxide, \(\text{H}_2\text{O}_2\text{(aq)}\), is represented by the following balanced equation:

\[ 2\text{ H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{ H}_2\text{O(l)} + \text{O}_2\text{(g)} \]

For this reaction, the standard enthalpy change is negative (\(\Delta H^\circ  0\)). Which of the following statements best predicts the temperature conditions under which the reaction is thermodynamically favored, and provides the correct justification?

The catalytic decomposition of hydrogen peroxide, \(\text{H}_2\text{O}_2\text{(aq)}\), is represented by the following balanced equation:

\[ 2\text{ H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{ H}_2\text{O(l)} + \text{O}_2\text{(g)} \]

For this reaction, the standard enthalpy change is negative (\(\Delta H^\circ < 0\)) and the standard entropy change is positive (\(\Delta S^\circ > 0\)). Which of the following statements best predicts the temperature conditions under which the reaction is thermodynamically favored, and provides the correct justification?

- **A.** The reaction is thermodynamically favored at all temperatures because \(\Delta H^\circ < 0\) and the \(-T\Delta S^\circ\) term is negative at all absolute temperatures, ensuring \(\Delta G^\circ < 0\).
- **B.** The reaction is thermodynamically favored at all temperatures because the negative \(\Delta H^\circ\) value is always larger in magnitude than the positive \(T\Delta S^\circ\) value regardless of temperature.
- **C.** The reaction is thermodynamically favored only at low temperatures because a small value of \(T\) is required to prevent the \(-T\Delta S^\circ\) term from making \(\Delta G^\circ\) positive.
- **D.** The reaction is thermodynamically favored only at high temperatures because a large value of \(T\) is required for the \(-T\Delta S^\circ\) term to overcome an unfavorable enthalpy change.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120347/*