---
title: "A student studies the reversible gas-phase isomerization  \\[ \\text{X(g)} \\rightleftharpoons \\text{Y(g)} \\]  The proposed pathway contains \\(2\\) elementary steps, and measured activation-energy differences are shown in the potential-energy profile. Based on the graph, which value and reasoning correctly give \\(\\Delta H_{\\text{forward}}\\)?"
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url: "https://nerd-notes.com/ubq/120051/"
date_modified: "2026-08-21T08:40:58+00:00"
---

# A student studies the reversible gas-phase isomerization

\[
\text{X(g)} \rightleftharpoons \text{Y(g)}
\]

The proposed pathway contains \(2\) elementary steps, and measured activation-energy differences are shown in the potential-energy profile. Based on the graph, which value and reasoning correctly give \(\Delta H_{\text{forward}}\)?

A student studies the reversible gas-phase isomerization

\[
\text{X(g)} \rightleftharpoons \text{Y(g)}
\]

The proposed pathway contains \(2\) elementary steps, and measured activation-energy differences are shown in the potential-energy profile. Based on the graph, which value and reasoning correctly give \(\Delta H_{\text{forward}}\)?

![Draw a grayscale potential-energy graph on bare axes with no gridlines. Label the vertical axis “Potential energy (\(\text{kJ/mol}\))” and the horizontal axis “Reaction coordinate.” A single solid black curve begins at a horizontal reactant plateau labeled \(\text{X(g)}\), rises to the first and highest peak labeled \(\text{TS}_1\), falls to a local minimum labeled \(\text{I}\), rises to a lower second peak labeled \(\text{TS}_2\), and ends at a horizontal product plateau labeled \(\text{Y(g)}\) below the reactant level. Add exactly \(4\) vertical double-headed arrows: reactant level to first-peak height, \(75\text{ kJ/mol}\); product level to first-peak height, \(105\text{ kJ/mol}\); intermediate level to second-peak height, \(40\text{ kJ/mol}\); product level to second-peak height, \(90\text{ kJ/mol}\). Use short dashed horizontal guides only at arrow endpoints. Place arrows outside the curve. No legend or other text appears.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787301657-bcznKV.jpg)

- **A.** \(-50\text{ kJ/mol}\), because the barriers around \(\text{TS}_2\) are \(40\text{ kJ/mol}\) and \(90\text{ kJ/mol}\), so \(\Delta H=40-90\).
- **B.** \(-30\text{ kJ/mol}\), because the overall forward and reverse activation energies are \(75\text{ kJ/mol}\) and \(105\text{ kJ/mol}\), respectively, and \(\Delta H=E_{a,\text{forward}}-E_{a,\text{reverse}}\).
- **C.** \(+30\text{ kJ/mol}\), because \(\Delta H=E_{a,\text{reverse}}-E_{a,\text{forward}}\).
- **D.** \(+180\text{ kJ/mol}\), because \(\Delta H\) is the sum of the overall forward and reverse activation energies.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120051/*