---
title: "The reaction energy profile for the conversion of reactant \\(\\text{R}\\) into two different isomeric products, \\(\\text{P}_1\\) and \\(\\text{P}_2\\), along competing elementary pathways is shown in the diagram.  A sample of \\(\\text{R}\\) is reacted under Condition X (\\(200\\text{ K}\\) for a short reaction time, followed by rapid cooling). A separate sample of \\(\\text{R}\\) is reacted under Condition Y (\\(550\\text{ K}\\) for an extended period until chemical equilibrium is established).  Based on the energy profile, which of the following identifies the predominant product formed under each condition and provides the best justification?"
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url: "https://nerd-notes.com/ubq/120365/"
date_modified: "2026-08-23T04:23:28+00:00"
---

# The reaction energy profile for the conversion of reactant \(\text{R}\) into two different isomeric products, \(\text{P}_1\) and \(\text{P}_2\), along competing elementary pathways is shown in the diagram.

A sample of \(\text{R}\) is reacted under Condition X (\(200\text{ K}\) for a short reaction time, followed by rapid cooling). A separate sample of \(\text{R}\) is reacted under Condition Y (\(550\text{ K}\) for an extended period until chemical equilibrium is established).

Based on the energy profile, which of the following identifies the predominant product formed under each condition and provides the best justification?

The reaction energy profile for the conversion of reactant \(\text{R}\) into two different isomeric products, \(\text{P}_1\) and \(\text{P}_2\), along competing elementary pathways is shown in the diagram.

A sample of \(\text{R}\) is reacted under Condition X (\(200\text{ K}\) for a short reaction time, followed by rapid cooling). A separate sample of \(\text{R}\) is reacted under Condition Y (\(550\text{ K}\) for an extended period until chemical equilibrium is established).

Based on the energy profile, which of the following identifies the predominant product formed under each condition and provides the best justification?

![A reaction energy profile graph is displayed on bare vertical and horizontal axes without gridlines. The vertical axis is labeled Gibbs Free Energy, G. The horizontal axis is labeled Reaction Coordinate. A single horizontal plateau labeled R is located at an intermediate energy level on the left. Two curves originate from R and branch to the right. Pathway 1 is drawn as a solid black curve that rises to a transition state peak labeled TS1 and then descends to a product plateau labeled P1 located below the energy level of R. Pathway 2 is drawn as a dashed black curve that rises to a transition state peak labeled TS2, positioned significantly higher than TS1, and then descends to a product plateau labeled P2 located significantly lower in energy than both R and P1. A legend in the upper right corner indicates: solid line = Pathway 1, dashed line = Pathway 2. No other particles, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787459007-oKJl0p.jpg)

- **A.** Under Condition X, \(\text{P}_1\) predominates because Pathway 1 has a lower activation energy, allowing \(\text{P}_1\) to form more rapidly; under Condition Y, \(\text{P}_2\) predominates because \(\text{P}_2\) is at a lower free energy, making it the thermodynamically more stable product.
- **B.** Under Condition X, \(\text{P}_2\) predominates because Pathway 2 has a larger driving force with a more negative \(\Delta G^\circ\); under Condition Y, \(\text{P}_1\) predominates because higher temperatures always favor the pathway with the lower activation energy.
- **C.** Under Condition X, \(\text{P}_1\) predominates because \(\text{P}_1\) is the thermodynamically favored product; under Condition Y, \(\text{P}_2\) predominates because the higher temperature allows the system to overcome the higher activation energy barrier to reach kinetic control.
- **D.** Under Condition X, \(\text{P}_2\) predominates because its formation is irreversible at low temperatures; under Condition Y, \(\text{P}_1\) predominates because reaching equilibrium favors the pathway that proceeds through the lower-energy transition state \(\text{TS}_1\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120365/*