---
title: "The kinetics of the gas-phase reaction represented by the equation below was investigated at two different temperatures, \\(T_1\\) and \\(T_2\\).  \\[ \\text{NO}_2(g) + \\text{CO}(g) \\rightarrow \\text{NO}(g) + \\text{CO}_2(g) \\]  The graph shows the distribution of collision energies for the reacting molecules at both temperatures, along with the activation energy, \\(E_a\\), for the reaction. Which of the following correctly compares \\(T_1\\) and \\(T_2\\) and justifies why the reaction rate is greater at the higher temperature?"
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url: "https://nerd-notes.com/ubq/123671/"
date_modified: "2026-09-28T12:01:57+00:00"
---

# The kinetics of the gas-phase reaction represented by the equation below was investigated at two different temperatures, \(T_1\) and \(T_2\).

\[ \text{NO}_2(g) + \text{CO}(g) \rightarrow \text{NO}(g) + \text{CO}_2(g) \]

The graph shows the distribution of collision energies for the reacting molecules at both temperatures, along with the activation energy, \(E_a\), for the reaction. Which of the following correctly compares \(T_1\) and \(T_2\) and justifies why the reaction rate is greater at the higher temperature?

The kinetics of the gas-phase reaction represented by the equation below was investigated at two different temperatures, \(T_1\) and \(T_2\).

\[ \text{NO}_2(g) + \text{CO}(g) \rightarrow \text{NO}(g) + \text{CO}_2(g) \]

The graph shows the distribution of collision energies for the reacting molecules at both temperatures, along with the activation energy, \(E_a\), for the reaction. Which of the following correctly compares \(T_1\) and \(T_2\) and justifies why the reaction rate is greater at the higher temperature?

![A 2D line graph with a horizontal axis labeled 'Kinetic Energy' and a vertical axis labeled 'Fraction of Collisions'. The origin is at the bottom left with no numerical tick marks or gridlines. Two smooth distribution curves originate near the origin. Curve 1 is drawn as a solid line labeled '\(T_1\)' with a relatively tall, narrow peak located toward the left side of the graph. Curve 2 is drawn as a dashed line labeled '\(T_2\)' with a lower, broader peak shifted toward higher kinetic energy to the right. Both curves decay toward the horizontal axis at high kinetic energy, with the dashed curve remaining above the solid curve at high energies. A vertical dotted line is positioned in the high-energy tail region to the right of both peaks and is labeled '\(E_a\)' at the top. The area under the dashed curve to the right of '\(E_a\)' is visibly larger than the area under the solid curve to the right of '\(E_a\)'. No other lines, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790596916-Eis3LT.jpg)

- **A.** \(T_1 > T_2\), because the taller peak at \(T_1\) indicates that a greater total number of reactant molecules are present in the sample.
- **B.** \(T_2 > T_1\), because a greater fraction of molecular collisions have kinetic energy greater than or equal to \(E_a\) at \(T_2\).
- **C.** \(T_2 > T_1\), because the value of \(E_a\) decreases as the temperature increases, allowing more collisions to be effective.
- **D.** \(T_1 > T_2\), because the peak of the distribution at \(T_1\) occurs at a lower kinetic energy, leading to more frequent collisions.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123671/*