---
title: "A sample of pure \\(\\text{N}_2\\text{O}_4\\text{(g)}\\) is placed into a rigid, evacuated container maintained at a constant temperature. The reversible reaction represented below occurs:  \\[\\text{N}_2\\text{O}_4\\text{(g)} \\rightleftharpoons 2\\,\\text{NO}_2\\text{(g)}\\]  The rates of the forward and reverse reactions are monitored over time and plotted in the graph. Which of the following correctly identifies the time at which dynamic equilibrium is first established in the container, along with the valid justification?"
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url: "https://nerd-notes.com/ubq/123843/"
date_modified: "2026-09-28T12:30:31+00:00"
---

# A sample of pure \(\text{N}_2\text{O}_4\text{(g)}\) is placed into a rigid, evacuated container maintained at a constant temperature. The reversible reaction represented below occurs:

\[\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}\]

The rates of the forward and reverse reactions are monitored over time and plotted in the graph. Which of the following correctly identifies the time at which dynamic equilibrium is first established in the container, along with the valid justification?

A sample of pure \(\text{N}_2\text{O}_4\text{(g)}\) is placed into a rigid, evacuated container maintained at a constant temperature. The reversible reaction represented below occurs:

\[\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\,\text{NO}_2\text{(g)}\]

The rates of the forward and reverse reactions are monitored over time and plotted in the graph. Which of the following correctly identifies the time at which dynamic equilibrium is first established in the container, along with the valid justification?

![A 2D line graph with a horizontal axis labeled Time and a vertical axis labeled Reaction Rate. The origin is at the bottom-left corner marked 0. The horizontal axis contains four evenly spaced tick marks labeled in order from left to right: \(t_1\), \(t_2\), \(t_3\), and \(t_4\). Bare axes with no gridlines. Two curves are plotted: A solid curve representing Forward Rate begins at a high value on the vertical axis at time 0, decreases smoothly with a decreasing downward slope until time \(t_3\), and becomes completely horizontal from \(t_3\) through \(t_4\). A dashed curve representing Reverse Rate begins at the origin at time 0, increases smoothly with a decreasing upward slope until time \(t_3\), where it merges with the solid curve at a non-zero height. From \(t_3\) through \(t_4\), the dashed curve and solid curve overlap as a single constant horizontal line. A legend at the top right indicates: solid line = Forward rate, dashed line = Reverse rate. No other lines, curves, points, labels, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790598631-vElmqH.jpg)

- **A.** At \(t_1\), because the forward reaction rate is at a high value while the reverse reaction has just begun.
- **B.** At \(t_2\), because the rate of the reverse reaction is increasing rapidly to counteract the forward reaction.
- **C.** At \(t_4\), because both the forward and reverse reaction rates have leveled off and the chemical reaction has completely stopped.
- **D.** At \(t_3\), because the forward reaction rate first becomes equal to the reverse reaction rate, and both processes continue at equal non-zero rates.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123843/*