---
title: "The synthesis of ammonia is carried out in a closed, variable-volume container at a constant temperature:  \\[ \\text{N}_2\\text{(g)} + 3\\,\\text{H}_2\\text{(g)} \\rightleftharpoons 2\\,\\text{NH}_3\\text{(g)} \\quad \\Delta H^\\circ < 0 \\]  The graph below displays the forward reaction rate (solid curve) and the reverse reaction rate (dashed curve) as a function of time. At time \\(t_1\\), the volume of the reaction container is rapidly halved at constant temperature, and the system is allowed to reach a new equilibrium at time \\(t_2\\).  Which of the following best explains the immediate effect observed at \\(t_1\\) and the subsequent changes that occur until \\(t_2\\)?"
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url: "https://nerd-notes.com/ubq/120300/"
date_modified: "2026-08-23T04:04:25+00:00"
---

# The synthesis of ammonia is carried out in a closed, variable-volume container at a constant temperature:

\[ \text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} \quad \Delta H^\circ < 0 \]

The graph below displays the forward reaction rate (solid curve) and the reverse reaction rate (dashed curve) as a function of time. At time \(t_1\), the volume of the reaction container is rapidly halved at constant temperature, and the system is allowed to reach a new equilibrium at time \(t_2\).

Which of the following best explains the immediate effect observed at \(t_1\) and the subsequent changes that occur until \(t_2\)?

The synthesis of ammonia is carried out in a closed, variable-volume container at a constant temperature:

\[ \text{N}_2\text{(g)} + 3\,\text{H}_2\text{(g)} \rightleftharpoons 2\,\text{NH}_3\text{(g)} \quad \Delta H^\circ < 0 \]

The graph below displays the forward reaction rate (solid curve) and the reverse reaction rate (dashed curve) as a function of time. At time \(t_1\), the volume of the reaction container is rapidly halved at constant temperature, and the system is allowed to reach a new equilibrium at time \(t_2\).

Which of the following best explains the immediate effect observed at \(t_1\) and the subsequent changes that occur until \(t_2\)?

![A 2D Cartesian line graph showing reaction rate versus time in grayscale. The horizontal axis is labeled 'Time' with two vertical dashed reference lines labeled \(t_1\) and \(t_2\). The vertical axis is labeled 'Reaction rate'. A solid line represents the forward rate, and a dashed line represents the reverse rate. Prior to \(t_1\), both lines are horizontal and superimposed at the same constant baseline value. At time \(t_1\), both curves jump vertically upward instantaneously: the solid line jumps to a significantly higher rate than the dashed line. Between \(t_1\) and \(t_2\), the solid curve slopes downward smoothly while the dashed curve slopes upward smoothly. At time \(t_2\) and beyond, the two curves merge into a single horizontal superimposed line at a constant rate higher than the initial baseline before \(t_1\). A legend in the upper right indicates: solid line = forward rate, dashed line = reverse rate. No other curves, gridlines, labels, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787457864-yBuMvY.jpg)

- **A.** At \(t_1\), both rates increase because all species concentrations increase, with the forward rate increasing more due to a higher collision frequency among reactant molecules; between \(t_1\) and \(t_2\), the forward rate decreases and the reverse rate increases as \(Q < K\).
- **B.** At \(t_1\), only the forward rate increases because the system shifts toward the side with fewer moles of gas; between \(t_1\) and \(t_2\), the reverse rate increases while the forward rate remains constant as \(Q < K\).
- **C.** At \(t_1\), both rates increase equally because the equilibrium constant \(K\) is unchanged at constant temperature; between \(t_1\) and \(t_2\), the forward rate increases further because \(Q > K\).
- **D.** At \(t_1\), the forward rate increases and the reverse rate decreases to relieve the pressure stress; between \(t_1\) and \(t_2\), the forward rate decreases until both rates return to the initial baseline rate observed before \(t_1\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120300/*