---
title: "A student investigates the kinetics of the decomposition of \\(\\text{XY(g)}\\) at \\(300\\text{ K}\\), as represented by the equation below.  \\[ 2\\,\\text{XY(g)} \\rightarrow \\text{X}_2\\text{(g)} + \\text{Y}_2\\text{(g)} \\]  The student measures the concentration of \\(\\text{XY(g)}\\) at regular time intervals and creates the graph shown.  Based on the graph, which of the following is the rate law for the reaction?"
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url: "https://nerd-notes.com/ubq/119945/"
date_modified: "2026-08-21T08:18:40+00:00"
---

# A student investigates the kinetics of the decomposition of \(\text{XY(g)}\) at \(300\text{ K}\), as represented by the equation below.

\[ 2\,\text{XY(g)} \rightarrow \text{X}_2\text{(g)} + \text{Y}_2\text{(g)} \]

The student measures the concentration of \(\text{XY(g)}\) at regular time intervals and creates the graph shown.

Based on the graph, which of the following is the rate law for the reaction?

A student investigates the kinetics of the decomposition of \(\text{XY(g)}\) at \(300\text{ K}\), as represented by the equation below.

\[ 2\,\text{XY(g)} \rightarrow \text{X}_2\text{(g)} + \text{Y}_2\text{(g)} \]

The student measures the concentration of \(\text{XY(g)}\) at regular time intervals and creates the graph shown.

Based on the graph, which of the following is the rate law for the reaction?

![A single graph on a set of coordinate axes. The vertical y-axis is labeled '\(\ln[\text{XY}]\)' and the horizontal x-axis is labeled '\(\text{Time (s)}\)'. The origin is at the bottom left. No numerical tick marks or gridlines are present on either axis. A single solid straight line begins near the top of the vertical axis and extends downward with a constant negative slope toward the lower-right area of the graph. No data points, curved lines, dashed lines, legends, or other annotations appear on the graph.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787300319-ghvWSf.jpg)

- **A.** \(\text{Rate} = k\), because \([\text{XY}]\) decreases at a constant rate over time.
- **B.** \(\text{Rate} = k[\text{XY}]\), because the plot of \(\ln[\text{XY}]\) versus time is linear.
- **C.** \(\text{Rate} = k[\text{XY}]^2\), because the plot of \(\ln[\text{XY}]\) versus time has a negative slope.
- **D.** \(\text{Rate} = k[\text{XY}]^2\), because the stoichiometric coefficient of \(\text{XY}\) in the balanced chemical equation is \(2\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119945/*