---
title: "A student investigates the kinetics of the reaction between a colored dye, \\(\\text{X(aq)}\\), and a colorless reactant, \\(\\text{Y(aq)}\\), represented by the equation below.  \\[ \\text{X(aq)} + \\text{Y(aq)} \\rightarrow \\text{Z(aq)} \\]  The rate law for the reaction is \\(\\text{Rate} = k[\\text{X}]^m[\\text{Y}]^n\\). The student uses a spectrophotometer set to the wavelength of maximum absorbance for \\(\\text{X}\\) to monitor \\([\\text{X}]\\) over time.  In a successful trial, the student mixes solutions such that \\([\\text{X}]_0 = 1.0 \\times 10^{-5} \\text{ M}\\) and \\([\\text{Y}]_0 = 0.50 \\text{ M}\\), allowing the order \\(m\\) to be determined from an integrated rate law plot of \\([\\text{X}]\\) versus time.  If the student repeats the experiment using \\([\\text{X}]_0 = 1.0 \\times 10^{-5} \\text{ M}\\) and \\([\\text{Y}]_0 = 2.0 \\times 10^{-5} \\text{ M}\\), which of the following best explains why an integrated rate law plot of \\([\\text{X}]\\) versus time can no longer be used to determine \\(m\\)?"
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url: "https://nerd-notes.com/ubq/123651/"
date_modified: "2026-09-28T12:01:54+00:00"
---

# A student investigates the kinetics of the reaction between a colored dye, \(\text{X(aq)}\), and a colorless reactant, \(\text{Y(aq)}\), represented by the equation below.

\[ \text{X(aq)} + \text{Y(aq)} \rightarrow \text{Z(aq)} \]

The rate law for the reaction is \(\text{Rate} = k[\text{X}]^m[\text{Y}]^n\). The student uses a spectrophotometer set to the wavelength of maximum absorbance for \(\text{X}\) to monitor \([\text{X}]\) over time.

In a successful trial, the student mixes solutions such that \([\text{X}]_0 = 1.0 \times 10^{-5} \text{ M}\) and \([\text{Y}]_0 = 0.50 \text{ M}\), allowing the order \(m\) to be determined from an integrated rate law plot of \([\text{X}]\) versus time.

If the student repeats the experiment using \([\text{X}]_0 = 1.0 \times 10^{-5} \text{ M}\) and \([\text{Y}]_0 = 2.0 \times 10^{-5} \text{ M}\), which of the following best explains why an integrated rate law plot of \([\text{X}]\) versus time can no longer be used to determine \(m\)?

A student investigates the kinetics of the reaction between a colored dye, \(\text{X(aq)}\), and a colorless reactant, \(\text{Y(aq)}\), represented by the equation below.

\[ \text{X(aq)} + \text{Y(aq)} \rightarrow \text{Z(aq)} \]

The rate law for the reaction is \(\text{Rate} = k[\text{X}]^m[\text{Y}]^n\). The student uses a spectrophotometer set to the wavelength of maximum absorbance for \(\text{X}\) to monitor \([\text{X}]\) over time.

In a successful trial, the student mixes solutions such that \([\text{X}]_0 = 1.0 \times 10^{-5} \text{ M}\) and \([\text{Y}]_0 = 0.50 \text{ M}\), allowing the order \(m\) to be determined from an integrated rate law plot of \([\text{X}]\) versus time.

If the student repeats the experiment using \([\text{X}]_0 = 1.0 \times 10^{-5} \text{ M}\) and \([\text{Y}]_0 = 2.0 \times 10^{-5} \text{ M}\), which of the following best explains why an integrated rate law plot of \([\text{X}]\) versus time can no longer be used to determine \(m\)?

- **A.** The molar absorptivity of \(\text{X}\) will change significantly during the reaction because \([\text{Y}]\) is no longer concentrated enough to maintain a constant solution environment.
- **B.** The rate constant \(k\) will decrease significantly as \(\text{Y}\) is consumed, preventing the reaction data from fitting any standard integrated rate law.
- **C.** The reaction will reach dynamic equilibrium almost instantaneously because the two reactants are present in comparable initial concentrations.
- **D.** The concentration of \(\text{Y}\) decreases by a substantial percentage as \(\text{X}\) reacts, so \([\text{Y}]^n\) is not constant and cannot be merged with \(k\) into a pseudo-rate constant.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123651/*