AP Chemistry
5.3 Concentration Changes Over Time
5.2 Introduction to Rate Law
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 \(m\)?
\[ \text{X(aq)} + \text{Y(aq)} \rightarrow \text{Z(aq)} \]
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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