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Chapter 12: Problem 16

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

Short Answer

Expert verified
The choice between using a differential or integrated rate law depends on the ease of data collection during experiments. Differential rate laws are more suitable when there is a direct method to measure the reactant concentrations over time, such as spectrophotometry or conductivity measurements. Integrated rate laws are convenient when data points of concentration change over time can be easily obtained for a known reaction order. The practical aspects of data collection determine which rate law is most efficient and useful for a given reaction.

Step by step solution

01

Introduction to Differential and Integrated Rate Laws

Differential rate laws show the relationship between the rate of a reaction and the concentrations of reactants. Integrated rate laws relate the concentration of a reactant at the initial time (t=0) to the concentrations at any time (t) during the reaction. Both differential and integrated rate laws can be used to describe the reaction rates. However, the choice of the rate law depends on the ease of collecting data during experiments.
02

Differential Rate Laws and Data Collection

Differential rate laws are expressed as a rate equation, such as: Rate \(= k[A]^m[B]^n\) Where Rate is the rate of the reaction, k is the rate constant, [A] and [B] are the concentrations of reactants A and B, and m and n are the order of the reaction with respect to each reactant. To determine the values of m and n, experimental data is required, which usually involves monitoring the change in the concentrations of reactants over time. Collecting this data can be easier if there are direct methods to measure the concentrations, such as spectrophotometry for colored solutions or conductivity measurements for ionic species.
03

Integrated Rate Laws and Data Collection

Integrated rate laws are derived from the differential rate equations by integrating them with respect to time. They can be represented as: For zero-order reactions: \([A]_t = [A]_0 - kt\) For first-order reactions: ln\([A]_t = ln[A]_0 - kt\) For second-order reactions: \(\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\) Where \([A]_t\) is the concentration of reactant A at time t, \([A]_0\) is the initial concentration of reactant A, and k is the rate constant. To analyze data using integrated rate laws, it is usually more convenient to collect data points of concentration or concentration change over time. The method depends on the reaction order, which should be determined through experimental observation.
04

Conclusion

In summary, the choice of using a differential or integrated rate law mainly depends on the practical aspects of data collection during experiments. Differential rate laws are more suitable when there is a direct method to measure the reactant concentrations over time, while integrated rate laws are convenient when data points of concentration change over time can be easily obtained for a known reaction order.

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Most popular questions from this chapter

A certain substance, initially at 0.10\(M\) in solution, decomposes by second- order kinetics. If the rate constant for this process is 0.40 $\mathrm{L} / \mathrm{mol} \cdot \min$ , how much time is required for the concentration to reach 0.020 \(\mathrm{M}\) ?

Consider the following initial rate data for the decomposition of compound AB to give A and B: Determine the half-life for the decomposition reaction initially having 1.00$M \mathrm{AB}$ present.

Consider a reaction of the type aA \(\longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 .\) s, what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text {Rate} =\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3} $$

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$
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