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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

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in A and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{s}\) has elapsed assuming \([\mathrm{B}]_{0}=0\)

Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{array}{l}{2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} A_{2}} \\ {2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2}}\end{array} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C},\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} M\) It was found that after each reaction had progressed for \(3.00 \mathrm{min},[\mathrm{A}]=3.00[\mathrm{B}]\) . In this case the rate laws are defined as $$ \begin{array}{l}{\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2}} \\ {\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2}}\end{array} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 \(\mathrm{min}\) . b. Calculate the value of \(k_{2}\) . c. Calculate the half-life for the experiment involving A.

The combustion of carbohydrates and the combustion of fats are both exothermic processes, yet the combustion of carbohydrates is a faster process. How can this be?

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.

A certain reaction has an activation energy of 54.0 \(\mathrm{kJ} / \mathrm{mol}\) . As the temperature is increased from \(22^{\circ} \mathrm{C}\) to a higher temperature, the rate constant increases by a factor of 7.00 . Calculate the higher temperature.
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