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Chapter 14: Problem 17

Consider the reaction: $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From the following data obtained a a certain temperature, determine the order of the reaction and calculate the rate constant. $$ \begin{array}{ccc} {[\mathbf{A}](\boldsymbol{M})} & {[\mathbf{B}](\boldsymbol{M})} & \text { Rate }(\boldsymbol{M} / \mathbf{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$

Short Answer

Expert verified
The reaction is first order with respect to A and zeroth order with respect to B; rate constant \( k = 2.13 \times 10^{-1} \text{ s}^{-1} \).

Step by step solution

01

Understand the Reaction and Initial Data

The reaction given is \( A + B \rightarrow ext{products} \). The task is to determine the order of the reaction with respect to both \( A \) and \( B \) using the provided experimental data, and then calculate the rate constant of the reaction.
02

Analyze Experiment 1 and 2 - Determine Order with Respect to B

Compare experiments 1 and 2, where \([A]\) is constant at 1.50 M. The rate does not change despite increasing \([B]\) from 1.50 M to 2.50 M. This indicates that the rate does not depend on \([B]\), meaning the reaction is zeroth order with respect to \( B \).
03

Analyze Experiment 1 and 3 - Determine Order with Respect to A

Compare experiments 1 and 3, where \([B]\) is constant at 1.50 M. Doubling \([A]\) from 1.50 M to 3.00 M doubles the rate from \(3.20 \times 10^{-1} \text{ M/s}\) to \(6.40 \times 10^{-1} \text{ M/s}\). This indicates a direct proportionality, meaning the reaction is first order with respect to \( A \).
04

Write the Rate Law

Since the reaction is zeroth order with respect to \( B \) and first order with respect to \( A \), the rate law is: \[ ext{Rate} = k[A]^1[B]^0 = k[A] \]
05

Calculate the Rate Constant (k)

Using the rate law from Step 4 and any of the provided data points: - Choosing experiment 1 (\([A] = 1.50 \text{ M}, \text{Rate} = 3.20 \times 10^{-1} \text{ M/s}\)):\[ 3.20 \times 10^{-1} = k(1.50) \]Solve for \( k \): \[ k = \frac{3.20 \times 10^{-1}}{1.50} = 2.13 \times 10^{-1} \text{ s}^{-1} \]
06

Conclusion

The reaction is first order with respect to \( A \) and zeroth order with respect to \( B \). The rate constant is calculated to be \( 2.13 \times 10^{-1} \text{ s}^{-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Constant
The rate constant, denoted as \( k \), is a proportionality factor in the rate law equation that links the rate of a chemical reaction to the concentrations of reactants. It is unique for each reaction and depends on the reaction conditions, particularly temperature. The units of the rate constant fluctuate depending on the overall order of the reaction, ensuring that the rate equation maintains consistent units.
  • In a first-order reaction, like the one provided, the rate constant has units of \( ext{s}^{-1} \).
  • It tells you how fast the reaction proceeds when all concentrations are at 1 M.
  • An essential factor for determining reaction speed and comparison of different reactions.
The calculated rate constant, from the given problem, is \( 2.13 \times 10^{-1} \, \text{s}^{-1} \). This confirms how quickly the reaction occurs per second under the given conditions. By understanding the rate constant, you can predict how changing reactant concentrations may affect the overall reaction rate.
Kinetics
Kinetics is the study of the speed, or rate, of chemical reactions and the factors affecting them. It allows chemists to understand how different conditions alter reaction pathways.
  • Important parameters include concentration, temperature, catalysts, and surface area.
  • Studying kinetics helps in determining the mechanism of a reaction.
  • Analyzing reaction data, as shown in the exercise, enables us to deduce reaction orders.
In the provided exercise, kinetics is used to determine the effect of reactant concentrations \([A]\) and \([B]\) on the rate. The experimental data shows that while \([A]\) affects the rate, \([B]\) does not, aiding in determining the reaction order with respect to each reactant. Understanding kinetics is crucial for optimizing chemical processes in industrial applications.
Rate Law
The rate law is an equation that summarises the relationship between the concentration of reactants and the rate of a reaction. It is expressed in the form: \[\text{Rate} = k[A]^m[B]^n\]where \( m \) and \( n \) are the orders of reaction with respect to each reactant. They reflect the degree to which the concentration of each reactant affects the rate.
  • A rate law must be determined experimentally and cannot be inferred from the balanced equation alone.
  • The rate law provides a clear representation of a reaction's kinetics.
  • For complex reactions, the rate law can include more terms accounting for additional reactants or steps.
In your exercise, the rate law was derived to be \( \text{Rate} = k[A] \), indicating that the reaction is first order in \( A \) and zeroth order in \( B \). This means the rate is directly proportional to \([A]\), while \([B]\) does not influence the rate. Mastery of rate laws allows one to predict the effect of varying concentrations on the reaction speed.

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Most popular questions from this chapter

The rate law for the decomposition of ozone to molecular oxygen: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) $$ is $$ \text { rate }=k \frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]} $$ The mechanism proposed for this process is: $$ \begin{array}{l} \mathrm{O}_{3} \stackrel{k_{1}}{\rightleftarrows} \mathrm{O}+\mathrm{O}_{2} \\\ \mathrm{O}+\mathrm{O}_{3} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2} \end{array} $$ Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing \(\mathrm{O}_{2}\) concentration.

Polyethylene is used in many items, including water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules together. (Ethylene is the basic unit, or monomer, for polyethylene.) The initiation step is: \(\mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot\) (initiation) The \(\mathrm{R}\). species (called a radical) reacts with an ethylene molecule (M) to generate another radical: $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} $$ The reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: \(\mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad\) (propagation) This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine: $$ \mathbf{M}^{\prime} \cdot+\mathbf{M}^{\prime \prime} \cdot \stackrel{k_{\mathrm{t}}}{\longrightarrow} \mathbf{M}^{\prime}-\mathbf{M}^{\prime \prime} \quad \text { (termination) } $$ The initiator frequently used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]\) : $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min}\). (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\), or \(438 \mathrm{~min}\), at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the preceding polymerization process, and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long, high-molar-mass polyethylenes?

The decomposition of \(\mathrm{N}_{2} \mathrm{O}\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is a first-order reaction. At \(730^{\circ} \mathrm{C}\) the half-life of the reaction is \(3.58 \times 10^{3}\) min. If the initial pressure of \(\mathrm{N}_{2} \mathrm{O}\) is 2.10 atm at \(730^{\circ} \mathrm{C},\) calculate the total gas pressure after one half-life. Assume that the volume remains constant.

What are the advantages of measuring the initial rate of a reaction?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ \(\begin{array}{ll}\text { fructose } & \text { glucose }\end{array}\) This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \text { Time (min) } & {\left[\mathbf{C}_{12} \mathbf{H}_{22} \mathbf{O}_{11}\right](\boldsymbol{M})} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \end{array} $$
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