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Chapter 11: Problem 35

In the gas phase, two butadiene molecules can dimerizes to give larger molecules according to the reaction: \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{~g})\) \(\rightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{~g})\). The rate law for this reac- tion is, \(r=K\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]^{2}\) with \(K=6.1 \times 10^{-2}\) \(1 \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at the temperature of reaction. The rate of formation of \(\mathrm{C}_{8} \mathrm{H}_{12}\), when the concentration of \(\mathrm{C}_{4} \mathrm{H}_{6}\) is \(0.02 \mathrm{M}\), is (a) \(2.44 \times 10^{-5} \mathrm{Ms}^{-1}\) (b) \(1.22 \times 10^{-5} \mathrm{Ms}^{-1}\) (c) \(1.22 \times 10^{-3} \mathrm{Ms}^{-1}\) (d) \(2.44 \times 10^{-6} \mathrm{Ms}^{-1}\)

Short Answer

Expert verified
1.22 x 10^-5 Ms^-1

Step by step solution

01

Determine the Rate of Reaction

Use the given rate law expression to calculate the rate of reaction for the dimerization. The rate law is given by r = K[C4H6]^2, where K is the rate constant and [C4H6] is the concentration of butadiene. Substitute the given values of K=6.1x10^-2 1 mol^-1 s^-1 and [C4H6]=0.02 M into the rate law expression to find r.
02

Calculate the Rate of Formation of C8H12

Since the stoichiometry of the reaction is 2 moles of C4H6 yielding 1 mole of C8H12, the rate of formation of C8H12 is half the rate of reaction. Therefore, divide the calculated rate of reaction by 2 to obtain the rate of formation of C8H12.
03

Choose the Correct Option

Compare the calculated rate of formation of C8H12 with the given options to find the correct answer.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the science that deals with understanding the speed or rate at which a chemical reaction occurs and the factors that affect this rate. It is pivotal to the study of how reactions happen and how to control them. A rate law expresses the relationship between the reaction rate and the concentration of reactors. For example, in the given dimerization reaction, the rate law is expressed as \( r = K[\mathrm{C}_4\mathrm{H}_6]^2 \). Here, \( K \) represents the rate constant, and \( [\mathrm{C}_4\mathrm{H}_6] \) is the concentration of butadiene. This quadratic dependence on butadiene’s concentration means that if the concentration is doubled, the reaction rate will increase by a factor of four, assuming temperature and other conditions remain constant.

The study of chemical kinetics also involves analyzing reaction mechanisms, which are the step-by-step sequences of elementary reactions by which an overall chemical change occurs. Understanding these mechanisms can be crucial in predicting the course of a reaction and designing processes and products in various industries.
Reaction Rate Calculation
The calculation of reaction rates is at the core of chemical kinetics, providing quantitative means of understanding and predicting the speed of chemical reactions. The rate of a reaction is often determined experimentally and mathematically described through the rate law, which for a reaction like the dimerization of butadiene is given by \( r = K[\mathrm{C}_4\mathrm{H}_6]^2 \). To calculate the rate, you must know the value of the rate constant \( K \) and the concentrations of reactants.

In the context of the exercise, knowing the values of \( K \) and the concentration of \( \mathrm{C}_4\mathrm{H}_6 \), the rate for the formation of \( \mathrm{C}_8\mathrm{H}_12 \) can be calculated by substituting these into the rate law and performing the calculation. It is important to note that the units of the rate constant depend on the order of the reaction and must be consistent with the units of concentration and time in the rate equation to get the correct units for the rate of reaction.
Dimerization Reaction
A dimerization reaction is a specific type of chemical reaction where two identical molecules (monomers) combine to form a new compound, which is a dimer. This process can occur in various phases, including the gas phase and solution. The dimerization of butadiene, \( 2\mathrm{C}_4\mathrm{H}_6 \rightarrow \mathrm{C}_8\mathrm{H}_12 \), is an example of such a reaction. It's considered a second-order reaction because two molecules of butadiene are involved.

Understanding the kinetics of dimerization reactions like this one can be crucial for industries such as polymer production, where the formation of larger molecules from monomeric units is a fundamental process. In addition to determining the rate law, analyzing dimerization reactions may also involve studying the reaction mechanism, temperature dependence, and influence of catalysts or inhibitors.

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Most popular questions from this chapter

For a first-order reaction, \(t_{0.75}=1386 \mathrm{~s}\). Its specific reaction rate is (a) \(10^{-3} \mathrm{~s}^{-1}\) (b) \(10^{-2} \mathrm{~s}^{-1}\) (c) \(10^{-4} \mathrm{~s}^{-1}\) (d) \(10^{-5} \mathrm{~s}^{-1}\)

For the consecutive unimolecular-type first-order reaction: \(\mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{R} \stackrel{k_{2}}{\longrightarrow} \mathrm{S}\), the concentration of component ' \(\mathrm{R}\) ', \(C_{\mathrm{R}}\), at any time, ' \(t\) ' is given by \(C_{\mathrm{R}}=C_{\mathrm{AO}} \cdot K_{1}\left[\frac{e^{-k_{1} t}}{\left(k_{2}-k_{1}\right)}+\frac{e^{-k_{2} t}}{\left(k_{1}-k_{2}\right)}\right]\) If \(C_{\mathrm{A}}=C_{\mathrm{AO}}, C_{\mathrm{R}}=C_{\mathrm{s}}=0\) at \(t=0\), the time at which the maximum concentration of 'R' occurs is (a) \(t_{\max }=\frac{k_{2}-k_{1}}{\ln \left(k_{2} / k_{1}\right)}\) (b) \(t_{\max }=\frac{\ln \left(k_{2} / k_{1}\right)}{k_{2}-k_{1}}\) (c) \(t_{\max }=\frac{e^{k_{2} / k_{1}}}{k_{2}-k_{1}}\) (d) \(t_{\max }=\frac{e^{k_{2}-k_{1}}}{k_{2}-k_{1}}\)

Azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}\), decomposes according to the following equation: \(\mathrm{CH}_{3}-\mathrm{N} \equiv \mathrm{N}-\mathrm{CH}_{3}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g})\) The initial concentration of azomethane was \(1.50 \times 10^{-2}\) M. After 10 min, the concentration decreased to \(1.29 \times 10^{-2} \mathrm{M}\). The average rate of reaction during this time interval is (a) \(3.5 \times 10^{-6} \mathrm{Ms}^{-1}\) (b) \(2.1 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3.5 \times 10^{-6} \mathrm{M} \mathrm{h}^{-1}\) (d) \(2.1 \times 10^{-3} \mathrm{Mmin}^{-1}\)

Which is correct about first-order reaction? (a) \(t_{0.5}=50 \mathrm{~s}, t_{0.75}=100 \mathrm{~s}\) (b) \(t_{0.5}=50 \mathrm{~s}, t_{0,75}=75 \mathrm{~s}\) (c) \(t_{0.5}=50 \mathrm{~s}, t_{0,75}=50 \mathrm{~s}\) (d) \(t_{0.5}=50 \mathrm{~s}, t_{0,25}=25 \mathrm{~s}\)

The reaction: \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{NO}_{2}+\mathrm{OH}^{-}\) \(\rightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}^{-}-\mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}\) obeys the rate law for pseudo first-order kinetics in the presence of a large excess of hydroxide ion. If \(1 \%\) of nitro ethane undergoes reaction in half a minute when the reactant concentration is \(0.002 \mathrm{M}\), what is the pseudo first-order rate constant? (a) \(2 \times 10^{-2} \mathrm{~min}^{-1}\) (b) \(6 \times 10^{-3} \mathrm{~min}^{-1}\) (c) \(4 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(1 \times 10^{-2} \mathrm{~min}^{-1}\)
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