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Chapter 14: Problem 44

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of the term "rate determining."

Short Answer

Expert verified
The rate-determining step is the slowest step in a reaction sequence, essentially determining the overall reaction rate. An everyday analogy would be a road trip where no matter your speed on highways, the overall time is largely determined by the slower, winding paths.

Step by step solution

01

Understanding Rate-Determining Step

The rate-determining step is the slowest step in a reaction. Because it is the slowest, it determines the speed or rate at which the overall reaction can occur. If this step is slow, even if the other steps are fast, the overall reaction can only proceed as fast as this slow step allows.
02

Providing an Analogy for Rate-Determining Step

An analogy for a rate-determining step could be traveling on a road trip. Imagine you're on a journey that involves multiple roads. Some are highways where you can drive fast, and some are slow, narrow, winding paths. No matter how fast you drive on the highways, the overall time of your journey is largely determined by the slow, winding paths. In this analogy, the slow paths are like the rate-determining step of a chemical reaction.

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Most popular questions from this chapter

Consider the following elementary steps for a consecutive reaction $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(\mathrm{A}\).

Thallium(I) is oxidized by cerium(IV) as follows: $$ \mathrm{Tl}^{+}+2 \mathrm{Ce}^{4+} \longrightarrow \mathrm{Tl}^{3+}+2 \mathrm{Ce}^{3+} $$ The elementary steps, in the presence of \(\mathrm{Mn}(\mathrm{II}),\) are as follows: $$ \begin{aligned} \mathrm{Ce}^{4+}+\mathrm{Mn}^{2+} & \longrightarrow \mathrm{Ce}^{3+}+\mathrm{Mn}^{3+} \\ \mathrm{Ce}^{4+}+\mathrm{Mn}^{3+} \longrightarrow \mathrm{Ce}^{3+}+\mathrm{Mn}^{4+} \\ \mathrm{Tl}^{+}+\mathrm{Mn}^{4+} \longrightarrow \mathrm{Tl}^{3+}+\mathrm{Mn}^{2+} \end{aligned} $$ (a) Identify the catalyst, intermediates, and the ratedetermining step if the rate law is given by rate \(=\) \(k\left[\mathrm{Ce}^{4+}\right]\left[\mathrm{Mn}^{2+}\right]\) (b) Explain why the reaction is slow without the catalyst. (c) Classify the type of catalysis (homogeneous or heterogeneous).

What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in 60 min? Assume first-order kinetics.

What is meant by the order of a reaction?

Polyethylene is used in many items such as 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 (the basic unit is called a monomer) together (see p. 369 ). The initiation step is $$ \mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot \quad \text { initiation } $$ The \(\mathrm{R}\) - species (called a radical) reacts with an ethylene molecule \((\mathrm{M})\) to generate another radical $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} \cdot $$ Reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: $$ \mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad \text { propagation } $$ This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine $$ \mathrm{M}^{\prime} \cdot+\mathrm{M}^{\prime \prime} \cdot \stackrel{k_{t}}{\longrightarrow} \mathrm{M}^{\prime}-\mathrm{M}^{\prime \prime} \quad \text { termination } $$ (a) The initiator used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]:\) $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} \cdot $$ 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 above polymerization process and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long high-molar-mass polyethylenes?
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