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Consider this three-step mechanism for a reaction: \(\mathrm{Cl}_{2}(g) \underset{k_{2}}{\stackrel{k_{1}}{\rightleftarrows}} 2 \mathrm{Cl}(g)\) Fast \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) Slow \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) Fast a. What is the overall reaction? b. Identify the intermediates in the mechanism. c. What is the predicted rate law?

Short Answer

Expert verified
a. The overall reaction is \(\mathrm{Cl}_{2}(g) + \mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g) + \mathrm{CCl}_{4}(g)\). b. The intermediates are \(\mathrm{Cl}(g)\) and \(\mathrm{CCl}_{3}(g)\). c. The predicted rate law is \(\text{rate} = k[\mathrm{Cl}][\mathrm{CHCl}_{3}]\), which after substitution for \(\mathrm{Cl}\) using the equilibrium expression from step 1 becomes \(\text{rate} = k' [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}]\).

Step by step solution

01

Determine the Overall Reaction

To find the overall reaction, add up all the steps of the mechanism and cancel out species that appear on both sides of the reaction arrow as reactants and products. These canceled species are intermediates, not present in the overall reaction.
02

Identify the Intermediates

Intermediates are species that are produced in one step of a mechanism and consumed in a subsequent step. They do not appear in the overall reaction formula.
03

Derive the Rate Law

The rate law for a reaction is determined by the slow step of the mechanism. Write the rate law by using the concentrations of the reactants in the slow step. In cases where intermediates are involved, use the fast pre-equilibrium step to express the intermediates in terms of only initial reactants.

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Key Concepts

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

Reaction Mechanism
Understanding the reaction mechanism is key to grasping how chemical reactions occur. It is essentially the microscopic pathway or series of steps through which reactants transform into products. Each step is characterized by its own speed: fast or slow. In the given exercise, the overall chemical transformation occurs through a three-step mechanism, involving two fast steps, and one slow, rate-determining step.

The significance of the mechanism lies in its ability to detail the order of bond breakage and formation, along with the changes in molecular structure during the reaction. In a typical mechanism, you will find that the actual sequence of events can be quite complex and may include the creation of unstable, short-lived species called intermediates. These mechanisms are frequently used to predict the rate law of a reaction, a critical aspect in the field of chemical kinetics.
Reaction Intermediates
In the context of a chemical mechanism, reaction intermediates are species that appear temporarily in the course of a reaction sequence. They are produced in one step and consumed in another and are distinct from reactants or products which are present at the start or end of the reaction, respectively. In our exercise, intermediates are identified by analyzing the mechanism and noting species that are both formed and subsequently consumed as the reaction proceeds.

These intermediates do not appear in the overall balanced chemical equation for the reaction, as seen in the exercise where you were asked to cancel out species appearing on both sides of the arrows when outlining the overall reaction. Understanding intermediates is crucial because their fleeting existence can impact the rate and outcome of a reaction and also aid in identifying the rate law.
Rate Law
The rate law provides a mathematical relationship between the concentration of reactants and the rate of the reaction. It is determined by the slowest step in the reaction mechanism, also known as the rate-determining or rate-limiting step. This step acts as a bottleneck to the overall reaction speed since it is the slowest to occur.

As outlined in the exercise, to deduce the rate law, focus solely on the reactants involved in this slow step, rather than intermediate species or products. However, if an intermediate appears in the slow step, it's essential to express it in terms of the initial reactants through the relationships established by earlier, faster steps. This is achieved through algebraic manipulation based on the fast pre-equilibrium conditions prior to the slow step. Consequently, the rate law derived from the mechanism not only tells us about the speed of a reaction but also hints at the concentration dependencies and the order of reaction, which are pivotal for any further kinetics studies.

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Most popular questions from this chapter

The reaction \(2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) is first order in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and under certain conditions has a rate constant of \(0.00752 \mathrm{~s}^{-1}\) at \(20.0^{\circ} \mathrm{C}\). A reaction vessel initially contains \(150.0 \mathrm{~mL}\) of \(30.0 \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass solution (the density of the solution is \(1.11 \mathrm{~g} / \mathrm{mL}\) ). The gaseous oxygen is collected over water at \(20.0^{\circ} \mathrm{C}\) as it forms. What volume of \(\mathrm{O}_{2}\) forms in \(\begin{array}{lllll}85.0 & \text { seconds at a barometric pressure of } & 742.5 & \mathrm{mmHg} ?\end{array}\) (The vapor pressure of water at this temperature is \(17.5 \mathrm{mmHg}\).)

The evaporation of a 120 -nm film of \(n\) -pentane from a single crystal of aluminum oxide is zero order with a rate constant of \(1.92 \times 10^{13} \mathrm{molecules} / \mathrm{cm}^{2} \cdot \mathrm{s}\) at \(120 \mathrm{~K}\) a. If the initial surface coverage is \(8.9 \times 10^{16}\) molecules \(/ \mathrm{cm}^{2}\), how long will it take for one-half of the film to evaporate? b. What fraction of the film is left after 10 s? Assume the same initial coverage as in part a.

Consider the tabulated data showing the initial rate of a reaction (A \(\longrightarrow\) products) at several different concentrations of A. What is the order of the reaction? Write a rate law for the reac- tion, including the value of the rate constant, \(k\). $$ \begin{array}{cc} {[\mathrm{A}](\mathrm{M})} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\\ 0.12 & 0.0078 \\ \hline 0.16 & 0.0104 \\ \hline 0.20 & 0.0130 \\ \hline \end{array} $$

Indicate the order of reaction consistent with each observation. a. A plot of the concentration of the reactant versus time yields a straight line. b. The reaction has a half-life that is independent of initial concentration. c. A plot of the inverse of the concentration versus time yields a straight line.

The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration. How long will it take for \(10 \%\) of the U- 238 atoms in a sample of U- 238 to decay? If a sample of U-238 initially contained \(1.5 \times 10^{18}\) atoms when the universe was formed 13.8 billion years ago, how many U- 238 atoms does it contain today?

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