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The following three-step mechanism has been proposed for the reaction of chlorine and chloroform. $$\begin{aligned} & \text { (1) } \quad \mathrm{Cl}_{2}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons_{k-1}} 2 \mathrm{Cl}(\mathrm{g})\\\ & \text { (2) } \quad \mathrm{Cl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(\mathrm{g})+\mathrm{CCl}_{3}(\mathrm{g})\\\ &\text { (3) } \quad \mathrm{CCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}(\mathrm{g}) \end{aligned}$$ The numerical values of the rate constants for these steps are \(k_{1}=4.8 \times 10^{3} ; \quad k_{-1}=3.6 \times 10^{3} ; \quad k_{2}=1.3 \times 10^{-2} ; k_{3}=2.7 \times 10^{2} .\) Derive the rate law and the magnitude of \(k\) for the overall reaction.

Short Answer

Expert verified
The rate law for the overall reaction given by this three-step mechanism is \(\text{Rate overall} = k[Cl_{2}]^{0.5}\), where \(k \approx 2.44 \times 10^{-2}\).

Step by step solution

01

Identify Slow Step

Identify the slowest step in the given reaction mechanism. The slowest (rate-determining step) guides the overall rate of a multi-step reaction. By given constants, the rate constant \(k_2\) is the smallest, therefore, step (2) is the slowest step.
02

Write Preliminary Rate Law

Write the rate law for the slow step (2). A rate law, as the name implies, expresses the rate of a reaction in terms of the concentration of reactants. For the second step we have: \(\text{Rate 2} = k_{2}[CHCl_{3}][Cl]\). \(CHCl_{3}\) concentration is constant because it isn't involved in any other step.
03

Adjust Preliminary Rate Law

We need to adjust the preliminary rate law as it involves [Cl], which is also involved in other steps. We write the expression for [Cl] from the equilibrium of step 1. The rate equilibrium for step 1 is: \(k_{1}[Cl_{2}] = k_{-1}[Cl]^2\). Therefore, we can derive that \([Cl] = \sqrt{k_{1}[Cl_{2}]/k_{-1}}\).
04

Substitute and Simplify

Substitute [Cl] in the rate law of step 2. This leads to: \(\text{Rate }= k_2 \cdot [CHCl_3] \cdot \sqrt{k_{1}[Cl_{2}/k_{-1}}\). Since [CHCl3] is constant, it can be merged with rate constants. Now, we get the rate law for overall reaction as: \(\text{Rate overall} = k_[Cl_{2}^{0.5}]\).
05

Calculate the Magnitude of k

Calculate the magnitude of \(k\) for the overall reaction. This can be obtained by multiplying all constants involved: \(k = k_2* \sqrt{k_{1}/k_{-1}}\). Substituting the given values in, we can find the value: \(k = 1.3 \times 10^{-2} \cdot \sqrt{4.8 \times 10^{3} / 3.6 \times 10^{3}}\), which gives \(k \approx 2.44 \times 10^{-2}\).

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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 area of chemistry that concerns the rates at which chemical reactions occur and the factors that influence these rates. Understanding kinetics provides insight into how reactions can be controlled and utilized in processes such as reactions in biological systems, chemical manufacturing, and the degradation of materials.

To determine the rate of a reaction, one must consider the concentrations of the reactants and the temperature of the system. These factors impact the frequency and energy of collisions between reactant molecules, which are necessary for chemical transformations to take place. For example, a higher concentration of reactants increases the likelihood of collisions, therefore speeding up the reaction.

Mathematically, we express the speed of a reaction with a rate law, which equates the reaction rate to the product of a rate constant and the concentrations of the reactants raised to their respective powers, indicating the order of the reaction for each reactant. The rate constant is a measurement of how quickly a reaction proceeds and is influenced by temperature and the presence of catalysts. The study of chemical kinetics allows chemists to design and predict the outcomes of reactions, making it an essential discipline in many scientific and industrial fields.
Rate-Determining Step
The rate-determining step often referred to as the 'bottleneck' or 'slow step', is the slowest step in a sequence of reactions that determines the overall rate of the reaction mechanism. This concept is pivotal in chemical kinetics because this step's rate law is representative of the entire mechanism's rate law.

For the reaction of chlorine with chloroform, Step (2) of the proposed mechanism is the rate-determining step, owing to its smallest rate constant, as indicated by the given values. When deriving a rate law, the reaction rate is primarily based upon the concentration of the reactants involved in this step. However, it's not always straightforward because intermediate species -- produced in one step and consumed in another -- might complicate the expression of the reaction rate. In such cases, it is often necessary to express the concentration of the intermediates in terms of the concentrations of stable species.

Identifying the rate-determining step is crucial because it can direct efforts in reaction optimization and control. By changing the conditions to lower the energy barrier or by adding a catalyst that specifically accelerates the rate-determining step, chemists can increase the overall reaction rate in industrial or laboratory settings.
Reaction Rate Equilibrium
Reaction rate equilibrium is the point in a chemical reaction at which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. This concept is particularly relevant in reversible reactions where reactants convert to products, which can then revert back to reactants.

In the provided scenario, Step (1) exhibits a reversible reaction where equilibrium conditions apply. The equilibrium constant for this step is expressed in terms of the rate constants for the forward and reverse reactions, showing how these constants relate to the concentrations of reactants and products at equilibrium. For Step (1), we arrive at an expression for the concentration of the intermediate chlorine atoms in terms of the equilibrium concentrations of chlorine molecules. This expression is integral when determining the overall reaction rate for the entire mechanism.

Understanding reaction rate equilibrium allows chemists to predict the extent of a reaction and the concentrations of different components in a reaction mixture at any given time. Furthermore, knowledge of equilibrium can be applied to push reactions toward product formation or reactant recovery, as desired for a particular application. This understanding is fundamental for chemical engineering design, pharmaceutical synthesis optimization, and environmental science, among others.

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Most popular questions from this chapter

For the reaction \(A \longrightarrow 2 B+C\), the following data are obtained for \([\mathrm{A}]\) as a function of time: \(t=0 \mathrm{min}\) \([\mathrm{A}]=0.80 \mathrm{M} ; 8 \mathrm{min}, 0.60 \mathrm{M} ; 24 \mathrm{min}, 0.35 \mathrm{M} ; 40 \mathrm{min}\) \(0.20 \mathrm{M}\) (a) By suitable means, establish the order of the reaction. (b) What is the value of the rate constant, \(k ?\) (c) Calculate the rate of formation of \(\mathrm{B}\) at \(t=30 \mathrm{min}\).

One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.

The following rates of reaction were obtained in three experiments with the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}).\) $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial Rate of } \\ \text { Expt } & \text { [NO], M } & \text { [Cl }_{2} \text { ], M } & \text { Reaction, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 1 & 0.0125 & 0.0255 & 2.27 \times 10^{-5} \\ 2 & 0.0125 & 0.0510 & 4.55 \times 10^{-5} \\ 3 & 0.0250 & 0.0255 & 9.08 \times 10^{-5} \\ \hline \end{array}$$ What is the rate law for this reaction?

For the reaction \(A \longrightarrow\) products, the data tabulated below are obtained. (a) Determine the initial rate of reaction (that is, \(-\Delta[\mathrm{A}] / \Delta t)\) in each of the two experiments. (b) Determine the order of the reaction. $$\begin{array}{ll} \hline \text { First Experiment } & \\ \hline[\mathrm{A}]=1.512 \mathrm{M} & t=0 \mathrm{min} \\ \begin{array}{l} | \mathrm{A}\rfloor=1.490 \mathrm{M} \\ {[\mathrm{A}]=1.469 \mathrm{M}} \end{array} & \begin{array}{l} t=1.0 \mathrm{min} \\ t=2.0 \mathrm{min} \end{array} \\ \hline & \\ \hline \text { Second Experiment } & \\ \hline[\mathrm{A}]=3.024 \mathrm{M} & t=0 \mathrm{min} \\ {[\mathrm{A}]=2.935 \mathrm{M}} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=2.852 \mathrm{M}} & t=2.0 \mathrm{min} \\ \hline \end{array}$$

Suppose that the reaction in Example 14-8 is first order with a rate constant of \(0.12 \mathrm{min}^{-1}\). Starting with \([\mathrm{A}]_{0}=1.00 \mathrm{M},\) will the curve for \([\mathrm{A}]\) versus \(t\) for the first-order reaction cross the curve for the second-order reaction at some time after \(t=0 ?\) Will the two curves cross if \([\mathrm{A}]_{0}=2.00 \mathrm{M} ?\) In each case, if the curves are found to cross, at what time will this happen?

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