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Chapter 17: Problem 75

The reaction of CO with Cl_gives phosgene (COCl_), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises: \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (fast, \(\mathrm{k}_{1}\) represents the forward rate constant, \(k_{-1}\) the reverse rate constant) \(\mathrm{CO}(g)+\mathrm{Cl}(g) \longrightarrow \operatorname{COCl}(g)\) (slow, \(k_{2}\) the rate constant) \(\operatorname{COCl}(g)+\operatorname{Cl}(g) \longrightarrow \operatorname{COCl}_{2}(g)\) (fast, \(k_{3}\) the rate constant) (a) Write the overall reaction. (b) Identify all intermediates. (c) Write the rate law for each elementary reaction. (d) Write the overall rate law expression.

Short Answer

Expert verified
The overall reaction is \(\mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{COCl}_{2}(g)\). The intermediates are \(\mathrm{Cl}(g)\) and \(\operatorname{COCl}(g)\). The rate laws for the elementary reactions are \(\text{Rate}_1 = k_1[\mathrm{Cl}_2] - k_{-1}[\mathrm{Cl}]^2\), \(\text{Rate}_2 = k_2[\mathrm{CO}][\mathrm{Cl}]\), and \(\text{Rate}_3 = k_3[\operatorname{COCl}][\mathrm{Cl}]\). The overall rate law is \(\text{Rate} = k_2[\mathrm{CO}][\mathrm{Cl}]\) since the second step is the slowest.

Step by step solution

01

- Writing the Overall Reaction

To write the overall reaction, add up all the elementary steps and cancel out any intermediates or species that appear on both sides of the reaction arrows.
02

- Identifying Intermediates

Intermediates are species that are produced in one step of a mechanism and consumed in a subsequent step. Look for species that are not present in the initial reactants or final products.
03

- Writing the Rate Law for Each Elementary Reaction

Each elementary step has its own rate law based on the molecularity of the reaction. Write each rate law using the appropriate rate constants and reactant concentrations.
04

- Writing the Overall Rate Law Expression

The overall rate law is determined by the rate of the slowest step, also known as the rate-determining step. Use the rate law for this step as the overall rate law.

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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 branch of physical chemistry that studies the rates of chemical reactions and the mechanisms by which they occur. Understanding kinetics allows us to predict how quickly a reaction will proceed, which is essential in fields like pharmaceuticals, material science, and environmental science.

For a given reaction, kinetics involves analyzing various factors, such as temperature, pressure, and concentration of reactants, to determine the speed or rate of the reaction. It is important to realize that kinetics is concerned with the 'how fast' of a reaction, not just the 'if' a reaction will happen, which is the domain of thermodynamics.
Rate Law
The rate law is an equation that relates the rate of a reaction to the concentration of its reactants. For an elementary reaction, the rate law can often be written directly from the stoichiometry of the reactants involved. It takes the form \[Rate = k[Reactant]^n\]where 'k' is the rate constant, '[Reactant]' is the concentration of the reactant, and 'n' is the order of the reaction with respect to that reactant.

Rate laws are crucial in helping us predict how changes in conditions affect the speed of the reaction. It is worth noting that the rate law can only be determined experimentally and not solely from the balanced chemical equation.
Reaction Intermediates
Reaction intermediates are species that are formed in one step of a reaction mechanism and consumed in another. They are not present in the overall balanced equation for the reaction but play a crucial role in the conversion of reactants to products.

Identifying intermediates helps us understand the sequence of steps that make up the complex pathway from reactants to products. In the provided exercise, we can see intermediates that are produced and then used up as the reaction proceeds. Clarifying their role can aid in designing catalysts and in controlling reaction pathways for desired outcomes.
Elementary Reactions
Elementary reactions are the simple steps that together constitute the overall mechanism of a complex reaction. These are individual reactions where reactants are converted to products in a single step, with no intermediates. They can vary in molecularity, which is the number of reactant molecules involved in the step.

Understanding elementary reactions gives insight into the detailed events at the molecular level during the course of a chemical reaction. For each elementary step, we can write a rate law based on the actual reacting species. This contrasts with the rate law for the overall reaction, which can differ because it reflects the rate of the slowest, rate-determining step.

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Most popular questions from this chapter

The rate constant for the first-order decomposition at \(45^{\circ} \mathrm{C}\) of dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), dissolved in chloroform, \(\mathrm{CHCl}_{3},\) is \(6.2 \times 10^{-4} \mathrm{min}^{-1}\). $$2 \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$$ What is the rate of the reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.40 \mathrm{M}\) ?

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same? (a) \(\mathrm{Cl}_{2}+\mathrm{CO} \longrightarrow \mathrm{Cl}_{2} \mathrm{CO}\) rate \(=k\left[\mathrm{Cl}_{2}\right]^{3 / 2}[\mathrm{CO}]\) (b) \(\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \longrightarrow \mathrm{PCl}_{5}\) rate \(\left.=k\left[\mathrm{PCl}_{3}\right] \mathrm{Cl}_{2}\right]\) (c) \(2 \mathrm{NO}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) rate \(=k[\mathrm{NO}]\left[\mathrm{H}_{2}\right]\) (d) \(2 \mathrm{NO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}\) rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\) (e) \(\mathrm{NO}+\mathrm{O}_{3} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\) rate \(=k[\mathrm{NO}]\left[\mathrm{O}_{3}\right]\)

Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.

Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of \(3 \times 10^{4} \mathrm{g} / \mathrm{mol}\) that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: \(1.0 \mathrm{L}\) of a solution containing \(0.15 \mu \mathrm{g}\left(0.15 \times 10^{-6} \mathrm{g}\right)\) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant. $$\begin{array}{|c|c|} \hline \text { [Penicillin] (M) } & \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}\right)\) } \\ \hline 2.0 \times 10^{-6} & 1.0 \times 10^{-10} \\ \hline 3.0 \times 10^{-6} & 1.5 \times 10^{-10} \\ \hline 4.0 \times 10^{-6} & 2.0 \times 10^{-10} \\\ \hline \end{array}$$

From the given data, use a graphical method to determine the order and rate constant of the following reaction: $$2 X \longrightarrow Y+Z$$ $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Time (S) } & 5.0 & 10.0 & 15.0 & 20.0 & 25.0 & 30.0 & 35.0 & 40.0 \\ \hline [X](M) & 0.0990 & 0.0497 & 0.0332 & 0.0249 & 0.0200 & 0.0166 & 0.0143 & 0.0125 \\ \hline \end{array}$$
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