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Chapter 11: Problem 111

Which of these mechanisms (more than one may be chosen) is compatible with the rate law? Rate \(=k\left[\mathrm{Cl}_{2}\right]^{3 / 2}[\mathrm{CO}]\) $$ \text { (a) } \frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl} $$ \(\mathrm{Cl}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}_{3}\) fast \(\mathrm{Cl}_{3}+\mathrm{CO} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \quad\) slow $$ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} $$ (b) \(\mathrm{Cl}_{2}+\mathrm{CO} \longrightarrow \mathrm{CCl}_{2}+\mathrm{O}\) slow \(\mathrm{O}+\mathrm{Cl}_{2} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}\) fast $$ \mathrm{Cl}_{2} \mathrm{O}+\mathrm{CCl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}_{2} $$ fast s (c) \(\frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}\) $$ \mathrm{Cl}+\mathrm{CO} \rightleftharpoons \mathrm{COCl} $$ fast $$ \begin{array}{l} \mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \\\ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} \end{array} $$ slow (d) \(\mathrm{Cl}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{COCl}+\mathrm{Cl}\) fast \(\mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}\) slow \(\mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_{2}\) fast

Short Answer

Expert verified
Mechanisms (a) and (d) are compatible with the rate law.

Step by step solution

01

Understand the Rate Law

The given rate law is \( k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}] \). It means the reaction rate is dependent on the concentration of \([\mathrm{Cl}_{2}]\) raised to the power of \(3/2\) and \([\mathrm{CO}]\) to the power of 1.
02

Analyze Mechanism (a)

Mechanism (a) involves the intermediary \(\mathrm{Cl}_{3}\). The rate-determining step is slow and involves \(\mathrm{Cl}_{3}\) and \(\mathrm{CO}\). However, this step does not produce a rate law with a \([\mathrm{Cl}_{2}]^{3/2}\) term, as it does not match intermediates or products expected by the rate law.
03

Analyze Mechanism (b)

In mechanism (b), the slow step is the formation of \(\mathrm{CCl}_{2}\) and \(\mathrm{O}\) from \(\mathrm{Cl}_{2}\) and \(\mathrm{CO}\). This slow step suggests a rate law \(k[\mathrm{Cl}_{2}][\mathrm{CO}]\), which does not match the given rate law's \([\mathrm{Cl}_{2}]^{3/2}\) term.
04

Analyze Mechanism (c)

Mechanism (c) has a rate-determining step forming \(\mathrm{COCl}\) from \(\mathrm{Cl}\) and \(\mathrm{CO}\) quickly, followed by a slow step. This doesn't match the given rate law \(k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}]\) since there isn't a mechanism for achieving the \([\mathrm{Cl}_{2}]^{3/2}\) exponent.
05

Analyze Mechanism (d)

In mechanism (d), the fast equilibrium step involving \(\mathrm{COCl}\) from \(\mathrm{Cl}_{2}\) and \(\mathrm{CO}\) can produce a rate law consistent with \(k[\mathrm{Cl}_{2}]^{1/2}[\mathrm{CO}]\) after considering intermediates and fast reactions. This partial dependence on intermediates and \([\mathrm{Cl}_{2}]\) aligns with the rate law requirement of \([\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}]\).

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Key Concepts

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

Rate Law
Understanding rate laws is crucial in chemical kinetics as they provide insights into how the concentrations of substances affect the speed of a reaction. A rate law is expressed using a mathematical equation that relates the rate of a reaction to the concentration of reactants. In this case, the rate law is \( k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}] \). This indicates that the reaction rate is directly proportional to the concentration of \([\mathrm{Cl}_{2}]\) raised to the power of \(3/2\) and the concentration of \([\mathrm{CO}]\) raised to the power of 1. The constants and exponents in a rate law are determined experimentally and provide crucial information about the reaction pathway and mechanism.

Rate laws can vary greatly depending on the mechanism of the reaction. Therefore, it is important to analyze the mechanism carefully to see which step determines the rate and matches with the given rate law. This often involves considering the slowest step in the reaction mechanism or the step involving intermediates.
Chemical Kinetics
Chemical kinetics is a field of chemistry that studies the rates of chemical processes. It delves into understanding how different conditions affect the speed of reactions. The goal is to identify the rate-determining step, which is often the slowest step in a series of reactions and controls the overall rate of the process.

In multi-step reactions, not all steps occur at the same rate. Some are fast, while others are slow. The slowest step acts as a bottleneck, determining the speed of the entire reaction. When analyzing mechanisms, attention should be paid to this rate-determining step as it often dictates the structure of the rate law. By understanding chemical kinetics, chemists can manipulate conditions to optimize reaction speeds, thereby improving efficiency in industrial and laboratory settings.
Intermediates
Intermediates are transient molecules that are formed in one step of a reaction mechanism and consumed in another. They play a crucial role in understanding reaction mechanisms as they often appear in the elementary steps of a reaction but not in the overall balanced equation.

Analyzing intermediates helps chemists verify the validity of proposed reaction mechanisms. For example, in mechanism (a), the intermediate \(\mathrm{Cl}_{3}\) does not contribute to forming a rate law consistent with the one provided in the problem. However, in mechanism (d), the presence of intermediates along with fast equilibrium steps involving them allows the creation of a rate law that matches \( k[\mathrm{Cl}_{2}]^{3/2}[\mathrm{CO}] \).

Understanding how intermediates interact within a reaction mechanism is key to linking the experimental rate law with the theoretical pathway. This connection provides critical insights into how reactions proceed and can assist in predicting catalysts or conditions to enhance reaction speed.

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Most popular questions from this chapter

The rate constant for decomposition of azomethane at $$ \begin{aligned} 425^{\circ} \mathrm{C} \text { is } 0.68 \mathrm{~s}^{-1} & \\ \mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \end{aligned} $$ (a) Based on the units of the rate constant, determine if this reaction is zeroth-, first-, or second-order. (b) If \(2.0 \mathrm{~g}\) azomethane is placed in a \(2.0-\mathrm{L}\) flask and heated to \(425^{\circ} \mathrm{C},\) calculate the mass of azomethane that remains after \(5.0 \mathrm{~s}\). (c) Calculate how long it takes for the mass of azomethane to drop from \(2.0 \mathrm{~g}\) to \(0.24 \mathrm{~g}\). (d) Calculate the mass of nitrogen that would be found in the flask after \(0.50 \mathrm{~s}\) of reaction.

Draw a reaction energy diagram for an exothermic process. Mark the positions of reactants, products, and activated complex. Indicate the activation energies of the forward and reverse processes and explain how \(\Delta_{\mathrm{r}} E\) for the reaction can be calculated from the diagram.

The transfer of an oxygen atom from \(\mathrm{NO}_{2}\) to CO has been studied at \(540 \mathrm{~K}\) : \(\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g})\) These data were collected: \begin{tabular}{ccc} \hline Initial Rate & \multicolumn{2}{c} { Initial Concentration (mol/L) } \\ \cline { 2 - 3 } (mol \(\mathrm{L}^{-1} \mathrm{~h}^{-1}\) ) & {\(\left[\mathrm{NO}_{2}\right]\)} \\ \hline \(5.1 \times 10^{-4}\) & \(0.35 \times 10^{-4}\) & \(3.4 \times 10^{-8}\) \\ \(5.1 \times 10^{-4}\) & \(0.70 \times 10^{-4}\) & \(1.7 \times 10^{-8}\) \\ \(5.1 \times 10^{-4}\) & \(0.18 \times 10^{-4}\) & \(6.8 \times 10^{-8}\) \\ \(1.0 \times 10^{-3}\) & \(0.35 \times 10^{-4}\) & \(6.8 \times 10^{-8}\) \\ \(1.5 \times 10^{-3}\) & \(0.35 \times 10^{-4}\) & \(10.2 \times 10^{-8}\) \\ \hline \end{tabular} (a) Write the rate law. (b) Determine the reaction order with respect to each reactant (c) Calculate the rate constant and express it in appropriate units.

Cyclopropane isomerizes to propene when heated. Rate constants for the reaction cyclopropane \(\longrightarrow\) propene are \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470.0^{\circ} \mathrm{C}\) and \(1.02 \times 10^{-3} \mathrm{~s}^{-1}\) at \(510.0^{\circ} \mathrm{C}\) (a) Calculate the activation energy, \(E_{\mathrm{a}},\) for this reaction. (b) Calculate how long it takes at \(500 .{ }^{\circ} \mathrm{C}\) for the concentration of cyclopropane to drop from \(0.10 \mathrm{M}\) to \(0.023 \mathrm{M}\).

In an automobile catalytic converter the catalysis is accomplished on a surface consisting of platinum and other precious metals. The metals are deposited as a thin layer on a ceramic support that is a fine grid (see the photo). (a) Why is the ceramic support arranged in the grid-like geometry? (b) Why are the metals deposited on the ceramic surface instead of being used as strips or rods?
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