/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 The mechanism for the reaction o... [FREE SOLUTION] | 91Ó°ÊÓ

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The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. $$ \text { Step 1: } \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {fast, endothermic } $$ $$ \text { Step } 2: \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} $$ slow (a) Write an equation for the overall reaction. (b) Draw a reaction energy diagram for this reaction. (c) Show that the rate law for this reaction is $$ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right] $$

Short Answer

Expert verified
Overall: \(\mathrm{CH}_3\mathrm{OH} + \mathrm{HBr} \rightarrow \mathrm{CH}_3\mathrm{Br} + \mathrm{H}_2\mathrm{O}\). Diagram shows exothermic profile. Rate is \(k[\mathrm{CH}_3\mathrm{OH}][\mathrm{H}^+][\mathrm{Br}^-]\).

Step by step solution

01

Identify Overall Reaction

The overall reaction involves the reactants and the final products of the reaction mechanism. From the provided steps: Step 1 and Step 2, we know that \(\mathrm{CH}_3\mathrm{OH}\) reacts with \(\mathrm{H}^{+}\) and \(\mathrm{Br}^{-}\) to form \(\mathrm{CH}_3\mathrm{Br}\) and \(\mathrm{H}_2\mathrm{O}\). Thus, the balanced equation for the overall reaction is: \[\mathrm{CH}_3\mathrm{OH} + \mathrm{HBr} \rightarrow \mathrm{CH}_3\mathrm{Br} + \mathrm{H}_2\mathrm{O}\].
02

Reaction Energy Diagram

Draw an energy diagram where the y-axis represents energy and the x-axis represents the reaction progress. Start with reactants \(\mathrm{CH}_3\mathrm{OH}\) and \(\mathrm{H}^{+}\) at a certain energy level. Show an energy increase for the endothermic step (Step 1). Indicate a transition to an intermediate state (\(\mathrm{CH}_3\mathrm{OH}_2^{+}\)). Next, show a second curve with a smaller energy barrier due to the slow, exothermic reaction (Step 2), resulting in the products \(\mathrm{CH}_3\mathrm{Br}\) and \(\mathrm{H}_2\mathrm{O}\). The final energy level of the products should be lower than the initial reactants, highlighting the overall exothermic nature.
03

Derive Rate Law from Mechanism

From the given mechanism, Step 1 is fast and reversible while Step 2 is slow and therefore rate-determining. Assuming a fast equilibrium in Step 1 for \(\mathrm{CH}_3\mathrm{OH}_2^{+}\), the concentration of \(\mathrm{CH}_3\mathrm{OH}_2^{+}\) can be expressed in terms of \(\mathrm{CH}_3\mathrm{OH}\) and \(\mathrm{H}^{+}\) using equilibrium constants. The rate of the overall reaction is determined by the rate of Step 2: \[\text{Rate} = k_2[\mathrm{CH}_3\mathrm{OH}_2^{+}][\mathrm{Br}^{-}]\]. Applying the steady-state assumption or assuming equilibrium in Step 1 allows us to express \([\mathrm{CH}_3\mathrm{OH}_2^{+}]\) in terms of \([\mathrm{CH}_3\mathrm{OH}]\) and \([\mathrm{H}^{+}]\), leading to: \[\text{Rate} = k[\mathrm{CH}_3\mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}]\].

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Key Concepts

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

Exothermic Reaction
In chemistry, an exothermic reaction is a process that releases energy, usually in the form of heat, to its surroundings. This is why such reactions often result in an increase in temperature. In the case of the reaction between \( \mathrm{CH}_3\mathrm{OH} \) and \( \mathrm{HBr} \), the overall process is exothermic.

This means that while the system progresses from reactants like \( \mathrm{CH}_3\mathrm{OH} \) and \( \mathrm{HBr} \) to the products \( \mathrm{CH}_3\mathrm{Br} \) and \( \mathrm{H_2O} \), it releases energy to the surrounding environment.
  • The energy released is often visualized in an energy diagram where the final energy state of the products is lower than that of the reactants.
  • This lower energy state is what defines the exothermic nature of the overall reaction.
Understanding the nature of exothermic reactions helps to predict the feasibility and conditions under which the reaction might occur spontaneously.
Reaction Kinetics
Reaction kinetics explores how and why reactions occur at certain rates. It involves analyzing the steps or mechanism the reaction undergoes to form the final products.

For the reaction of \( \mathrm{CH}_3\mathrm{OH} \) and \( \mathrm{HBr} \), it happens in two steps. The first step is a fast, endothermic process forming an intermediate \( \mathrm{CH}_3\mathrm{OH}_2^{+} \). The second step is slower and therefore rate-determining, meaning it governs the reaction's overall rate.
  • The speed of each step contributes to the overall kinetics of the reaction.
  • Understanding which step is slower helps in identifying the rate-limiting factor and provides insight for increasing the reaction rate through catalysts or temperature changes if necessary.
Studying these kinetics is essential to manipulate and control reactions in chemical processes and industrial applications.
Rate Laws
Rate laws describe how the concentration of reactants affects the speed of a reaction. They are usually expressed as equations that relate the rate of a reaction to the concentration of its reactants.

For the reaction between \( \mathrm{CH}_3\mathrm{OH} \) and \( \mathrm{HBr} \), the rate law is given by: \[ \text{Rate} = k[\mathrm{CH}_3\mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \] where \( k \) is the rate constant, and the exponents indicate the order of the reaction with respect to each reactant.
  • This rate law shows a direct dependence of the reaction rate on the concentration of all three reactants involved.
  • It can be deduced from the mechanism, considering that the second, slow step determines the rate.
Knowing this relationship allows chemists to predict how varying concentrations will affect the reaction speed, which is useful for controlling reaction conditions.
Energy Diagrams
Energy diagrams are graphical representations that show the change in energy as a chemical reaction progresses. These diagrams illustrate the energy levels of reactants, intermediates, and products.

In our example of \( \mathrm{CH}_3\mathrm{OH} \) and \( \mathrm{HBr} \), the energy diagram would begin with the initial reactants at a higher energy level. As the reaction proceeds through the endothermic step, there is a rise in energy—an energy barrier that the reactants must overcome.
  • This is followed by a drop in energy representing the exothermic nature of the final step leading to the formation of products.
  • The final energy level is lower than the initial, marking the overall reaction as exothermic.
These diagrams are helpful to visualize the energy changes occurring during a reaction, allowing for insights into stability and reaction spontaneity.
Molecular Interactions
Molecular interactions are forces that act between molecules, affecting their reactions. These interactions are crucial in understanding the mechanism of how reactants transform into products.

During the reaction of \( \mathrm{CH}_3\mathrm{OH} \) and \( \mathrm{HBr} \), initial molecular interactions enable the formation of \( \mathrm{CH}_3\mathrm{OH}_2^{+} \). This intermediate is stabilized through interactions with \( \mathrm{H}^{+} \). In the subsequent step, interactions between \( \mathrm{CH}_3\mathrm{OH}_2^{+} \) and \( \mathrm{Br}^{-} \) lead to the final product formation.
  • These interactions determine the course and speed of the reaction.
  • They also help in the stabilization of transition states and intermediates.
Comprehending these molecular interactions provides a deeper insight into the steps and energies involved in chemical reactions, aiding in designing better reaction conditions and new synthetic pathways.

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Most popular questions from this chapter

Define the terms "unimolecular elementary reaction" and "bimolecular elementary reaction," and give an example of each.

A biological catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ} / \mathrm{mol}\) to \(206 \mathrm{~kJ} / \mathrm{mol}\). Calculate by what factor the rate constant, \(k,\) would increase at \(25^{\circ} \mathrm{C}\). Assume that the frequency factors \((A)\) are the same for the uncatalyzed and catalyzed reactions.

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

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic

Two mechanisms are proposed for the reaction $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ Mechanism $$ \text { 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} $$ fast $$ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} $$ slow $$ \text { Mechanism } 2: \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} $$ fast Show that each mechanism is consistent with the observed rate law: Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\).

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