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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. Step 1: Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2: Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is Rate \(=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

Short Answer

Expert verified
(a) \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \) (b) Energy rises then drops due to slow, exothermic steps. (c) Rate law: \( k[\mathrm{CH}_{3} \mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \).

Step by step solution

01

Write Overall Reaction

To find the overall reaction, add the two elementary reactions given: \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \) and \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \). Cancel out intermediate species \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \). The overall reaction is: \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \).
02

Draw Reaction Coordinate Diagram

A reaction coordinate diagram requires plotting the energy changes over the reaction pathway. For this mechanism:1. Begin with the energy of reactants \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} + \mathrm{Br}^{-} \).2. First, depict an endothermic step as a rise in energy for \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \), as the given step 1 is endothermic.3. Follow this with a peak representing the slow step (rate-determining step) of \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \).4. Finally, show an overall decrease in energy to indicate the reaction is exothermic.
03

Derive Rate Law

Since Step 2 is the slowest, it is the rate-determining step. Rate is determined by slowest step: \( \text{Rate} = k[\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}][\mathrm{Br}^{-}] \). From \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \), the rate law for formation of \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \) can be approximated as \( [\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}] \approx [\mathrm{CH}_{3} \mathrm{OH}][\mathrm{H}^{+}] \) at equilibrium. Substitute in step 2's rate equation: \( \text{Rate} = k[\mathrm{CH}_{3} \mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \).

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Key Concepts

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

Elementary Steps
In chemical reaction mechanisms, understanding elementary steps is crucial as they describe the individual stages of a chemical process. An **elementary step** is a single event or reaction on the molecular level. It showcases how reactant molecules transform into products in a single kinetic event.
  • For instance, in the reaction of \( \mathrm{CH}_{3} \mathrm{OH} \) and \( \mathrm{HBr} \), we observe two elementary steps. The first is the interaction of \( \mathrm{CH}_{3} \mathrm{OH} \) with \( \mathrm{H}^{+} \) to form \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \).
  • The second step, which involves \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \) reacting with \( \mathrm{Br}^{-} \), results in the end products \( \mathrm{CH}_{3} \mathrm{Br} \) and \( \mathrm{H}_{2} \mathrm{O} \).
The distinction between these steps helps in identifying how catalysts or changes in conditions might affect the reaction. Notice that each elementary step corresponds to a molecular event that contributes to the overall conversion of reactants to products. It's like viewing the reaction process in "snapshots," which when added together give a complete picture.
Rate Law
The rate law for a chemical reaction specifies the relationship between the reaction rate and the concentrations of reactants. For the two-step reaction mechanism we're discussing, the rate law is derived from the rate-determining step, which is the slowest elementary step.Consider these key points:
  • The second step, \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \), is slower, making it the **rate-determining step**.
  • Because this step controls the reaction rate, the rate law is derived from it. Initially, it can be expressed as \( \text{Rate} = k[\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}][\mathrm{Br}^{-}] \).
  • However, since \( [\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}] \) is an intermediate not present as a reactant, we substitute its concentration using equilibrium expressions from the first step: \( [\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}] \approx [\mathrm{CH}_{3} \mathrm{OH}][\mathrm{H}^{+}] \).
Ultimately, the rate law for the overall reaction becomes \( \text{Rate} = k[\mathrm{CH}_{3} \mathrm{OH}][\mathrm{H}^{+}][\mathrm{Br}^{-}] \). This highlights how the concentration of each reactant influences the speed of the reaction based on the rate-determining step.
Reaction Coordinate Diagram
A reaction coordinate diagram offers a visual representation of the energy changes that occur during a chemical reaction. It helps students and chemists to understand the progression from reactants to products in terms of energy changes.Visualizing the Diagram:
  • Plot the energy of the system on the vertical axis, and the reaction progress on the horizontal axis.
  • For the **first step** of the reaction, \( \mathrm{CH}_{3} \mathrm{OH} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \), you will see a rise in the curve, indicating an **endothermic** process (energy is absorbed).
  • The second step, \( \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} + \mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br} + \mathrm{H}_{2} \mathrm{O} \), follows and features a high peak, signifying it as the **slow, rate-determining** step.
  • The final part of the diagram shows an overall decline in energy, illustrating that the entire reaction is **exothermic** (energy is released).
This diagram helps to intuitively understand both the energy barriers that must be overcome at each step and why certain steps limit the speed of the reaction. Use this visualization to better grasp why intermediate species might momentarily form and how energy transformations occur, providing insight into the mechanism's efficiency and feasibility.

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Most popular questions from this chapter

The decomposition of HOF occurs at \(25^{\circ} \mathrm{C}\) $$\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ Using the data in the table below, determine the rate law, and then calculate the rate constant. $$\begin{array}{cc}{[\mathrm{HOF}](\mathrm{mol} / \mathrm{L})} & \mathrm{Time}(\mathrm{min}) \\ \hline 0.850 & 0 \\\0.810 & 2.00 \\\0.754 & 5.00 \\\0.526 & 20.0 \\\0.243 & 50.0 \\\\\hline\end{array}$$

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures, a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

The radioactive isotope \(^{64} \mathrm{Cu}\) is used in the form of copper(II) acetate to study Wilson's disease. The isotope has a half-life of 12.70 hours. What fraction of radioactive copper(II) acetate remains after 64 hours?

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is a first-order reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ The rate constant for the reaction is \(2.8 \times 10^{-3} \mathrm{min}^{-1}\) at \(600 \mathrm{K}\). If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.24 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) how long will it take for the concentration to drop to \(0.31 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

The dimerization of butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) to form 1,5-cyclooctadiene is a second-order process that occurs when the diene is heated. In an experiment, a sample of 0.0087 mol of \(\mathrm{C}_{4} \mathrm{H}_{6}\) was heated in a \(1.0-\mathrm{L}\) flask. After 600 seconds, \(21 \%\) of the butadiene had dimerized. Calculate the rate constant for this reaction.

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