/*! 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 70 You have studied the gas-phase o... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 70

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \cdots & \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Indicate how the elementary reactions add to give the overall reaction. (Hint: You will need to multiply the coefficients of one of the equations by 2.) (b) Based on the rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

Short Answer

Expert verified
In this exercise, we found that: (a) The given elementary reactions can add to give the overall reaction when their coefficients are properly adjusted. (b) The rate-determining step is the first elementary reaction (\(\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\)), as it matches the rate law. (c) The intermediates in this mechanism are \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\). (d) The absence of detected intermediates does not disprove the mechanism, as intermediates are typically unstable and short-lived. However, further studies are needed to validate the proposed mechanism.

Step by step solution

01

Analyzing the given mechanism and finding the overall reaction

Based on the proposed mechanism, we have three elementary reactions: 1. \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\) 2. \(\mathrm{HOOBr}(g)+\mathrm{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)\) 3. \(\mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)\) Notice that the coefficients on equation 2 and equation 3 need to be multiplied by 2 to match the stoichiometry of the overall reaction: 2(2). \(2\mathrm{HOOBr}(g)+2\mathrm{HBr}(g) \longrightarrow 4 \mathrm{HOBr}(g)\) 2(3). \(2\mathrm{HOBr}(g)+2\mathrm{HBr}(g) \longrightarrow 2\mathrm{H}_{2} \mathrm{O}(g)+2\mathrm{Br}_{2}(g)\) Now, adding equations 1, 2(2), and 2(3) together, we obtain: \[ 4 \mathrm{HBr}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(g) + 2 \mathrm{Br}_{2}(g) \] Which is the overall reaction. So, the elementary reactions can add to give the overall reaction.
02

Determining the rate-determining step

Now, we know that the rate law for the overall reaction is first order with respect to both \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\): Rate \(= k[\mathrm{HBr}][\mathrm{O}_{2}]\) Since the rate law matches the reaction 1 (\(\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\)), this step is the rate-determining step because it controls the rate of the overall reaction.
03

Identifying the intermediates

The intermediates in the reaction mechanism are the species that are produced in one elementary step and consumed in another elementary step, not appearing in the overall reaction. In this mechanism, the intermediates are: 1. \(\mathrm{HOOBr}(g)\) - formed in step 1 and consumed in step 2. 2. \(\mathrm{HOBr}(g)\) - formed in step 2 and consumed in step 3.
04

Analyzing the absence of detected intermediates

As for the question regarding the absence of detected intermediates, it does not disprove the mechanism. Intermediates are typically unstable and short-lived, meaning they may not be detected among the final products. However, further studies could provide more evidence to validate the proposed mechanism.

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Key Concepts

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

Rate-Determining Step
In any multi-step chemical reaction, the rate-determining step is like the slowest runner in a relay race. It is the slowest step in a reaction mechanism. This step limits the overall speed of the reaction and determines the rate law.
In the gas-phase oxidation of \(\mathrm{HBr}\), the rate law is first order with respect to both \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\). This tells us that the rate-determining step involves one molecule of \(\mathrm{HBr}\) and one molecule of \(\mathrm{O}_{2}\).
  • Step 1 of the elementary steps, \[\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\], fits this description.
  • Since this step involves \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\), it matches the first-order dependence of each in the rate law, making it the rate-determining step.
Understanding which step is rate-determining helps chemists improve reaction efficiency by focusing on speeding up this particular step.
Chemical Kinetics
Chemical kinetics is the study of reaction rates. It shows us how different factors affect the speed at which a reaction reaches completion. Several aspects of chemical kinetics are useful when analyzing the oxidation of \(\mathrm{HBr}\).
For instance, determining rate laws provides insight into how reactants interact. If a reaction is first-order concerning a particular reactant, doubling its concentration will double the reaction rate.
  • The overall rate of a reaction is influenced by the concentration and the nature of the substances involved.
  • Temperature can also impact reaction rates, typically increasing rates with higher temperatures.
  • Catalysts are another kinetic factor, speeding up reactions without being consumed.
Understanding these factors helps in controlling reactions, especially in industrial applications where efficiency and yield are crucial.
Reaction Intermediates
Intermediates in a chemical reaction are fleeting species formed and consumed during the different steps of a reaction mechanism. They play a pivotal but temporary role in the transition from reactants to products.
In the gas-phase oxidation of \(\mathrm{HBr}\), intermediates such as \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\) are crucial for the mechanism's progression.
  • \(\mathrm{HOOBr}\): Formed in the first step and reacts in the second.
  • \(\mathrm{HOBr}\): Produced in the second step, consumed in the third.
Even though they don't appear in the net reaction, the presence and understanding of these intermediates are vital. Because they are short-lived and unstable, intermediates might not be detected at the end of the reaction. However, their roles provide essential clues to how a reaction proceeds and offer insights for optimizing reaction conditions.

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Most popular questions from this chapter

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .\) This rapid reaction gives the following rate data: \begin{tabular}{lll} \hline \(\left.\mathrm{OCl}^{-}\right](\mathrm{M})\) & {\(\left[\mathrm{I}^{-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline \(1.5 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(1.36 \times 10^{-4}\) \\ \(3.0 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \(1.5 \times 10^{-3}\) & \(3.0 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=\) \(2.0 \times 10^{-3} M\) and \(\left[1^{-}\right]=5.0 \times 10^{-4} M\)

A flask is charged with \(0.100 \mathrm{~mol}\) of \(\mathrm{A}\) and allowed to react to form B according to the hypothetical gas-phase reaction \(\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})\). The following data are collected: \begin{tabular}{lccccc} \hline Time (s) & 0 & 40 & 80 & 120 & 160 \\ \hline Moles of A & \(0.100\) & \(0.067\) & \(0.045\) & \(0.030\) & \(0.020\) \\ \hline \end{tabular} (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 40 -s interval, in units of \(\mathrm{mol} / \mathrm{s}\). (c) What additional information would be needed to calculate the rate in units of concentration per time?

Consider two reactions. Reaction (1) has a constant halflife, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose. When the concentration of invertase is \(4.2 \times 10^{-7} \mathrm{M}\) and the concentration of sucrose is \(0.0077 \mathrm{M}\) invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s}\). When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme-substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?
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