Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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: $$ \begin{aligned} \operatorname{HBr}(g)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{HOOBr}(g) \\\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined 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
(a) Yes, the elementary reactions add up to the overall reaction: \(4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(g)+2 \mathrm{Br}_{2}(g)\). (b) The rate-determining step is the first step: \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{HOOBr}(g)\). (c) The intermediates in this mechanism are \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\). (d) Not detecting \(\mathrm{HOBr}\) or \(\mathrm{HOOBr}\) among the products does not necessarily disprove the proposed mechanism because they are intermediates and might be present in very low concentrations or decompose into other products during the reaction.

Step by step solution

01

(a) Confirming the elementary reactions add to the overall reaction

To confirm that the given elementary reactions add up to the overall reaction, we can simply add the equations for each reaction and then cancel out any species that appear on both the reactant and product sides. Let's do that: Step 1: HBr(g) + O2(g) → HOOBr(g) Step 2: HOOBr(g) + HBr(g) → 2 HOBr(g) Step 3: HOBr(g) + HBr(g) → H2O(g) + Br2(g) Adding all three equations: [HBr(g) + O2(g)] + [HOOBr(g) + HBr(g)] + [HOBr(g) + HBr(g)] → [HOOBr(g)] + [2 HOBr(g)] + [H2O(g) + Br2(g)] After canceling out the species that appear on both sides: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g) The overall reaction obtained by adding the elementary reactions is the same as the given overall reaction. So the answer is: yes, the elementary reactions add up to the overall reaction.
02

(b) Determining the rate-determining step

The reaction is found to be first order with respect to HBr and first order with respect to O2. The rate law can be expressed as: Rate = k[HBr]^1[O2]^1 To find the rate-determining step, we need to look at the proposed mechanism and identify the step that involves HBr and O2 as reactants. In this case, it is the first step: HBr(g) + O2(g) → HOOBr(g) Since the rate law agrees with this step's reactants, it is the rate-determining step.
03

(c) Identifying intermediates in the mechanism

Intermediates are the species that are produced in one step and consumed in a subsequent step. In the proposed mechanism, we can identify the following intermediates: 1. HOOBr(g): It is produced in Step 1 and consumed in Step 2. 2. HOBr(g): It is produced in Step 2 and consumed in Step 3. Hence, the intermediates in this mechanism are HOOBr(g) and HOBr(g).
04

(d) Impact of not detecting HOBr or HOOBr on the proposed mechanism

Not being able to detect HOBr or HOOBr among the products does not necessarily disprove the proposed mechanism. The reason is that these two species are intermediates, which means they are produced and consumed during the reaction process, and they might be present in very low concentrations. They could be consumed faster than they are produced or might decompose into other products during the reaction. Therefore, their absence in the final products does not disprove the proposed mechanism. Additional experiments and analyses would be required to confirm or disprove the mechanism.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Gas-phase Reactions
Gas-phase reactions occur when the reactants involved are gases. These types of reactions are often encountered in atmospheric and industrial processes. The advantage of gas-phase reactions is that the reactants can mix more thoroughly, allowing for more uniform reactions compared to reactions in the solid or liquid phase.

In the case of the oxidation of hydrogen bromide ( HBr) by oxygen ( O_2), the reaction produces water (H_2O) and bromine (Br_2), both in the gaseous phase. The ability of gases to move and collide freely results in potentially faster reactions, but it also requires appropriate conditions such as temperature and pressure to be maintained.
  • Gases have high kinetic energy, allowing molecules to mix and collide effectively.
  • Collisions between gas molecules can lead to chemical reactions when there is enough energy.
  • Conditions like temperature and pressure influence the rate and success of these reactions.
Reaction Mechanisms
Reaction mechanisms provide a step-by-step sequence of elementary reactions by which the overall chemical change occurs. Each elementary reaction involves a few reactants and products, and collectively, these steps explain how the overall reaction proceeds.

Understanding the mechanism helps in identifying the steps where reactants are converted into products. In our example, the oxidation of HBr involves multiple elementary steps, where each step represents a simple reaction that may involve the formation of intermediate species. This mechanistic view allows chemists to comprehend not just how the reactants produce the final products, but also what happens in between.
  • Mechanisms are a series of elementary reactions.
  • They offer insight into the path of a reaction.
  • Each step in a mechanism usually involves only two molecules interacting at once.
Reaction Intermediates
Reaction intermediates are species that are formed in one step of a reaction mechanism and consumed in a subsequent step. They are crucial for understanding the full mechanism but do not appear in the overall chemical equation because they are not present at the start or end of the reaction.

In our specific reaction, the intermediates identified are HOOBr and HOBr. These intermediates are important because they transform into other species or enable further reactions to take place. Often, intermediates are unstable, which makes them challenging to detect experimentally, and they usually exist only transiently during the reaction sequence.
  • Intermediates do not exist in the final reaction equation.
  • They appear and vanish throughout the process.
  • Detection of intermediates can be difficult due to their transient nature.
Rate-determining Step
The rate-determining step is the slowest step in a reaction mechanism and essentially controls the overall rate of the chemical reaction. It acts as a bottleneck, and the rate of the overall reaction cannot surpass this step.

For the oxidation of HBr by O_2, the first step involving the combination of HBr and O_2 into HOOBr is the rate-determining step, as indicated by the rate law. Since the reaction is first order concerning both reactants ( HBr and O_2), this supports the evidence that this initial step limits how fast the overall reaction can occur.
  • The slowest step is the rate-determining step.
  • This step dictates the speed of the entire reaction.
  • Analyzing the rate law helps to identify the rate-determining step.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O}\). At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1} .\) Calculate the half- life at this temperature. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

The human body is characterized by an extremely complex system of interrelated chemical reactions. A large number of enzymes are necessary for many of these reactions to occur at suitable rates. Enzymes are very selective in the reactions they catalyze, and some are absolutely specific. Use the lock-and- key model to account for the specificity of an enzyme.

(a) In which of the following reactions would you expect the orientation factor to be more important in leading to reaction: \(\mathrm{O}_{3}+\mathrm{O} \longrightarrow 2 \mathrm{O}_{2}\) or \(\mathrm{NO}+\mathrm{NO}_{3} \longrightarrow 2 \mathrm{NO}_{2} ?\) (b) What is related to the orientation factor? Which, smaller or larger ratio of effectively oriented collisions to all possible collisions, would lead to a smaller orientation factor?

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. Step \(1: \quad \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (fast) Step \(2: \quad \mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)\) (slow) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is \(\mathrm{O}\) a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) $$ \text { is } 8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} $$ (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free