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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: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \longrightarrow \operatorname{HOOBr}(g) \\\ \mathrm{HOOBr}(g)+\operatorname{HBr}(g) & \longrightarrow 2 \operatorname{HOBr}(g) \\ \mathrm{HOBr}(g)+\operatorname{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (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
In summary, the given mechanism's elementary reactions do add up to the overall reaction, and the rate-determining step is the first step involving both HBr and O2. The intermediates are HOOBr and HOBr. Non-detection of these intermediates does not necessarily disprove the mechanism, as they could be reacting quickly or have low concentrations, and further experiments would be needed for a conclusive decision.

Step by step solution

01

Confirm that the elementary reactions add to give the overall reaction

To verify that the elementary reactions given in the mechanism add up to the overall reaction, we will add them up and see if the result matches the overall equation. The elementary reactions are: 1. HBr(g) + O2(g) -> HOOBr(g) 2. HOOBr(g) + HBr(g) -> 2 HOBr(g) 3. HOBr(g) + HBr(g) -> H2O(g) + Br2(g) To add these reactions, let's cancel out any species that do not appear in the overall reaction: HBr(g) + O2(g) + HBr(g) + HBr(g) -> H2O(g) + Br2(g) Now combining the species on the left-hand side: 4 HBr(g) + O2(g) -> H2O(g) + Br2(g) This matches the given overall reaction, so the elementary reactions do add up to the overall reaction.
02

Identify the rate-determining step

The experimentally determined rate law for this reaction is first-order with respect to HBr and first-order with respect to O2. This implies that the rate-determining step must involve both HBr and O2: rate = k[HBr][O2] Comparing the given elementary reactions, we can see that the first step involves both HBr and O2: 1. HBr(g) + O2(g) -> HOOBr(g) Thus, the first step is the rate-determining step.
03

Identify the intermediates

Intermediates are species that are produced in one step of a reaction mechanism and consumed in a later step. To identify the intermediates, we look at the given steps and find which species get canceled out when added up. In this mechanism, the intermediates are: 1. HOOBr(g): Produced in step 1 and consumed in step 2. 2. HOBr(g): Produced in step 2 and consumed in step 3.
04

Discuss if non-detection of HOBr or HOOBr disproves the mechanism

If we are unable to detect HOBr or HOOBr among the products, it does not necessarily disprove the mechanism. It is possible that these two species are being formed as intermediates and that they react quickly enough to form the final products that we are unable to detect them experimentally. It is also possible that the concentrations of these intermediates are too low to be detected. Non-detection of intermediates is only one piece of evidence, and more experiments and data would be needed to conclusively disprove the proposed mechanism.

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Key Concepts

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

Elementary Reactions
Elementary reactions are the individual steps or stages that combine to form the complete mechanism of a complex chemical reaction. Each elementary reaction represents a single molecular event, such as the collision between molecules, which leads to the formation or breaking of bonds.
In the context of the gas-phase oxidation of hydrogen bromide (HBr) by oxygen (O₂), we have three proposed elementary reactions:
  • 1. HBr(g) + O₂(g) → HOOBr(g)
  • 2. HOOBr(g) + HBr(g) → 2 HOBr(g)
  • 3. HOBr(g) + HBr(g) → H₂O(g) + Br₂(g)
These reactions collectively sum up to the overall reaction: 4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g).
Verification involves ensuring that the sum of these individual reactions removes any intermediate species, leaving only the reactants and products of the overall reaction.
Rate-Determining Step
The rate-determining step (RDS) is the slowest step in a reaction mechanism. It acts like a bottleneck, significantly influencing the overall reaction rate. For a given mechanism, the experimentally observed rate law can guide us in identifying the RDS.
In this exercise, the reaction has been observed to be first-order with respect to both HBr and O₂, suggesting that both species are involved in the rate-determining step. When examining the proposed sequence of elementary reactions:
  • 1. HBr(g) + O₂(g) → HOOBr(g)
  • 2. HOOBr(g) + HBr(g) → 2 HOBr(g)
  • 3. HOBr(g) + HBr(g) → H₂O(g) + Br₂(g)
Only the first step involves both HBr and O₂ directly. Therefore, the first step is likely to be the rate-determining step. This is because it matches the observed rate law: rate = k[HBr][O₂], indicating both reactants are involved in the formation of the product, influencing the speed of the whole process.
Reaction Intermediates
Reaction intermediates are transient species formed during the course of a chemical reaction. They appear in the reaction pathways but not in the overall equation, usually because they are consumed in subsequent steps.
In the discussed mechanism, two intermediates appear:
  • **HOOBr(g):** Formed in the first step and consumed in the second.
  • **HOBr(g):** Produced in the second step and consumed in the third.
Intermediates are crucial for understanding the detailed pathway of reactions, although they are often difficult to detect experimentally. The inability to detect intermediates such as HOBr or HOOBr in the product mixture does not necessarily invalidate the mechanism. This may be because:
  • They convert quickly into products,
  • Their concentrations remain very low, making them hard to pinpoint with conventional techniques.
Thus, further experiments may be needed to determine the presence and role of these intermediates.

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Most popular questions from this chapter

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C}\) ? (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a \(1.00\)-mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00-mg sample of each isotope remains after 4 days?

You perform a series of experiments for the reaction \(A \longrightarrow B+C\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of 8 .

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) \(\mathrm{Do}\) we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

(a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is \(160 \mathrm{~kJ} / \mathrm{mol}\). Calculate the fraction of methyl isonitrile molecules that has an energy of \(160.0 \mathrm{~kJ}\) or greater at \(500 \mathrm{~K}\). (b) Calculate this fraction for a temperature of \(520 \mathrm{~K}\). What is the ratio of the fraction at \(520 \mathrm{~K}\) to that at \(500 \mathrm{~K} ?\)

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