Question

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 HBr and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (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

(a) The elementary reactions do add up to give 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: \(\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \rightarrow \mathrm{HOOBr}(g)\). (c) The intermediates in the mechanism are HOOBr(g) and HOBr(g). (d) Not detecting HOBr or HOOBr among the products does not necessarily disprove the proposed mechanism, as they are intermediates that could be quickly converted into the final products. The mechanism is considered valid until an inconsistency is found or a more accurate mechanism is proposed.

Step-by-step solution

a) Confirming that the elementary reactions result in the overall reaction

Add all the elementary reactions together and simplify by canceling out any species present on both sides of the reactions: Step 1: \( \mathrm{HBr} + \mathrm{O}_{2} \longrightarrow \mathrm{HOOBr} \) Step 2: \( \mathrm{HOOBr} + \mathrm{HBr} \longrightarrow 2 \mathrm{HOBr} \) Step 3: \(\mathrm{HOBr} + \mathrm{HBr} \longrightarrow \mathrm{H}_{2}\mathrm{O} + \mathrm{Br}_{2} \) Adding these reactions, we have: \( \cancel{4}\mathrm{HBr} + \mathrm{O}_{2} \cancel{\longrightarrow} \cancel{\mathrm{HOOBr}} + \cancel{2}\mathrm{HOBr} \cancel{\longrightarrow} 2\mathrm{H}_{2}\mathrm{O} + 2\mathrm{Br}_{2} \) After cancellation, we get the overall reaction: \( 4 \mathrm{HBr}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(g) + 2 \mathrm{Br}_{2}(g) \) The elementary reactions do add up to give the overall reaction.

b) Determining the rate-determining step

Since the overall reaction is first-order with respect to HBr and first-order with respect to O2, the rate equation can be written as: Rate = k [HBr][O2] Rate-determining step is the slowest step in the reaction mechanism. We will assume one of the steps as the slowest step and compare its rate equation with the overall rate equation. If we assume Step 1 is the rate-determining step, the rate equation would be: Rate = k_[1] [HBr][O2] Since this matches the overall rate equation, the rate-determining step is the first step: HBr(g) + O2(g) → HOOBr(g)

c) Identifying the intermediates in the mechanism

Intermediates are species that are formed during the reaction and then consumed before the reaction completes. In the given mechanism, the intermediates are: 1. HOOBr(g) - formed in Step 1 and consumed in Step 2 2. HOBr(g) - formed in Step 2 and consumed in Step 3

d) Discussing if not detecting HOBr or HOOBr among the products disproves the mechanism

Not detecting HOBr or HOOBr among the products does not necessarily disprove the proposed mechanism. The reason is that these species are intermediates, and they are formed and consumed during the reaction. It is possible that they get quickly converted into the final products and, thus, are not detected. It is also worth noting that there might be other possible mechanisms for the given reaction. The given mechanism is considered valid until an inconsistency is found, or a more accurate (experimentally verified) mechanism is proposed.

Key concepts

Chemical Kinetics

Understanding how chemical reactions occur requires diving into the realm of chemical kinetics, which is the study of reaction rates and the factors that affect them. When examining the gas-phase oxidation of HBr by O₂, we see that kinetics plays a crucial role in determining the speed and outcome of the reaction. The fact that the reaction between HBr and O₂ is first order with respect to each reactant indicates how the concentration of these substances influences the reaction rate.

In practical terms, the rate law can be thought of as a recipe for the reaction rate, with each ingredient (reactant) contributing to the speed at which the reaction proceeds. For our HBr oxidation reaction, the rate law would be written as: Rate = k [HBr][O₂], where k is the rate constant, and the concentrations are signified by the square brackets. It's this precise, quantifiable relationship that allows chemists to predict how quickly a reaction will go under varying concentrations of reactants.

Reaction Intermediates

Dive into the chemical reaction mechanism, and you'll encounter a bustling microscopic metropolis of molecules and atoms rapidly forming, then disappearing - these elusive entities are known as reaction intermediates. In the given oxidation mechanism of HBr, the intermediates, specifically HOOBr and HOBr, play a pivotal but transient role, acting as the 'middle-men' of the reaction sequence.

As they are not present in the initial reactants or the final products, intermediates are the chemical equivalents of undercover agents in a spy film - essential to the narrative but vanishing before the end. Not being able to detect HOBr or HOOBr in the final product mix doesn't throw our mechanism off its rails, because their fleeting existence is part of the mechanism's design - to facilitate the transformation without sticking around for the applause.

Rate-Determining Step

In the dramatic unfolding of a chemical reaction, the rate-determining step is akin to the slowest runner in a relay race - it sets the pace for the whole sequence. It is the most time-consuming step that controls the rate at which the reaction progresses. In our specific case of HBr oxidation, we anchor our understanding by matching the experimentally determined rate law with the rate of the presumed slowest step in the proposed mechanism.

By scrutinizing the steps of the mechanism and comparing them with the Rate = k [HBr][O₂], we deduce that the initial step, HBr(g) + O₂(g) → HOOBr(g), is indeed the rate-determining step. This bottleneck step governs how fast the reaction will go, and, in essence, is the director of the reaction's tempo.

Elementary Reactions

Elementary reactions are the building blocks of a complex chemical equation, and each one is a single-step process involving a specific, straightforward change from reactants to products. These little leaps, whether they involve the collision between two molecules, the dissociation of a molecule, or a molecule capturing an electron, are the steps that choreograph the dance of the complete reaction mechanism.

In the case of gas-phase oxidation of HBr, the proposed sequence is made up of three elementary reactions, each with its own discreet reactant-to-product journey. What makes these steps truly elementary is that their stoichiometry directly corresponds to their rate law. Their simplicity allows us to piece together the full picture of the intricate chemical process that transforms humble HBr and O₂ into H₂O and Br₂.