Question

You have studied the gas-phase oxidation of 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: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (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) Adding the elementary reactions gives 4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g), which is the overall reaction. (b) The rate-determining step is the first step, HBr(g) + O₂(g) → HOOBr(g), as it aligns with the observed rate law. (c) The intermediates in this mechanism are HOOBr and HOBr. (d) Not detecting HOBr or HOOBr among the products does not necessarily disprove the mechanism, as they are intermediates and can react to form other products before detection.

Step-by-step solution

(a) Confirm that the elementary reactions add to give the overall reaction.

To confirm that the elementary reactions give the overall reaction, we need to add up the elementary reactions and check if they equal the overall reaction. 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) Adding these reactions: HBr(g) + O₂(g) + HOOBr(g) + HBr(g) + 2 HOBr(g) → HOOBr(g) + 2 HOBr(g) + H₂O(g) + Br₂(g) Now, eliminate the intermediate species (HOOBr and HOBr) from both sides: HBr(g) + O₂(g) + HBr(g) → H₂O(g) + Br₂(g) Simplifying: 4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g) which is the overall reaction.

(b) Rate law determination

The experimentally determined rate law is first order for HBr and O₂. Thus, the rate law can be represented as: Rate = k[HBr][O₂] To determine which step is the rate-determining step, we need to find the step with the slowest rate. In most cases, the rate-determining step aligns with the observed rate law. Since the rate law is showing first order dependence on both reactants (HBr and O₂), it's most likely the first step is rate-determining, as it involves both reactants: HBr(g) + O₂(g) → HOOBr(g)

(c) Intermediates in the mechanism

Intermediates are species that are formed during the mechanism and are consumed before the final products are produced. To identify the intermediates, we look for species that appear on both sides of the combined elementary reaction equations. In this case, the intermediates are: 1. HOOBr (formed in the first step and consumed in the second step) 2. HOBr (formed in the second step and consumed in the third step)

(d) Detection of HOBr or HOOBr among the products

If we cannot detect HOBr or HOOBr among the products, it does not necessarily disprove our mechanism. As pointed out earlier, these species are intermediates, which means they can react and form other products, including the final products. Therefore, detecting them in the final products may not be straightforward. However, lack of detection might encourage further investigation to find alternative mechanisms or experimental techniques to identify these intermediates.

Key concepts

Gas-Phase Oxidation

Gas-phase oxidation refers to a chemical process where a compound, typically an organic or inorganic molecule, is oxidized by oxygen in the gas phase. This type of reaction is fundamental in fields such as atmospheric chemistry, combustion, and industrial processes.

In the context of the exercise, the gas-phase oxidation of hydrogen bromide (HBr) by oxygen (O_{2}) is described. Understanding gas-phase reactions is crucial since they can involve a complex sequence of steps, making them more challenging than their liquid-phase counterparts. The reactions involve the molecular collisions and the formation of transient reaction intermediates which are not found in the final products of the reaction. This means that, while we may observe the starting materials and the end products, the intermediates are typically too short-lived to be isolated under normal conditions.

Rate-Determining Step

The concept of the rate-determining step is critical in understanding chemical kinetics and reaction mechanisms. It's the slowest step in a sequence of reactions that determines the overall rate at which a reaction proceeds. This step, often compared to a bottleneck in a process, acts as the limiting factor for the reaction rate.

In our exercise, we are told that the gas-phase oxidation reaction is first order with respect to both HBr and O_{2}. From the provided steps, we infer that the first step – the formation of HOOBr from HBr and O_{2} – is the rate-determining step since it's the only step that involves both reactants directly. Therefore, under these conditions, the reaction rate is dependent on how fast this step occurs, with the subsequent steps proceeding at a faster rate.

Reaction Intermediates

In a chemical reaction mechanism, reaction intermediates are the species that form in one step and are consumed in subsequent steps before the formation of the final products. These are crucial for understanding how reactions proceed on a molecular level but are often challenging to identify experimentally because of their typically transient nature.

Within the solution to the exercise, we identify HOOBr and HOBr as intermediates. As the exercise explains, even if we can't detect them in the final products, this doesn't rule out their presence in the reaction mechanism. Instead, it reinforces the nature of intermediates: present during the reaction but not in the end products due to further reaction steps. In our example, HOOBr is formed first and then consumed by reaction with HBr to form HOBr, which is, in turn, consumed to finally form H_{2}O and Br_{2}. The investigation into such intermediates is vital for the complete understanding of the reaction kinetics and mechanism.