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

(a) When the coefficients in the second elementary reaction are multiplied by 2, 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 elementary reaction, \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HOOBr}(g)\). (c) The intermediates in the mechanism are \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\). (d) Absence of \(\mathrm{HOBr}\) or \(\mathrm{HOOBr}\) among the products does not necessarily disprove the proposed mechanism, as they may rapidly decompose, react with other species, or be present at very low concentrations.

Step-by-step solution

a. Adding Elementary Reactions

To see how the elementary reactions add up to the overall reaction, we first write the given elementary reactions: 1) \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HOOBr}(g)\) 2) \(\mathrm{HOOBr}(g)+\mathrm{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)\) 3) \(\mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)\) Following the hint, we will multiply the coefficients of the second equation by 2: 2) \(2(\mathrm{HOOBr}(g)+\mathrm{HBr}(g)) \longrightarrow 4 \mathrm{HOBr}(g)\) Now, we sum up the modified elementary equations: 1) + 2) \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HOOBr}(g)\) 2) \(2(\mathrm{HOOBr}(g)+\mathrm{HBr}(g)) \longrightarrow 4 \mathrm{HOBr}(g)\) 3) \(\mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)\) \(\,\) \(\rightarrow 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)\) Thus, we find that these elementary reactions add up to the overall reaction when we multiply the coefficients in the second elementary reaction by 2.

b. Rate-Determining Step

The reaction is first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). Therefore, the rate law is given by: Rate = k[\(\mathrm{HBr}\)][\(\mathrm{O}_{2}\)] Since the reaction is given to be first order with respect to both reactants, the rate-determining step must be the one involving \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\). Looking at the elementary reactions, we see that the first reaction, \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HOOBr}(g)\), involves both reactants. Therefore, the first elementary reaction is the rate-determining step.

c. Intermediates in the Mechanism

Intermediates are species that are formed in one reaction of the mechanism and consumed in a following reaction. In this mechanism, we have: \(\mathrm{HOOBr}(g)\): Produced in the first reaction and consumed in the second reaction. \(\mathrm{HOBr}(g)\): Produced in the second reaction and consumed in the third reaction. So, the intermediates in the mechanism are \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\).

d. Absence of Intermediates in the Products

If we are unable to detect \(\mathrm{HOBr}\) or \(\mathrm{HOOBr}\) among the products, it does not necessarily disprove the proposed mechanism. The absence of intermediates in the products could mean that they either rapidly decompose or react with other species present in the system. The intermediates might also be present at very low concentrations, making it difficult to detect them. Therefore, the absence of intermediates in the products does not disprove the proposed mechanism.

Key concepts

Rate-Determining Step

The rate-determining step in a chemical mechanism is akin to the slowest runner in a relay race. Just as the entire team's time depends on this runner, the speed of the overall reaction hinges on this single step. It controls the reaction rate and determines how fast the whole process occurs.

In the scenario of The oxidation of oPrimitive steps in our mechanism, the first step stands out. It involves OPrations undergone by on Systematic reactions indicate that the combination of Ostep as the rate-determining step because it both reactants

Understanding which step is the slowest is essential for chemists because altering the conditions of this step can accelerate or decelerate the entire reaction. It provides vital insights for refining industrial chemical processes to enhance efficiency.

Chemical Intermediates

Chemical intermediates are temporary species formed during a reaction but do not appear in the final products. Imagine them as travelers passing through stations—the intermediate species appear and disappear as the reaction progresses.

In our Ointermediates in the given mechanism are Formation of these species temporarily stabilizes elsewhere in the reaction pathway.

• HOOBr(g): It is formed in the first step and then consumed in the subsequent second step.
• HOBr(g): This intermediate is created in the second step and used up in the third.

Their presence is crucial as they reflect the ongoing reaction's pathway. However, because intermediates are consumed in subsequent steps, they are often challenging to detect and study.

Detecting or not detecting these intermediates doesn't necessarily validate or disprove a reaction mechanism. Their transient nature means they might exist in concentrations too low for detection or quickly transform into other products, thus escaping observation.

Elementary Reactions

Elementary reactions form the backbone of complex chemical processes. They represent single events where reactants directly convert into products in a single step. Think of them as the building blocks of a complete reaction, each representing a distinct leap.

In our given example:

• The first elementary reaction involves the combination of The overall reaction starts with this basic move.
• The second step, modified to double the coefficients, shows that two subsequent occurrences of intermediate formation and conversion lead to the next species.
• The third step then finalizes the transformation leading to the ultimate products of the reaction.

By examining these small steps, chemists can build the entire picture of how a chemical reaction unfolds. Understanding elementary reactions allows scientists to deduce the overall reaction's behavior and predict its dynamics.