Problem 70 - Chemistry: The Central Science

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: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \cdots & \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (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?

Step-by-Step Solution

Answer

In this exercise, we found that: (a) The given elementary reactions can add to give the overall reaction when their coefficients are properly adjusted. (b) The rate-determining step is the first elementary reaction (\(\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\)), as it matches the rate law. (c) The intermediates in this mechanism are \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\). (d) The absence of detected intermediates does not disprove the mechanism, as intermediates are typically unstable and short-lived. However, further studies are needed to validate the proposed mechanism.

Step 1: Analyzing the given mechanism and finding the overall reaction

Based on the proposed mechanism, we have three elementary reactions: 1. \(\mathrm{HBr}(g)+\mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\) 2. \(\mathrm{HOOBr}(g)+\mathrm{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)\) 3. \(\mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)\) Notice that the coefficients on equation 2 and equation 3 need to be multiplied by 2 to match the stoichiometry of the overall reaction: 2(2). \(2\mathrm{HOOBr}(g)+2\mathrm{HBr}(g) \longrightarrow 4 \mathrm{HOBr}(g)\) 2(3). \(2\mathrm{HOBr}(g)+2\mathrm{HBr}(g) \longrightarrow 2\mathrm{H}_{2} \mathrm{O}(g)+2\mathrm{Br}_{2}(g)\) Now, adding equations 1, 2(2), and 2(3) together, we obtain: \[ 4 \mathrm{HBr}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(g) + 2 \mathrm{Br}_{2}(g) \] Which is the overall reaction. So, the elementary reactions can add to give the overall reaction.

Step 2: Determining the rate-determining step

Now, we know that the rate law for the overall reaction is first order with respect to both \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\): Rate \(= k[\mathrm{HBr}][\mathrm{O}_{2}]\) Since the rate law matches the reaction 1 (\(\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\)), this step is the rate-determining step because it controls the rate of the overall reaction.

Step 3: Identifying the intermediates

The intermediates in the reaction mechanism are the species that are produced in one elementary step and consumed in another elementary step, not appearing in the overall reaction. In this mechanism, the intermediates are: 1. \(\mathrm{HOOBr}(g)\) - formed in step 1 and consumed in step 2. 2. \(\mathrm{HOBr}(g)\) - formed in step 2 and consumed in step 3.

Step 4: Analyzing the absence of detected intermediates

As for the question regarding the absence of detected intermediates, it does not disprove the mechanism. Intermediates are typically unstable and short-lived, meaning they may not be detected among the final products. However, further studies could provide more evidence to validate the proposed mechanism.

Key Concepts

Rate-Determining Step

In any multi-step chemical reaction, the rate-determining step is like the slowest runner in a relay race. It is the slowest step in a reaction mechanism. This step limits the overall speed of the reaction and determines the rate law.
In the gas-phase oxidation of \(\mathrm{HBr}\), the rate law is first order with respect to both \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\). This tells us that the rate-determining step involves one molecule of \(\mathrm{HBr}\) and one molecule of \(\mathrm{O}_{2}\).

• Step 1 of the elementary steps, \[\mathrm{HBr}(g) + \mathrm{O}_{2}(g) \cdots \mathrm{HOOBr}(g)\], fits this description.
• Since this step involves \(\mathrm{HBr}\) and \(\mathrm{O}_{2}\), it matches the first-order dependence of each in the rate law, making it the rate-determining step.

Understanding which step is rate-determining helps chemists improve reaction efficiency by focusing on speeding up this particular step.

Chemical Kinetics

Chemical kinetics is the study of reaction rates. It shows us how different factors affect the speed at which a reaction reaches completion. Several aspects of chemical kinetics are useful when analyzing the oxidation of \(\mathrm{HBr}\).
For instance, determining rate laws provides insight into how reactants interact. If a reaction is first-order concerning a particular reactant, doubling its concentration will double the reaction rate.

• The overall rate of a reaction is influenced by the concentration and the nature of the substances involved.
• Temperature can also impact reaction rates, typically increasing rates with higher temperatures.
• Catalysts are another kinetic factor, speeding up reactions without being consumed.

Understanding these factors helps in controlling reactions, especially in industrial applications where efficiency and yield are crucial.

Reaction Intermediates

Intermediates in a chemical reaction are fleeting species formed and consumed during the different steps of a reaction mechanism. They play a pivotal but temporary role in the transition from reactants to products.
In the gas-phase oxidation of \(\mathrm{HBr}\), intermediates such as \(\mathrm{HOOBr}(g)\) and \(\mathrm{HOBr}(g)\) are crucial for the mechanism's progression.

• \(\mathrm{HOOBr}\): Formed in the first step and reacts in the second.
• \(\mathrm{HOBr}\): Produced in the second step, consumed in the third.

Even though they don't appear in the net reaction, the presence and understanding of these intermediates are vital. Because they are short-lived and unstable, intermediates might not be detected at the end of the reaction. However, their roles provide essential clues to how a reaction proceeds and offer insights for optimizing reaction conditions.