Problem 117 - Chemistry The Molecular Science

Question

For each reaction listed with its rate law, propose a reasonable mechanism. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{l} \text { CH }_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH} \text { (aq) } \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \\ \text {(b) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \end{array} $$ For each reaction listed with its rate law, propose a reasonable mechanism. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{l} \text { CH }_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq}) \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \\ \text {(b) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \end{array} $$

Step-by-Step Solution

Answer

(a) Involves acid-catalyzed ester hydrolysis; (b) is single-step bimolecular.

Step 1: Understanding Rate Law for Part (a)

The rate law given is \( \text{Rate} = k \left[ \mathrm{CH}_{3}\mathrm{COOCH}_{3} \right]\left[ \mathrm{H}^{+} \right] \). This suggests that the reaction involves the methyl acetate \( \mathrm{CH}_{3}\mathrm{COOCH}_{3} \) and is catalyzed by the hydrogen ion \( \mathrm{H}^{+} \). The reaction is first order in each of these species.

Step 2: Proposing Mechanism for Reaction (a)

Given the rate law, a reasonable mechanism is:1. \( \mathrm{CH}_{3}\mathrm{COOCH}_{3} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{C^(+)OOCH}_{3} \) \, (fast equilibrium, protonation)2. \( \mathrm{CH}_{3} \mathrm{C^(+)OOCH}_{3} + \mathrm{H_2O} \rightarrow \mathrm{CH}_{3} \mathrm{OH}+ \mathrm{CH}_{3} \mathrm{C(OH)OO} \) \, (slow, rate determining step)3. \( \mathrm{CH}_{3} \mathrm{C(OH)OO} \rightarrow \mathrm{CH}_{3} \mathrm{COOH} \) \, (fast)This mechanism accounts for the rate law because the overall rate is determined by the slow step, which involves \( \mathrm{CH}_{3}\mathrm{COOCH}_{3} \) and the equilibrium concentration of \( \mathrm{H}^{+} \).

Step 3: Understanding Rate Law for Part (b)

The rate law given is \( \text{Rate} = k \left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \), indicating that the reaction is first-order with respect to both \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \). This suggests a single-step, bimolecular mechanism.

Step 4: Proposing Mechanism for Reaction (b)

Given the rate law, a reasonable mechanism is a simple one-step process:\[ \mathrm{H}_{2} + \mathrm{I}_{2} \rightarrow 2 \mathrm{HI} \]This single-step mehanism directly reflects the rate law, as it involves both \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \) reacting in a bimolecular elementary step.

Key Concepts

Rate Law

In chemical kinetics, the rate law is an equation that relates the rate of a reaction to the concentration of reactants. It's typically written as \( ext{Rate} = k[ ext{A}][ ext{B}] \), where \( k \) is the rate constant, and \([\text{A}]\) and \([\text{B}]\) are the concentrations of the reactants.
The rate law indicates how the concentration of a substance affects the rate of a chemical reaction. In our given exercise, the rate for reaction (a) is influenced by both \( \mathrm{CH}_{3}\mathrm{COOCH}_{3} \) and \( \mathrm{H}^{+} \), whereas in reaction (b), it depends on \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \).
By determining the rate law, chemists can infer the involvement of substances in the reaction process and propose mechanisms consistent with the observed data.

Reaction Mechanism

Reaction mechanisms provide a step-by-step pathway through which reactants transform into products. They are crucial for explaining the observed rate laws and suggest how and why species in a reaction behave the way they do.
The mechanism for reaction (a) involves three steps, beginning with the protonation of methyl acetate, forming an intermediate. This is followed by a reaction with water, which forms methanol and another intermediate that eventually turns into acetic acid. Each step correlates with observed reaction speeds.
For reaction (b), the mechanism is simpler, assumed to be a single step where hydrogen and iodine gases react directly to form hydrogen iodide, consistent with the bimolecular rate law.

Bimolecular Reactions

Bimolecular reactions involve two reactant molecules coming together to form products. These reactions are defined by their elementary steps, where exactly two molecules interact.
In the exercise, the reaction between hydrogen and iodine (reaction (b)) is a bimolecular reaction evident from its rate law \( \text{Rate} = k[ ext{H}_2][\text{I}_2] \). This indicates that the reaction proceeds through a single collision event between one \( \text{H}_2 \) and one \( \text{I}_2 \) molecule.
Bimolecular reactions are essential for understanding simple reaction processes and serve as the building blocks for more complex mechanisms.

Catalysis

Catalysis is an acceleration of a chemical reaction by a catalyst, a substance that is not consumed in the reaction. Catalysts provide an alternative reaction pathway with lower activation energy.
In reaction (a), the hydrogen ion \( \mathrm{H}^{+} \) acts as a catalyst, facilitating the breakdown of methyl acetate into methanol and acetic acid by speeding up the reaction without being consumed.
Understanding how catalysts work in the reaction pathway helps chemists devise faster and more efficient processes in industrial and laboratory settings.

Reaction Order

The order of a reaction refers to the power to which a reactant concentration is raised in the rate law. It helps describe how the rate is affected by the concentration of reactants.
For example, in reaction (a), the order is first with respect to both \( \mathrm{CH}_{3}\mathrm{COOCH}_{3} \) and \( \mathrm{H}^{+} \), which makes it a second-order overall. However, in reaction (b), the order is first for each \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \).
Knowing the reaction order is crucial for predicting how changes in concentration can influence the overall reaction rate, allowing for better control in practical applications.