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Chapter 13: Problem 99

Methyl acetate reacts in acidic solution. $$ \mathrm{CH}_{3} \mathrm{COOCH}_{3}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} $$ methyl acetate methanol acetic acid The rate law is first order in methyl acetate in acidic solution, and the rate constant at \(25^{\circ} \mathrm{C}\) is \(1.26 \times 10^{-4} / \mathrm{s}\). How long will it take for \(65 \%\) of the methyl acetate to react?

Short Answer

Expert verified
It takes approximately 2.31 hours for 65% of the methyl acetate to react.

Step by step solution

01

Understand the Reaction

The problem describes the hydrolysis of methyl acetate in an acidic solution. The balanced chemical reaction is given as \( \mathrm{CH}_3\mathrm{COOCH}_3 + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}^+} \mathrm{CH}_3\mathrm{OH} + \mathrm{CH}_3\mathrm{COOH} \). The reaction is first order in methyl acetate.
02

Determine the Rate Law

Since the reaction is first order with respect to methyl acetate, the rate law can be expressed as:\[ \text{Rate} = k [\mathrm{CH}_3\mathrm{COOCH}_3] \]where \( k \) is the rate constant.
03

Use First Order Reaction Formula

For a first order reaction, the integrated rate law is given by:\[ \ln \frac{[A]_0}{[A]} = kt \]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant.
04

Define the Percent Completion

We are asked for the time at which 65% of the methyl acetate reacts. This means that 35% of the initial concentration remains. Therefore, \([A]_0 = 1\) and \([A] = 0.35\).
05

Substitute Values and Solve for Time

Substitute the known values into the integrated rate law equation:\[ \ln \frac{1}{0.35} = 1.26 \times 10^{-4} \times t \]First calculate:\[ \ln\left(\frac{1}{0.35}\right) = \ln(2.857) \approx 1.048 \]Substitute back to find \( t \):\[ 1.048 = 1.26 \times 10^{-4} \times t \]Now solve for \( t \):\[ t = \frac{1.048}{1.26 \times 10^{-4}} \approx 8323.8 \] seconds.
06

Convert Time to Appropriate Units

Since the time is calculated in seconds, convert it to a more convenient unit if necessary. To convert seconds to hours:\[ \text{Hours} = \frac{8323.8}{3600} \approx 2.31 \] hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law
A rate law is a mathematical equation that describes how the concentration of reactants affects the rate of a chemical reaction. In the case of first order reactions, the rate depends directly on the concentration of a single reactant. For the hydrolysis of methyl acetate, the rate law can be expressed simply as:
  • Rate = k [CH鈧僀OOCH鈧僝
Here, k is the rate constant, which gives an idea of how fast the reaction proceeds under given conditions. The rate law shows that as the concentration of methyl acetate decreases, the reaction rate decreases proportionally.
Methyl Acetate Hydrolysis
Methyl acetate hydrolysis involves its reaction with water in an acidic environment. In this reaction, methyl acetate (CH鈧僀OOCH鈧�) converts into methanol (CH鈧僌H) and acetic acid (CH鈧僀OOH). The process is facilitated by the presence of a catalyst, represented by protons (H鈦�), from the acid in the solution. During the hydrolysis:
  • Water breaks the ester bond in methyl acetate.
  • Methanol and acetic acid are formed as products.
This reaction is common in organic chemistry and is important for creating various commercial products and intermediaries.
Reaction Kinetics
The study of reaction kinetics involves understanding the speed or rate at which chemical reactions occur. For methyl acetate, the kinetics is such that it follows a first order reaction pattern. In first order kinetics, time plays a crucial role in determining how much reactant remains at any given moment. The relationship can be understood using an integrated rate equation given by:
  • \( \ln \frac{[A]_0}{[A]} = kt \)
Where \([A]_0\) is the initial concentration of the reactant, \([A]\) is the concentration remaining at time t, and k is the rate constant specific to the reaction at a given temperature.
Acidic Hydrolysis
Acidic hydrolysis is a type of reaction where water and an acid are used to break chemical bonds, such as the ester bond in methyl acetate. The acid (providing H鈦� ions) acts as a catalyst, speeding up the reaction without being consumed. During acidic hydrolysis:
  • The presence of H鈦� ions helps stabilize transition states, making it easier for bonds to break.
  • The stronger the acid, the more effective it is as a catalyst.
  • This method is widely used to prepare carboxylic acids and alcohols from esters.
The reaction of methyl acetate in acidic hydrolysis results in methanol and acetic acid, highlighting the importance of acids in catalyzing such transformations.

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Most popular questions from this chapter

Cyclopropane, \(\mathrm{C}_{3} \mathrm{H}_{6},\) is converted to its isomer propylene, \(\mathrm{CH}_{2}=\mathrm{CHCH}_{3}\), when heated. The rate law is first order in cyclopropane, and the rate constant is \(6.0 \times 10^{-4} / \mathrm{s}\) at \(500^{\circ} \mathrm{C}\). If the initial concentration of \(\mathrm{cy}-\) clopropane is \(0.0226 \mathrm{~mol} / \mathrm{L},\) what is the concentration after \(525 \mathrm{~s}\) ?

To obtain the rate of the reaction $$ 5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ you might follow the \(\mathrm{Br}^{-}\) concentration or the \(\mathrm{BrO}_{3}^{-}\) concentration. How are the rates in terms of these species related?

The rate law for reaction \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{O}_{2}(g)+\) \(\mathrm{NO}_{2}(g)\) is first order in \(\mathrm{O}_{3}\) and first order in NO. The reaction is exothermic, and the temperature dependence of the rate constant can be modeled by the Arrhenius equation. Which of the following statements regarding this reaction are true as the temperature of the reaction is increased (choose all that apply)? a. The reaction rate would increase. b. The activation energy would increase. c. A greater percentage of the molecular collisions between \(\mathrm{O}_{3}\) and \(\mathrm{NO}\) in a reaction mixture would exceed the activation energy. d. The reaction would become more exothermic. e. The magnitude of the rate constant would increase.

The decomposition of nitrogen dioxide, $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ has a rate constant of \(0.498 \mathrm{M} / \mathrm{s}\) at \(319^{\circ} \mathrm{C}\) and a rate constant of \(1.81 \mathrm{M} / \mathrm{s}\) at \(354^{\circ} \mathrm{C}\). What are the values of the activation energy and the frequency factor for this reaction? What is the rate constant at \(420^{\circ} \mathrm{C} ?\)

Write a rate equation, showing the dependence of rate on reactant concentrations, for each of the following elementary reactions. a. \(\mathrm{CS}_{2} \longrightarrow \mathrm{CS}+\mathrm{S}\) b. $\mathrm{CH}_{3} \mathrm{Br}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{Br}^
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