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Chapter 12: Problem 34

A certain reaction has the following general form: \(\mathrm{aA} \longrightarrow \mathrm{bB}\) At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M\), concentration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{~min}^{-1}\). a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of \(\mathrm{A}\) to decrease to \(2.50 \times 10^{-3} M ?\)

Short Answer

Expert verified
a. The rate law is \(Rate = k[A]\), the integrated rate law is \(ln[A] = -kt + ln[A]_0\), and the rate constant \(k = 2.97 \times 10^{-2} \mathrm{min}^{-1}\). b. The half-life of the reaction is approximately 23.37 minutes. c. Approximately 85.94 minutes are required for the concentration of A to decrease to \(2.50 \times 10^{-3} M\).

Step by step solution

01

Identifying the rate law and the integrated rate law for the reaction based on the graph

Since there is a straight line in a plot of ln[A] versus time, the reaction rate can be described by a first-order rate law: Rate = k[A] where k is the rate constant and [A] is the concentration of A at any time. The integrated rate law for a first-order reaction is: ln[A] = -kt + ln[A]₀ where [A]₀ is the initial concentration of A and t is the time elapsed.
02

Calculate the rate constant k using the given slope

Since the slope of the plot of ln[A] vs time is -2.97 × 10⁻² min⁻¹, we can equate the slope to -k and solve for k: k = 2.97 × 10⁻² min⁻¹
03

Calculate the half-life of the reaction

The half-life t₁/₂ of a first-order reaction is given by the formula: t₁/₂ = ln(2) / k Using the calculated k value, we can find the half-life: t₁/₂ = ln(2) / (2.97 × 10⁻² min⁻¹) t₁/₂ ≈ 23.37 minutes
04

Calculate the time required for [A] to decrease to 2.50 × 10⁻³ M

We can use the integrated rate law to find the time required for the concentration of A to decrease to a specific value: ln([A] / [A]₀) = -kt Using the given [A]₀, k, and target [A] value: ln((2.50 × 10⁻³ M) / (2.00 × 10⁻² M)) = -(2.97 × 10⁻² min⁻¹) * t Solving for t: t ≈ 85.94 minutes In summary: a. The rate law is Rate = k[A], the integrated rate law is ln[A] = -kt + ln[A]₀, and the rate constant k is 2.97 × 10⁻² min⁻¹. b. The half-life of the reaction is approximately 23.37 minutes. c. Approximately 85.94 minutes are required for the concentration of A to decrease to 2.50 × 10⁻³ M.

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Most popular questions from this chapter

Provide a conceptual rationale for the differences in the half-lives of zero-, first-, and second-order reactions.

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

The activation energy for a reaction is changed from \(184 \mathrm{~kJ} / \mathrm{mol}\) to \(59.0 \mathrm{~kJ} / \mathrm{mol}\) at \(600 . \mathrm{K}\) by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor \(A\) is constant and assume the initial concentrations are the same.

Consider a reaction of the type \(\mathrm{aA} \longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{~A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 . \mathrm{s}\), what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text { Rate }=\frac{-d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]^{3} $$

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.
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