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Chapter 12: Problem 38

A certain reaction has the following general form: \(\mathrm{aA} \longrightarrow \mathrm{bB}\) At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} \mathrm{M}\), 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 \(Rate = k[\mathrm{A}]\), the integrated rate law is \(\ln [\mathrm{A}] = -kt + \ln[\mathrm{A}]_0\), and the rate constant is \(k = 2.97 \times 10^{-2}\; \mathrm{min}^{-1}\). b. The half-life is \(t_{1/2} = 23.3\; \mathrm{min}\). c. It will take 137 minutes for the concentration of A to decrease to \(2.50 \times 10^{-3} M\).

Step by step solution

01

a. Determine the rate law, integrated rate law, and value of the rate constant.

Since we are given that a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line, this indicates that the reaction is first-order with respect to A. Hence, the rate law can be written as: \[Rate = k[\mathrm{A}]^1\] The integrated rate law for a first-order reaction is given by: \(\ln [\mathrm{A}] = -kt + \ln[\mathrm{A}]_0\) Given that the slope of the straight line when plotting \(\ln [\mathrm{A}]\) vs time is\(-2.97 \times 10^{-2}\; \mathrm{min}^{-1}\), we can identify the rate constant as: \[k = 2.97 \times 10^{-2}\; \mathrm{min}^{-1}\]
02

b. Calculate the half-life for this reaction.

The half-life for a first-order reaction is calculated using the formula: \[t_{1/2} = \frac{0.693}{k}\] Substitute the rate constant we found in part a: \[t_{1/2} = \frac{0.693}{2.97 \times 10^{-2}\; \mathrm{min}^{-1}}\] Calculate the half-life: \[t_{1/2} = 23.3\; \mathrm{min}\]
03

c. Calculate the time required for the concentration of A to decrease to \(2.50 \times 10^{-3} M\).

The integrated rate law is: \(\ln [\mathrm{A}] = -kt + \ln[\mathrm{A}]_0\) Substitute the values \( [\mathrm{A}] = 2.50 \times 10^{-3} M\) and \([\mathrm{A}]_0 = 2.00 \times 10^{-2} M \), and the rate constant, k: \(\ln \frac{2.50 \times 10^{-3}}{2.00 \times 10^{-2}} = - (2.97 \times 10^{-2}\; \mathrm{min}^{-1})t\) Now solve for the time, t: \[t = 137\; \mathrm{min}\] So, it will take 137 minutes for the concentration of A to decrease to \(2.50 \times 10^{-3} M\).

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Most popular questions from this chapter

The rate law for a reaction can be determined only from experiment and not from the balanced equation. Two experimental procedures were outlined in Chapter \(12 .\) What are these two procedures? Explain how each method is used to determine rate laws.

A certain substance, initially present at \(0.0800 M\), decomposes by zero-order kinetics with a rate constant of \(2.50 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). Calculate the time (in seconds) required for the system to reach a concentration of \(0.0210 \mathrm{M}\).

The reaction $$ \left(\mathrm{CH}_{\mathrm{3}}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). In several experiments, the rate constant \(k\) was determined at different temperatures. \(\mathrm{A}\) plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{~K}\) and \(y\) -intercept of 33.5. Assume \(k\) has units of \(\mathrm{s}^{-1}\). a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\). c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\).

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

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \mathrm{~kJ} / \mathrm{mol}\). In the presence of a catalyst at \(37^{\circ} \mathrm{C}\), the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.
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