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Chapter 12: Problem 105

A reaction of the form $$ \mathrm{aA} \longrightarrow $$ gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach 22.9\(\%\) completion.

Short Answer

Expert verified
For a first-order reaction, the integrated rate equation is \(\ln [\mathrm{A}] = -kt + \ln [\mathrm{A}]_0\). Given the initial concentration \([\mathrm{A}]_0 = 0.0100 M\) and the rate constant \(k = 7.35 \times 10^{-3} s^{-1}\), we can find the concentration of A at 22.9% completion, \([\mathrm{A}] = 0.00771 M\), and then solve for the time \(t\). The equation becomes \(\ln (0.00771) = -(7.35 \times 10^{-3} s^{-1})t + \ln (0.0100)\), which gives us \(t \approx 429.5s\). Thus, the time required for the reaction to reach 22.9% completion is approximately 429.5 seconds.

Step by step solution

01

Write down the integrated rate equation for a first-order reaction

For a first-order reaction, the integrated rate equation is given by: \[ \ln [\mathrm{A}] = -kt + \ln [\mathrm{A}]_0 \] where \([\mathrm{A}]\) is the concentration of A at time \(t\), \([\mathrm{A}]_0\) is the initial concentration of A, \(k\) is the rate constant, and \(t\) is the time in seconds.
02

Calculate the rate constant (k) using the given slope

We are given that the slope of the plot of \(\ln [\mathrm{A}]\) versus time is \(-7.35 \times 10^{-3}\). Since the slope of this plot is equal to the negative of the rate constant (-k), we can find the value of k as: \[ k = 7.35 \times 10^{-3} s^{-1} \]
03

Write down the equation for the percentage completion of the reaction

The percentage completion of the reaction is given by the fraction of A that has reacted. To find the time when the reaction reaches 22.9% completion, we can write the equation for the percentage completion as: \[ \frac{[\mathrm{A}]_0 - [\mathrm{A}]}{[\mathrm{A}]_0} = 0.229 \]
04

Solve the equation for the concentration of A at 22.9% completion

We know that the initial concentration \([\mathrm{A}]_0\) is \(0.0100 M\). We can now use the equation from Step 3 to find the concentration of A at 22.9% completion: \[ [\mathrm{A}] = [\mathrm{A}]_0(1 - 0.229) = 0.0100 M (1 - 0.229) = 0.00771 M \]
05

Substitute the known values into the integrated rate equation and solve for the time (t)

We can now substitute the known values of \([\mathrm{A}]\), \([\mathrm{A}]_0\), and \(k\) into the integrated rate equation from Step 1: \[ \ln [\mathrm{A}] = -kt + \ln [\mathrm{A}]_0 \] \[ \ln (0.00771) = -(7.35 \times 10^{-3} s^{-1})t + \ln (0.0100) \] Now, rearrange the equation and solve for \(t\): \[ t = \frac{\ln (0.00771) - \ln (0.0100)}{-7.35 \times 10^{-3} s^{-1}} \approx 429.5s \] Therefore, the time required for the reaction to reach 22.9% completion is approximately 429.5 seconds.

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Most popular questions from this chapter

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

Assuming that the mechanism for the hydrogenation of $\mathrm{C}_{2} \mathrm{H}_{4}$ given in Section 12.7 is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be $\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\( or \)\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?$ How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section 12.7\(?\)

The rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text {Rate} =-\frac{\Delta[\text { NOBr }]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to 0.100\(M ?\)

Define what is meant by unimolecular and bimolecular steps. Why are termolecular steps infrequently seen in chemical reactions?

Describe at least two experiments you could perform to determine a rate law.
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