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Chapter 12: Problem 47

A first-order reaction is \(75.0 \%\) complete in \(320 . \mathrm{s}\). a. What are the first and second half-lives for this reaction? b. How long does it take for \(90.0 \%\) completion?

Short Answer

Expert verified
a. The first and second half-lives for this reaction are both 255 seconds. b. It takes 448 seconds for the reaction to reach 90.0% completion.

Step by step solution

01

(Determine k from the given information)

As the reaction is 75% complete, it means that 25% is left. So, \(\frac{[A]}{[A]_0} = 0.25\). We can use this information to find the rate constant from the given time. \(k = \frac{1}{320 s} \ln{\frac{1}{0.25}} = 0.00272 s^{-1}\)
02

(Find the First Half-life)

Now that we have the rate constant, we can use the half-life equation to find the first half-life: \(t_{1/2} = \frac{0.693}{0.00272 s^{-1}} = 255 s\) The first half-life is 255 seconds.
03

(Find the Second Half-life)

Since this is a first-order reaction, the half-life remains constant throughout the reaction. So, the second half-life is equal to the first half-life. The second half-life is 255 seconds. #a. Answer:# The first and second half-lives for this reaction are both 255 seconds. #b. Time for 90.0% completion#
04

(Calculate the remaining concentration)

For 90% completion, it means that 10% is left. We can set up the equation to find the time it takes for the reaction to reach that point: \(k = \frac{1}{t} \ln{\frac{[A]_0}{[A]}}\) \(\frac{1}{t} = \frac{0.00272 s^{-1}}{\ln{\frac{1}{0.1}}}\)
05

(Solve for time)

Next, we solve for time: \(t = \frac{1}{0.00272 s^{-1}} \times \ln{\frac{1}{0.1}} = 448 s\) #b. Answer:# It takes 448 seconds for the reaction to reach 90.0% completion.

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Most popular questions from this chapter

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \mathrm{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$ \mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a halflife of \(667 \mathrm{~s}\). What is the concentration of \(\mathrm{In}^{+}(a q)\) after \(1.25 \mathrm{~h}\) if the initial solution of \(\mathrm{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{~g} \operatorname{InCl}(s)\) in \(5.00 \times 10^{2} \mathrm{~mL}\) dilute HCl? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{~h}\) ?

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0\), and \(40.0\) min for an experiment in which \([\mathrm{A}]_{0}=\) \(0.10 \mathrm{M} .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{~min}\) b. \(30.0 \mathrm{~min}\)

One of the concerns about the use of Freons is that they will migrate to the upper atmosphere, where chlorine atoms can be generated by the following reaction: $$ \mathrm{CCl}_{2} \mathrm{~F}_{2} \stackrel{\mathrm{hv}}{\longrightarrow} \mathrm{CF}_{2} \mathrm{Cl}+\mathrm{Cl} $$ Chlorine atoms can act as a catalyst for the destruction of ozone. The activation energy for the reaction $$ \mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{ClO}+\mathrm{O}_{2} $$ is \(2.1 \mathrm{~kJ} / \mathrm{mol}\). Which is the more effective catalyst for the destruction of ozone, \(\mathrm{Cl}\) or \(\mathrm{NO}\) ? (See Exercise \(69 .\) )

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.

Which of the following reactions would you expect to proceed at a faster rate at room temperature? Why? (Hint: Think about which reaction would have the lower activation energy.) \(\begin{aligned} 2 \mathrm{Ce}^{4+}(a q)+\mathrm{Hg}_{2}^{2+}(a q) & \longrightarrow 2 \mathrm{Ce}^{3+}(a q)+2 \mathrm{Hg}^{2+}(a q) \\\ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}\)
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