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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

Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, \(0.0048 \mathrm{~mol}\) \(\mathrm{PH}_{3}\) is consumed in a 2.0-L container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?

A certain reaction has an activation energy of \(54.0 \mathrm{~kJ} / \mathrm{mol}\). As the temperature is increased from \(22^{\circ} \mathrm{C}\) to a higher temperature, the rate constant increases by a factor of \(7.00 .\) Calculate the higher temperature.

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\), \([\mathrm{B}]_{0}=3.0 \mathrm{M}\), and \([\mathrm{C}]_{0}=2.0 \mathrm{M} .\) The reaction is started, and after \(8.0\) seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\). a. Calculate the value of \(k\) for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after \(13.0\) seconds. d. Calculate the concentration of \(\mathrm{C}\) after \(13.0\) seconds.

The rate law for the reaction $$ \mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \operatorname{HCl}(g)+\mathrm{CCl}_{4}(g) $$ is $$ \text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right] $$ What are the units for \(k\), assuming time in seconds and concentration in \(\mathrm{mol} / \mathrm{L}\) ?

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