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Chapter 11: Problem 50

A first-order reaction is \(75.0 \%\) complete in \(320 .\) 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 \(319\,s\). b. It takes \(1010\,s\) for the reaction to reach \(90.0\%\) completion.

Step by step solution

01

Calculate the rate constant (k)

Using the first-order reaction formula, we can find the rate constant k: \[ ln \frac{[A]_0}{[A]_{320}} = kt_{320} \] Since the reaction is 75% complete after 320 seconds, we have 25% remaining: \[ ln \frac{100\%}{25\%} = k(320) \] \[ ln(4) = k(320) \] Now, we can find k: \[ k = \frac{ln(4)}{320} = 0.00217\,s^{-1} \]
02

Calculate the first half-life

Using the half-life formula for a first-order reaction: \[ t_{1/2} = \frac{0.693}{k} \] We can now find the first half-life: \[ t_{1/2} = \frac{0.693}{0.00217} = 319\,s \]
03

Calculate the second half-life

For a first-order reaction, the half-life is constant, meaning that the second half-life is equal to the first half-life. Therefore, the second half-life is also 319 seconds.
04

Calculate the time for 90% completion

We are asked to find the time it takes to reach 90% completion. At this point, only 10% of the reactants remains. We can use the first-order reaction formula: \[ ln \frac{[A]_0}{[A]_{t_{0.9}}} = kt_{0.9} \] \[ ln \frac{100\%}{10\%} = (0.00217\, s^{-1})(t_{0.9}) \] \[ ln(10) = (0.00217\, s^{-1})(t_{0.9}) \] Now, we can find \(t_{0.9}\): \[ t_{0.9} = \frac{ln(10)}{0.00217} = 1010\,s \] So, the reaction takes 1010 seconds to reach 90% completion. In conclusion: a. The first and second half-lives for this reaction are both 319 seconds. b. It takes 1010 seconds for the reaction to reach 90% completion.

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Most popular questions from this chapter

The rate constant \((k)\) depends on which of the following (there may be more than one answer)? a. the concentration of the reactants b. the nature of the reactants c. the temperature d. the order of the reaction Explain.

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

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

A reaction of the form $$aA \longrightarrow Products$$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.

Consider a reaction of the type 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 \(40 .\) 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: $$\text { Rate }=\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3}$$
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