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Chapter 12: Problem 52

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
In conclusion, the first and second half-lives of the first-order reaction are both \(226 \,s\), and it takes approximately \(763 \,s\) for the reaction to reach \(90\%\) completion.

Step by step solution

01

Calculate the rate constant

For a first-order reaction, the integrated rate law is given by the equation: \[ln\frac{A_0}{A_t} = kt\] Where \(A_0\) is the initial concentration of the reactant, \(A_t\) is the concentration at time \(t\), \(k\) is the rate constant and \(t\) is time. In our problem, the reaction is 75% complete at 320 s which means 25% of the reactant is still present. Therefore, \(\frac{A_t}{A_0} = 0.25\), and we have \[ln\frac{1}{0.25} = k \times 320\] Now, let's solve for \(k\).
02

Solve for k

We can rearrange the equation to isolate k: \[k = \frac{ln(1/0.25)}{320}\] Now, calculate the value of \(k\): \[k \approx 0.00307 \,s^{-1}\] Now that we have the rate constant, we can calculate the half-lives and the time required for 90% completion.
03

Calculate the first half-life

For first-order reactions, the half-life is given by the equation: \[t_{1/2} = \frac{0.693}{k}\] Plugging in the value of k that we calculated in step 2: \[t_{1/2} =\frac{0.693}{0.00307} \approx 226 \,s\] Thus, the first half-life of the reaction is 226 s.
04

Calculate the second half-life

In first-order reactions, the half-life remains constant. Hence, the second half-life will also be 226 s.
05

Calculate the time for 90% completion

To find the time required for 90% completion, we can use the integrated rate law again: \[ln\frac{A_0}{A_t} = kt\] In this case, 90% completion means 10% reactant is still present, so \(\frac{A_t}{A_0} = 0.10\). We have the rate constant \(k\), so let's solve for the time \(t\): \[ln\frac{1}{0.10} = (0.00307) \times t\] Now, solve for t: \[t = \frac{ln(1/0.10)}{0.00307} \approx 763\] This means it takes approximately 763 s for the reaction to reach 90% completion. In conclusion, the first and second half-lives of the reaction are both 226 s, and it takes approximately 763 s for the reaction to reach 90% completion.

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Most popular questions from this chapter

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: $$ \text {Rate} =\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3} $$

Consider the following initial rate data for the decomposition of compound AB to give A and B: Determine the half-life for the decomposition reaction initially having 1.00$M \mathrm{AB}$ present.

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.80 \times 10^{-3} M,\) con- centration versus time data were collected for this reaction, and a plot of 1\(/[\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(+3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) . 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 A to decrease to $7.00 \times 10^{-4} M ?$

The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate $=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )$

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in A and to have a rate constant of $5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\( at \)25^{\circ} \mathrm{C}$ . An experiment was run at \(25^{\circ} \mathrm{C}\) where $[\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .$ a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after $5.0 \times 10^{-3} \mathrm{s}\( has elapsed assuming \)[\mathrm{B}]_{0}=0$
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