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Chapter 12: Problem 50

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 \(199.7 \, s\). b. It takes approximately \(646.6\, s\) for the reaction to reach \(90.0\%\) completion.

Step by step solution

01

Write the given information

It is given that the reaction is 75.0% complete in 320 seconds. That means 25.0% of the reactant remains at that time.
02

Determine the concentration of reactant at time t

As the reaction is 75.0% complete, the remaining percentage of reactant is 100 - 75 = 25.0%. Thus, the concentration of the reactant at time t is: \[A_t = 0.25A_0 \]
03

Calculate the rate constant (k)

We can use the first-order reaction formula given earlier: \[ k = \frac{1}{t} \ln{ \frac{A_0}{A_t} }\] Substitute the given values, A₀ and Aₜ: \[ k = \frac{1}{320s} \ln{ \frac{A_0}{0.25A_0} }\] \[ k = \frac{1}{320s} \ln{4} \] Now, calculate the value of k: \[ k ≈ 0.00347 \, s^{-1} \]
04

Calculate the first half-life

For a first-order reaction, the half-life (t₁/₂) is given by: \[t_{1/2} = \frac{0.693}{k}\] Substitute the value of k: \[t_{1/2} = \frac{0.693}{0.00347\, s^{-1}}\] Now, calculate the value of t₁/₂: \[t_{1/2} ≈ 199.7\, s\]
05

Calculate the second half-life

For first-order reactions, the half-life is constant. Therefore, the second half-life will be equal to the first half-life: \[t_{2nd\, 1/2} = t_{1st\, 1/2} = 199.7\, s\]
06

Calculate the time for 90% completion

We are asked to find the time it takes for the reaction to reach 90.0% completion. That means 10.0% of the reactant remains: \[A_t = 0.1A_0\] Use the first-order reaction formula: \[ t = \frac{1}{k} \ln{ \frac{A_0}{A_t} }\] Substitute the given values, A₀ and Aₜ: \[ t = \frac{1}{0.00347\, s^{-1}} \ln{ \frac{A_0}{0.1A_0} }\] \[ t = \frac{1}{0.00347\, s^{-1}} \ln{10} \] Now, calculate the value of t: \[ t ≈ 646.6\, s \] Therefore, the reaction will reach 90.0% completion in approximately 646.6 seconds. #Summary# a. The first and second half-lives for this reaction are both 199.7 seconds. b. It takes approximately 646.6 seconds for the reaction to reach 90.0% completion.

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Most popular questions from this chapter

One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is \(210 \mathrm{~kJ} / \mathrm{mol}\), which is less than either the \(\mathrm{Si}-\) Si or the \(\mathrm{Si}-\mathrm{H}\) bond energy. Why would a similar mechanism not be expected to play a very important role in the decomposition of long chains of carbon atoms as seen in organic compounds?

The rate law for the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) is $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$ It takes 120. s for \(1.00 M \mathrm{PH}_{3}\) to decrease to \(0.250 \mathrm{M}\). How much time is required for \(2.00 M \mathrm{PH}_{3}\) to decrease to a concentration of \(0.350 M ?\)

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.

In the Haber process for the production of ammonia, $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{aligned} &\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \\ &\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ Overall reaction \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{a}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ is \(14.0 \mathrm{~kJ} . E_{a}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.
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