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Chapter 12: Problem 110

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

Short Answer

Expert verified
The time for the second half-life of the termolecular reaction can be found using the integrated rate law and the rate constant, k, which was determined from the given first half-life. After deriving the integrated rate law and solving for the time for the second half-life, we find the following expression for the second half-life: $$ t_2 = \frac{80\,\mathrm{s} \left(\frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2}\right)}{1 - \frac{1}{[0.5\mathrm{A}]_0^2}} $$ By calculating this expression, we can find the time for the second half-life of the reaction.

Step by step solution

01

Derive the integrated rate law

Start with the given differential rate law for a termolecular reaction: $$ -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^{3} $$ Separate the variables: $$ -\frac{d[\mathrm{A}]}{[\mathrm{A}]^{3}} = k \, dt $$ Integrate both sides: $$ \int_{[\mathrm{A}]_0}^{[\mathrm{A}]}\frac{-d[\mathrm{A}]}{[\mathrm{A}]^{3}} = \int_{0}^{t}k \, dt $$ Calculate the integral: $$ \frac{1}{2[\mathrm{A}]^2} - \frac{1}{2[\mathrm{A}]_0^2} = kt $$ Rearrange the equation to solve for A: $$ \frac{1}{[\mathrm{A}]^2} - \frac{1}{[\mathrm{A}]_0^2} = 2kt $$ This is the integrated rate law for a termolecular reaction.
02

Find the rate constant, k

For the first half-life, 50% of A would be remaining, so we have: $$ \frac{1}{[0.5\mathrm{A}]_0^2} - \frac{1}{[\mathrm{A}]_0^2} = 2k(40\,\mathrm{s}) $$ Solving for k: $$ k = \frac{1 - \frac{1}{[0.5\mathrm{A}]_0^2}}{80\,\mathrm{s}} $$
03

Find the second half-life

Now we need to find the time for the second half-life, when 75% of A has reacted (25% remaining): $$ \frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2} = 2k(t_2) $$ Now, plug in the k value we found earlier: $$ \frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2} = \left(\frac{1 - \frac{1}{[0.5\mathrm{A}]_0^2}}{80\,\mathrm{s}}\right)(2t_2) $$ Solve for t2, the time for the second half-life: $$ t_2 = \frac{80\,\mathrm{s} \left(\frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2}\right)}{1 - \frac{1}{[0.5\mathrm{A}]_0^2}} $$ After computing this expression, we will find the time for the second half-life of the reaction.

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Most popular questions from this chapter

In the gas phase, the production of phosgene from chlorine and carbon monoxide is assumed to proceed by the following mechanism: $$ \mathrm{Cl}_{2} \stackrel{k_{1}}{\rightleftharpoons_{k_{1}}} 2 \mathrm{Cl} $$ $$ \mathrm{Cl}+\mathrm{CO} \stackrel{k_{2}}{\leftrightharpoons_{k-2}} \mathrm{COCl} $$ $$ \mathrm{COCl}+\mathrm{Cl}_{2} \stackrel{k_{3}}{\longrightarrow} \mathrm{COCl}_{2}+\mathrm{Cl} $$ $$ 2 \mathrm{Cl} \stackrel{k}{\longrightarrow} \mathrm{Cl}_{2} $$ Overall reaction: $\mathrm{CO}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}$ a. Write the rate law for this reaction. b. Which species are intermediates?

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}(g) $$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{s}^{-1} ;\) at $720 . \mathrm{K}, k=1.7 \times 10^{-2} \mathrm{s}^{-1}$ What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C},\) what is the pressure of iodoethane after three half-lives?

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

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.

For the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C},\) explain at least two ways in which the rate law could be zero order in chemical A.
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