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Chapter 14: Problem 23

What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in 60 min? Assume first-order kinetics.

Short Answer

Expert verified
The half-life of the compound, to the nearest minute, is approximately 21 minutes.

Step by step solution

01

Understand the principles of first-order kinetics

The general formula for first-order decay is given by: \( N = N_0 e^{-kt} \), where \(N\) is the amount remaining, \(N_0\) is the initial amount, \(k\) is the rate constant, and \(t\) is time. For this problem, we need to rearrange this equation to solve for \(k\), which we can then use to find the half-life
02

Setting up the equation

Given that 75% of the sample decomposes in 60 minutes, meaning that 25% remains. We can substitute this information into the decay formula: \(0.25N_0 = N_0 e^{-k(60)}\)
03

Solve for the rate constant (k)

After getting rid of the \(N_0\) on both sides and taking the natural logarithm (ln) we obtain: \(-ln(4) = -60k\), rearranging this to solve for \(k\) gives \(k = ln(4) / 60\)
04

Derive the half-life from the decay constant

The formula connecting the half-life (t1/2) with the rate constant in a first-order reaction is \(t1/2 = ln(2) / k\). Substituting our calculated \(k\) value into this formula will yield the half-life.

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Most popular questions from this chapter

(a) What can you deduce about the activation energy of a reaction if its rate constant changes significantly with a small change in temperature? (b) If a bimolecular reaction occurs every time an A and a B molecule collide, what can you say about the orientation factor and activation energy of the reaction?

A flask contains a mixture of compounds \(A\) and \(B\). Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for \(A\) and 18.0 min for \(B\). If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, how long will it take for the concentration of \(\mathrm{A}\) to be four times that of \(\mathrm{B} ?\)

What is the molecularity of a reaction?

In a certain industrial process using a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is \(10.0 \mathrm{~cm}^{3}\). Calculate the surface area of the catalyst. If the sphere is broken down into eight spheres, each of which has a volume of \(1.25 \mathrm{~cm}^{3}\), what is the total surface area of the spheres? Which of the two geometric configurations of the catalyst is more effective? Explain. (The surface area of a sphere is \(4 \pi r^{2},\) in which \(r\) is the radius of the sphere.)

Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons_{-1}} \mathrm{ES} \quad \text { (fast equilbrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)
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