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Chapter 13: Problem 25

What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in \(60 \mathrm{~min} ?\) Assume first-order kinetics.

Short Answer

Expert verified
The half-life of the compound is calculated using the formula \( t_{1/2} = \frac{ln(2)}{k} \) where k was calculated earlier. Substitute the value of k to get the answer.

Step by step solution

01

Calculate the rate constant

First, determine the rate constant (k). Using the equation \( t = \frac{ln([A]_0 / [A])}{k} \), we can replace [A]_0 with 100 (since our starting quantity is 100 percent), [A] with 25 (since 75% is decomposed, 25% is remaining), and t with 60 (since 75% decomposed in 60 minutes). This gives us the equation \(60 = \frac{ln(100/25)}{k}\). Solving for k gives \(k = \frac{ln(100/25)}{60}\).
02

Calculate half-life using the rate constant

With the calculated value of k, we can now determine the half-life of the compound using the formula \( t_{1/2} = \frac{ln(2)}{k} \). This will give us the half-life of the compound.
03

Evaluate the result

Ensure that your final result makes sense in the context of the problem. Since half-life is a measure of how quickly a substance decays, it should be a positive number.

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Most popular questions from this chapter

The following scheme in which A is converted to B, which is then converted to \(\mathrm{C}\) is known as a consecutive reaction. $$ \mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} $$ Assuming that both steps are first order, sketch on the same graph the variations of [A], [B], and [C] with time.

When methyl phosphate is heated in acid solution, it reacts with water: $$ \mathrm{CH}_{3} \mathrm{OPO}_{3} \mathrm{H}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{3} \mathrm{PO}_{4} $$ If the reaction is carried out in water enriched with \({ }^{18} \mathrm{O},\) the oxygen- 18 isotope is found in the phosphoric acid product but not in the methanol. What does this tell us about the mechanism of the reaction?

Consider the following elementary steps for a consecutive reaction: $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(A\).

Consider the reaction $$ \mathrm{A} \longrightarrow \mathrm{B} $$ The rate of the reaction is \(1.6 \times 10^{-2} M / \mathrm{s}\) when the concentration of \(\mathrm{A}\) is \(0.35 \mathrm{M}\). Calculate the rate constant if the reaction is (a) first order in A, and (b) second order in A.

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\)
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