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Chapter 14: Problem 80

The thermal decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) obeys firstorder kinetics. At \(45^{\circ} \mathrm{C}\), a plot of \(\ln \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) versus \(t\) gives a slope of \(-6.18 \times 10^{-4} \mathrm{~min}^{-1}\). What is the half-life of the reaction?

Short Answer

Expert verified
The half-life of the reaction at \( 45 ^{\circ} \mathrm{C} \) is 1121 minutes.

Step by step solution

01

Understanding Given Information

Even before starting to solve the task, it's necessary to understand the given information. The exercise mentions that the thermal decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) follows first-order kinetics. Therefore, the rate law of this reaction can be represented as \( r = k[\mathrm{N}_{2} \mathrm{O}_{5}]\), where \( k \) is the rate constant. The task also provides information that the slope of the graph \(\ln \left[\mathrm{N}_{2}\mathrm{O}_{5}\right]\) versus \( t \) is \(-6.18 \times 10^{-4} \mathrm{~min}^{-1}\). In terms of first-order kinetics, this slope is equivalent to \( -k \). Thus, the rate constant \( k \) is \( 6.18 \times 10^{-4} \mathrm{~min}^{-1}\).
02

Apply First-Order Kinetic Formula to Find Half-Life

The formula to find the half-life of a first-order reaction is \( t_{1/2} = \frac{0.693}{k} \). Substituting the value of \( k \) in this formula gives the half-life. Therefore, the half-life \( t_{1/2} \) of this reaction is \( \frac{0.693}{6.18 \times 10^{-4} \mathrm{~min}^{-1}} \).
03

Perform Calculation

By proceeding with calculation, we get \( t_{1/2} = 1121 \mathrm{~min} \). Therefore, at \( 45 ^{\circ} \mathrm{C} \), the half-life of the reaction is 1121 minutes.

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Most popular questions from this chapter

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.

The reaction \(\mathrm{H}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2}+\mathrm{H}\) has been studied for many years. Sketch a potential-energy-versusreaction-progress diagram for this 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.)

For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in terms of the half-life \(\left(t_{\frac{1}{2}}\right)\) and in terms of the rate constant \(k\)

Write the Arrhenius equation and define all terms.
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