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Chapter 15: Problem 90

Cyclopropane \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) reacts to form propene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) in the gas phase. The reaction is first order in cyclopropane and has a rate constant of \(5.87 \times 10^{-4 / \mathrm{s}}\) at \(485^{\circ} \mathrm{C}\) . If a 2.5 L reaction vessel initially contains 722 torr of cyclopropane at \(485^{\circ} \mathrm{C}\) , how long will it take for the par- tial pressure of cyclopropane to drop to below \(1.00 \times 10^{2}\) torr?

Short Answer

Expert verified
The time it will take for the partial pressure of cyclopropane to drop below 100 torr is calculated using the integrated first-order rate law after converting pressures to molar quantities.

Step by step solution

01

Convert initial pressure to moles using the ideal gas law

First, convert the initial pressure of cyclopropane to moles using the ideal gas law, which is stated as PV = nRT. Here, P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the gas constant (\(0.0821 \text{L atm K}^{-1}\text{mol}^{-1}\)), and T is the temperature in Kelvins. First, convert the pressure from torr to atm by dividing by 760. Convert the temperature from Celsius to Kelvins by adding 273. Calculate the initial number of moles of cyclopropane.
02

Write the first-order rate law

Write down the first-order rate law, which relates the rate of reaction to the concentration of reactant: rate = k[A], where k is the rate constant and [A] is the concentration of reactant A. For a first-order reaction, the integrated rate law is ln([A]0/[A]) = kt, where [A]0 is the initial concentration and [A] is the concentration at time t.
03

Calculate the time for pressure to drop below 100 torr

Use the integrated first-order rate law to calculate the time it will take for the partial pressure of cyclopropane to drop to below 100 torr. Convert the target pressure to moles using the same method as in step one. Substitute the values into the integrated rate law and solve for t.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that describes the relationship between the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. It is usually stated as \(PV = nRT\), where R is the universal gas constant with a value of \(0.0821 \text{L atm K}^{-1}\text{mol}^{-1}\).To apply the ideal gas law for calculating moles from pressure, as in the exercise, you must ensure that the pressure is in atmospheres (atm) and the temperature is in Kelvins (K). Once you have these values, you can rearrange the equation to solve for \(n\), the number of moles of the gas: \(n = \frac{PV}{RT}\). This step is crucial in converting pressure measurements into a more useful form for reaction kinetics calculations, where mole quantities are often required.
Integrated Rate Laws
Integrated rate laws are equations that describe the concentration of reactants or products in a chemical reaction as a function of time. They are derived from rate laws, which provide the speed at which a reaction proceeds. For a first-order reaction, one such integrated rate law is the expression \(\ln(\frac{[A]_0}{[A]}) = kt\), where:
  • \([A]_0\) is the initial concentration of the reactant A
  • \([A]\) is the concentration of A at time \(t\)
  • \(k\) is the first-order rate constant
  • \(t\) is the time elapsed
This law shows that the natural logarithm of the ratio of initial to remaining reactant concentration is directly proportional to time, allowing you to calculate the time it takes for the reaction to reach a certain point, such as a specific concentration of reactant.
Reaction Rates
Reaction rates indicate how fast a chemical reaction occurs and are defined as the rate of change in concentration of a reactant or product over time. For a first-order reaction, the rate is directly proportional to the concentration of a single reactant. This is often represented as \(\text{rate} = k[A]\), where:
  • \(k\) is the rate constant specific to the reaction at a particular temperature
  • \([A]\) is the concentration of the reactant
First-order reactions have a constant half-life, meaning it takes the same amount of time for half of any quantity of reactant to be consumed, regardless of its initial concentration. Understanding how reaction rates are affected by factors like concentration and temperature is crucial for controlling the speed of industrial chemical reactions and for studying reaction mechanisms in the lab.
Chemical Kinetics
Chemical kinetics is the study of the speed (rate) or velocity of chemical reactions and the factors that affect this speed. This field of chemistry is not just concerned with how fast chemical reactions occur, but also with the steps (mechanism) by which they occur.Kinetics allows chemists to determine the order of a reaction, which is an essential step in the process of finding the rate law. Additionally, it provides insight into the effects of various conditions on reaction rates, such as temperature, pressure, and the presence of catalysts. In the context of the exercise, understanding kinetics is essential to predict how long it will take for cyclopropane to reach a certain pressure (which is related to its concentration) during its transformation into propene. Such calculations are fundamental in the design of chemical reactors and the manufacturing of chemicals in safe, efficient ways.

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Most popular questions from this chapter

The evaporation of a 120 nm film of \(n\) -pentane from a single crystal of aluminum oxide is zero order with a rate constant of \(1.92 \times 10^{13}\) molecules/cm \(^{2} \cdot \mathrm{s}\) at 120 \(\mathrm{K}.\) \begin{equation} \begin{array}{l}{\text { a. If the initial surface coverage is } 8.9 \times 10^{16} \text { molecules/cm', how long }} \\ {\text { will it take for one- half of the film to evaporate? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. What fraction of the film is left after } 10 \text { s? Assume the same initial }} \\ {\text { coverage as in part a. }}\end{array} \end{equation}

This reaction was monitored as a function of time: $$\mathrm{AB} \longrightarrow \mathrm{A}+\mathrm{B}$$ A plot of 1\(/[\mathrm{AB}]\) versus time yields a straight line with slope \(-0.055 / \mathrm{M} \cdot \mathrm{s} .\) \begin{equation} \begin{array}{l}{\text { a. What is the value of the rate constant }(k) \text { for this reaction at this }} \\ {\text { temperature? }}\\\\{\text { b. Write the rate law for the reaction. }} \\ {\text { c. What is the half-life when the initial concentration is } 0.55 \mathrm{M} \text { ? }}\\\\{\text { d. If the initial concentration of } \mathrm{AB} \text { is } 0.250 \mathrm{M}, \text { and the reaction mixture }} \\ {\text { initially contains no products, what are the concentrations of } \mathrm{A} \text { and } \mathrm{B}} \\ {\text { after } 75 \mathrm{s} \text { ? }}\end{array} \end{equation}

The first-order integrated rate law for a reaction \(\mathrm{A} \longrightarrow\) products is derived from the rate law using calculus. \begin{equation} \begin{aligned} \text { Rate } &=k[\mathrm{A}] \quad \text { (first-order rate law) } \\ \text { Rate } &=\frac{d[\mathrm{A}]}{d t} \\\ \frac{d[\mathrm{A}]}{d t} &=-k[\mathrm{A}] \end{aligned} \end{equation} The equation just given is a first-order, separable differential equa- tion that can be solved by separating the variables and integrating: \begin{equation} \begin{aligned} \frac{d[\mathrm{A}]}{[\mathrm{A}]} &=-k d t \\\ \int_{[\mathrm{A}]_{0}}^{[\mathrm{A]}} \frac{d[\mathrm{A}]}{[\mathrm{A}]} &=-\int_{0}^{t} k d t \end{aligned} \end{equation} In the integral just given, \([\mathrm{A}]_{0}\) is the initial concentration of \(\mathrm{A} . \mathrm{We}\) then evaluate the integral: \begin{equation} \begin{aligned}[\ln [\mathrm{A}]]_{[\mathrm{A}]_{0}}^{[\mathrm{Al}} &=-k[t]_{0}^{t} \\ \ln [\mathrm{A}]-\ln [\mathrm{A}]_{0} &=-k t \end{aligned} \end{equation} \begin{equation} \ln [\mathrm{A}]=-k t+\ln [\mathrm{A}]_{0}(\text { integrated rate law }) \end{equation} \begin{equation} \begin{array}{l}{\text { a. Use a procedure similar to the one just shown to derive an inte- }} \\ {\text { grated rate law for a reaction } A \longrightarrow \text { products, which is one-half- }} \\ {\text { order in the concentration of } A \text { (that is, Rate }=k[A]^{1 / 2} )}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Use the result from part a to derive an expression for the half-life }} \\ {\text { of a one-half-order reaction. }}\end{array} \end{equation}

For a reaction with multiple reactants, how is the overall order of the reaction defined?

Explain the difference between a normal chemical equation for a chemical reaction and the mechanism of that reaction.
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