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Chapter 15: Problem 44

When heated, cyclopropane is converted to propene (see Example 15.5 ). Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=1.02 \times10^{-3} \mathrm{s}^{-1},\) respectively. Determine the activation energy, \(E_{\mathrm{a}}\) from these data.

Short Answer

Expert verified
The activation energy, \( E_a \), is approximately 260.5 kJ/mol.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation relates the rate constant \( k \) to the activation energy \( E_a \), temperature \( T \), and the gas constant \( R \): \[ k = A e^{-\frac{E_a}{RT}} \] where \( A \) is the pre-exponential factor, \( R = 8.314 \text{ J/mol K} \), and \( T \) is in Kelvin.
02

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \( T(K) = T(掳C) + 273.15 \). For \( 470^{\circ} \mathrm{C} \), \( T_1 = 470 + 273.15 = 743.15 \text{ K} \). For \( 510^{\circ} \mathrm{C} \), \( T_2 = 510 + 273.15 = 783.15 \text{ K} \).
03

Use Arrhenius Equation Form

We can rewrite the Arrhenius equation to solve for \( E_a \) using two different rate constants and temperatures: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \].
04

Substitute Values

Substitute the known values into the Arrhenius form: \( k_1 = 1.10 \times 10^{-4} \text{ s}^{-1} \), \( k_2 = 1.02 \times 10^{-3} \text{ s}^{-1} \), \( T_1 = 743.15 \text{ K} \), \( T_2 = 783.15 \text{ K} \), and \( R = 8.314 \text{ J/mol K} \).
05

Calculate Activation Energy

Calculate \( E_a \) by first evaluating \( \ln \left( \frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}} \right) \). This equals \( \ln(9.273) \approx 2.226 \). Then, calculate \( \frac{1}{743.15} - \frac{1}{783.15} \approx -6.166 \times 10^{-5} \text{ K}^{-1} \). Plug these into \( 2.226 = \frac{E_a}{8.314} \times -6.166 \times 10^{-5} \) and solve for \( E_a \): \[ E_a \approx \left(\frac{2.226}{-6.166 \times 10^{-5}}\right) \times 8.314 \text{ J/mol} \approx 260.5 \text{ kJ/mol} \].

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Key Concepts

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

Arrhenius Equation
The Arrhenius equation is a key formula in understanding how reaction rates are influenced by temperature. This equation is given by: \[ k = A e^{-\frac{E_a}{RT}} \] where:
  • \( k \) is the rate constant.
  • \( E_a \) is the activation energy.
  • \( R \) is the universal gas constant \(8.314\text{ J/mol K}\).
  • \( T \) is the temperature in Kelvin.
  • \( A \) represents the pre-exponential factor or frequency factor. This accounts for the frequency of collisions that result in a reaction.

This equation suggests that, at higher temperatures, the exponential term decreases, leading to a larger rate constant, meaning the reaction happens faster. The activation energy \( E_a \) serves as a threshold energy barrier that reactants need to overcome to form products.
Rate Constants
The rate constant \( k \) is central to the study of chemical kinetics. It defines the speed at which a reaction proceeds under a specific set of conditions. For reactions with smaller activation energies, rate constants tend to be larger because the energy barrier is easier to overcome.

In the given exercise, we have two rate constants at different temperatures:
  • At \( 470^{\circ} \text{C} \), \( k_1 = 1.10 \times 10^{-4} \text{ s}^{-1} \).
  • At \( 510^{\circ} \text{C} \), \( k_2 = 1.02 \times 10^{-3} \text{ s}^{-1} \).
Notice how \( k_2 \) is larger than \( k_1 \), indicating an increased reaction speed at the higher temperature. Rate constants are temperature-dependent, and understanding their relationship with temperature is crucial for predicting how a reaction will behave under varying conditions.
Temperature Conversion
Converting temperatures from Celsius to Kelvin is an essential step in any calculation involving the Arrhenius equation. The conversion is straightforward: add 273.15 to the Celsius temperature to get Kelvin. This is crucial because the Arrhenius equation requires temperatures to be in Kelvin to properly relate thermal energy to reaction rates.

In our scenario:
  • The temperature \( 470^{\circ} \text{C} \) becomes \( 743.15 \text{ K} \).
  • Likewise, \( 510^{\circ} \text{C} \) turns into \( 783.15 \text{ K} \).
Converting temperatures ensures that calculated reaction dynamics are accurate, as the Kelvin scale starts at absolute zero where thermal motion ceases.
Chemical Kinetics
Chemical kinetics focuses on the study of reaction rates and the factors that affect these rates. It helps in predicting how a change in conditions, like temperature or concentration, influences the speed at which reactants are converted into products.

Key components in chemical kinetics include:
  • The reaction mechanism, which outlines the detailed step-by-step pathway by which reactants become products.
  • The rate law, which links the reaction rate with reactant concentrations and rate constants.
  • The activation energy \( E_a \), a critical threshold energy that must be overcome for a reaction to occur.
In our exercise, we use chemical kinetic principles to determine the activation energy from rate constants at different temperatures. This demonstrates how kinetics aids in understanding and controlling chemical reactions, offering insights into reaction mechanisms and effectiveness.

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Most popular questions from this chapter

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

A proposed mechanism for the reaction of \(\mathrm{NO}_{2}\) and \(\mathrm{CO}\) is Step 1 Slow, endothermic $$2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})$$ Step 2 \(\quad\) Fast, exothermic $$\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$$ Overall Reaction Exothermic $$\mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$$ (a) Identify each of the following as a reactant, product, or intermediate: \(\mathrm{NO}_{2}(\mathrm{g}), \mathrm{CO}(\mathrm{g}), \mathrm{NO}_{3}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{g})\) \(\mathrm{NO}(\mathrm{g})\) (b) Draw a reaction coordinate diagram for this reaction. Indicate on this drawing the activation energy for each step and the overall reaction enthalpy.

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) (b) \(\mathbf{O}(\mathrm{g})+\mathbf{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\)

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a first-order reaction to elemental Xe with a half-life of 30. min. If you place \(7.50 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) in a flask, how long must you wait until only 0.25 mg of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

Data for the reaction $$\begin{aligned}\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{CH}_{3}\mathrm{CN}\right)\right]^{+}+\mathrm{NC}_{5}\mathrm{H}_{5} \longrightarrow & \\\&\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{NC}_{5}\mathrm{H}_{5}\right)\right]^{+}+\mathrm{CH}_{3} \mathrm{CN}\end{aligned}$$ are given in the table. Calculate \(E_{\mathrm{a}}\) from a plot of \(\ln k\) versus \(1 / T\) $$\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\min ^{-1}\right) \\\\\hline 298 & 0.0409 \\\308 & 0.0818 \\\318 & 0.157 \\\\\hline\end{array}$$
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