Problem 40 - Chemistry and Chemical Reactivity

Question

When heated, cyclopropane is converted to propene (Example \(14.5) .\) 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 \(k=1.02 \times 10^{-3} \mathrm{s}^{-1},\) respec- tively. Determine the activation energy, \(E_{a}\), from these data.

Step-by-Step Solution

Answer

The activation energy, \(E_a\), is approximately 272 kJ/mol.

Step 1: Understand the Arrhenius Equation

The Arrhenius equation is often used to calculate the activation energy of a reaction. It states that the rate constant (k) is related to the temperatures (T) and activation energy (E_a) by the equation: \( k = A e^{-E_a/(RT)} \), where \( A \) is the pre-exponential factor and \( R \) is the universal gas constant \( (8.314 \text{ J/mol K}) \).

Step 2: Use the Arrhenius Equation for Two Temperatures

Since we have the rate constants at two different temperatures, we can use the following relation derived from the Arrhenius equation:\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]This eliminates the need for the pre-exponential factor \( A \).

Step 3: Convert Temperature to Kelvin

Convert the given temperatures from Celsius to Kelvin by adding 273.15. So, \( T_1 = 470 + 273.15 = 743.15 \, K \) and \( T_2 = 510 + 273.15 = 783.15 \, K \).

Step 4: Substitute Known Values into the Modified Arrhenius Equation

With \( k_1 = 1.10 \times 10^{-4} \, \text{s}^{-1} \) at \( T_1 = 743.15 \, K \) and \( k_2 = 1.02 \times 10^{-3} \, \text{s}^{-1} \) at \( T_2 = 783.15 \, K \), substitute into the equation:\[ \ln\left(\frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}}\right) = \frac{-E_a}{8.314}\left(\frac{1}{783.15} - \frac{1}{743.15}\right) \]

Step 5: Solve for Activation Energy, E_a

Calculate \( \ln\left(\frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}}\right) \) which gives approximately 2.208. Solve the right-hand side for each part, then solve for \( E_a \):\[ 2.208 = \frac{-E_a}{8.314}\left(\frac{1}{783.15} - \frac{1}{743.15}\right) \]Calculate \( \left(\frac{1}{783.15} - \frac{1}{743.15}\right) \) to find the value and isolate \( E_a \). You will find \( E_a \) is approximately 272 kJ/mol.

Key Concepts

Arrhenius Equation

The Arrhenius equation forms the backbone of chemical kinetics, helping chemists understand how reaction rates change with temperature. It is given by:

• \( k = A e^{-E_a/(RT)} \)

Here, \( k \) is the reaction rate constant, \( A \) is the pre-exponential factor, and \( E_a \) is the activation energy. \( R \) is the universal gas constant with a value of 8.314 J/mol K.
This equation shows that the rate constant increases exponentially as the temperature rises. The pre-exponential factor \( A \) represents the frequency of collisions that lead to a reaction. Understanding the Arrhenius equation helps predict how a reaction's speed changes with different temperatures.

Reaction Rate Constant

The reaction rate constant, symbolized as \( k \), is a vital component in understanding how fast a reaction occurs. It varies with temperature and depends on factors like the nature of the reactants and the presence of a catalyst.
In the context of the Arrhenius equation, \( k \) reveals the relationship between the temperature and the rate at which a chemical reaction proceeds. Higher values of \( k \) at elevated temperatures indicate quicker reactions. This makes the calculation of \( k \) crucial when devising chemical processes or conducting experiments.

Temperature Conversion

Converting temperatures from Celsius to Kelvin is a critical step in kinetics calculations because the Arrhenius equation requires temperature to be in Kelvin. To convert:

• Add 273.15 to the Celsius temperature.

For example:

• \( 470^{\circ} \text{C} = 470 + 273.15 = 743.15 \text{ K} \)
• \( 510^{\circ} \text{C} = 510 + 273.15 = 783.15 \text{ K} \)

Using Kelvin ensures consistent calculations and reflects the absolute scale of temperature, where zero represents absolute zero, the point of no thermal energy.

Kinetics Calculation

Kinetics calculations allow us to figure out the activation energy \( E_a \) using the Arrhenius equation. For reactions with rate constants at two different temperatures, a simplified formula is:

• \( \ln\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \)

This relation helps find \( E_a \) without needing the pre-exponential factor \( A \). By substituting the known values, calculating the natural logarithm of the rate constant ratio, and solving for \( E_a \), you determine the energy required to initiate the reaction. This makes the Arrhenius-based kinetics calculation pivotal in understanding reaction mechanisms and designing better chemical processes.