Question

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 \(k=\) \(1.02 \times 10^{-3} \mathrm{s}^{-1},\) respectively. Determine the activation energy, \(E_{\omega},\) from these data.

Short answer

The activation energy, \( E_a \), is 268.1 kJ/mol.

Step-by-step solution

Understanding the Arrhenius Equation

The Arrhenius equation, which describes the relationship between the rate constant (k) and temperature (T), is given by:\[ k = A e^{-E_a / (RT)} \]where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.

Converting Temperatures to Kelvin

First, convert the given temperatures from degrees Celsius to Kelvin:- For \(470^{\circ}\text{C}\): \[ T_1 = 470 + 273.15 = 743.15 \text{ K} \]- For \(510^{\circ}\text{C}\): \[ T_2 = 510 + 273.15 = 783.15 \text{ K} \]

Using the Two-Point Form of the Arrhenius Equation

We use the formula derived from the Arrhenius equation:\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]Substitute the given values: \(k_1 = 1.10 \times 10^{-4} \text{ s}^{-1} \) and \(k_2 = 1.02 \times 10^{-3} \text{ s}^{-1} \).

Calculating the Natural Logarithm of the Rate Constants Ratio

Calculate \( \ln\left(\frac{k_2}{k_1}\right) \):\[ \ln\left(\frac{k_2}{k_1}\right) = \ln\left(\frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}}\right) \approx \ln(9.273) \approx 2.227 \]

Calculating the Activation Energy

Plug in the values into the formula:\[ 2.227 = \frac{-E_a}{8.314} \left(\frac{1}{783.15} - \frac{1}{743.15}\right) \]\[ 2.227 = \frac{-E_a}{8.314} \times (0.001277 - 0.001346) \]\[ 2.227 = \frac{-E_a}{8.314} \times -0.000069 \rightarrow E_a = \frac{2.227 \times 8.314}{0.000069} \]Calculate the value:\[ E_a \approx 2.681 \times 10^5 \text{ J/mol} \]

Converting Activation Energy to kJ/mol

Convert the activation energy from J/mol to kJ/mol by dividing by 1000:\[ E_a = 268.1 \text{ kJ/mol} \]

Key concepts

Arrhenius equation

Chemical reactions are sensitive to temperature changes. The Arrhenius equation illustrates this relationship with the formula \( k = A e^{-E_a / (RT)} \). Here, \( k \) is the rate constant, which tells us how fast the reaction occurs. \( A \) is the pre-exponential factor, representing the frequency of collisions with proper orientation. \( E_a \) is the activation energy, or energy barrier that must be overcome for reactants to convert to products. \( R \) is the gas constant, 8.314 J/mol K, a constant used in chemistry equations pertaining to temperature and energy. Finally, \( T \) is the temperature measured in Kelvin. By analyzing how \( k \) changes with \( T \), scientists can determine \( E_a \) and better understand the reaction's dynamics.

Using the Arrhenius equation often involves logarithmic manipulation, especially when comparing two different rate constants at different temperatures. This becomes vital in calculating \( E_a \) using experimental data.

rate constant

The rate constant \( k \) is an essential parameter in chemical kinetics, often observed as \( s^{-1} \) for first-order reactions. It provides insight into the speed of a reaction under certain conditions. For the reaction involving cyclopropane and propene, \( k \) values are given at two temperatures: \( 1.10 \times 10^{-4} \text{ s}^{-1} \) at 470°C and \( 1.02 \times 10^{-3} \text{ s}^{-1} \) at 510°C.

Notice how the increase in temperature from 470°C to 510°C causes a significant rise in the rate constant. This indicates that reactions generally proceed faster at higher temperatures, partly due to the increased kinetic energy allowing more reactants to surpass the activation energy barrier. The Arrhenius equation uses these rate constants to help determine activation energy, one of the key insights into the mechanism and characteristics of the reaction.

• Rate constants depend heavily on environmental conditions, especially temperature.
• They allow comparative analysis of reaction speeds under variable parameters.

temperature conversion

Temperature conversion, particularly from Celsius to Kelvin, is vital when dealing with gas laws and thermodynamic equations like the Arrhenius equation. The Kelvin scale starts at absolute zero, the point where particles theoretically have no kinetic energy.

To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. For example:

• 470°C becomes 743.15 K
• 510°C becomes 783.15 K

This conversion ensures that temperature is consistent with the units of the gas constant \( R \), which is derived based on Kelvin and Joules.
Always remember:

• Converting temperatures incorrectly can result in large errors in calculations involving the Arrhenius equation.
• Kelvin is the preferred scale for scientific calculations involving thermodynamic properties.

natural logarithm

The natural logarithm, denoted as \( \ln \), plays a crucial role in the mathematical manipulation of the Arrhenius equation. It transforms the exponential form into a linear one, making it easier to deduce \( E_a \) from rate constant data.

When comparing rate constants at two temperatures, we utilize the relation:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
In the given exercise, \( \ln\left(\frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}}\right) \approx 2.227 \). This logarithmic transformation simplifies solving for \( E_a \), effectively revealing the relationship between temperature and reaction rate constants.

• The natural logarithm provides a way to linearize relationships in exponential equations.
• This simplification aids in deriving activation energy from experimental kinetic data.

Understanding this concept is key to mastering reaction kinetics and utilizing the Arrhenius equation effectively.