Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![In this exercise, you will learn how to calculate the activation energy (Ea) of a reaction using the Arrhenius equation. The rate constant \( k \) for a certain reaction is measured at two different temperatures, as shown in the table below:
\[
\begin{array}{|c|c|}
\hline
\text{temperature} & k \\
\hline
248.0 \, ^\circ \text{C} & 9.9 \times 10^8 \\
169.0 \, ^\circ \text{C} & 9.8 \times 10^7 \\
\hline
\end{array}
\]
To calculate the activation energy \( E_a \) for this reaction, we will assume the rate constant obeys the Arrhenius equation:
\[
k = A e^{-\frac{E_a}{RT}}
\]
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)),
- \( T \) is the temperature in Kelvin.
Given:
- The temperatures are 248.0°C and 169.0°C, which need to be converted to Kelvin (K):
- \( 248.0°C + 273.15 = 521.15 \, K \)
- \( 169.0°C + 273.15 = 442.15 \, K \)
- The rate constants \( k \) at these temperatures are:
- \( k_1 = 9.9 \times 10^8 \) at \( T_1 = 521.15 \, K \)
- \( k_2 = 9.8 \times 10^7 \) at \( T_2 = 442.15 \, K \)
Using the logarithmic form of the Arrhenius equation:
\[
\ln \left( \frac{k_2}{k_1} \right) = - \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
\]
Rearranging to solve for \( E_a \):
\[
E_a = - R \frac{\ln \](https://content.bartleby.com/qna-images/question/e361a5c2-e3a6-4cd8-bee8-5dcb720cb10b/de4ae162-2e68-49a1-b31a-f570b777c306/hw1y99l_processed.jpeg)
Transcribed Image Text:In this exercise, you will learn how to calculate the activation energy (Ea) of a reaction using the Arrhenius equation. The rate constant \( k \) for a certain reaction is measured at two different temperatures, as shown in the table below:
\[
\begin{array}{|c|c|}
\hline
\text{temperature} & k \\
\hline
248.0 \, ^\circ \text{C} & 9.9 \times 10^8 \\
169.0 \, ^\circ \text{C} & 9.8 \times 10^7 \\
\hline
\end{array}
\]
To calculate the activation energy \( E_a \) for this reaction, we will assume the rate constant obeys the Arrhenius equation:
\[
k = A e^{-\frac{E_a}{RT}}
\]
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)),
- \( T \) is the temperature in Kelvin.
Given:
- The temperatures are 248.0°C and 169.0°C, which need to be converted to Kelvin (K):
- \( 248.0°C + 273.15 = 521.15 \, K \)
- \( 169.0°C + 273.15 = 442.15 \, K \)
- The rate constants \( k \) at these temperatures are:
- \( k_1 = 9.9 \times 10^8 \) at \( T_1 = 521.15 \, K \)
- \( k_2 = 9.8 \times 10^7 \) at \( T_2 = 442.15 \, K \)
Using the logarithmic form of the Arrhenius equation:
\[
\ln \left( \frac{k_2}{k_1} \right) = - \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
\]
Rearranging to solve for \( E_a \):
\[
E_a = - R \frac{\ln \
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