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:\[\begin{array}{|c|c|}\hline\text{temperature} & k \\\hline248.0 \, ^\circ \text{C} & 9.9 \times 10^8 \\169.0 \, ^\circ \text{C} & 9.8 \times 10^7 \\\hline\end{array}\]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 \

Name: In this exercise, you will learn how to calculate the activation ener
Uploaded: 2022-11-30T19:17:17.000Z
Channel: Bartleby
Description: 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