### Example Problem on Calculating Activation EnergyThe rate constant \( k \) for a certain reaction is measured at two different temperatures:| Temperature (\( \degree C \)) | \( k \) (s\(^{-1}\)) ||-------------------------------|-----------------------------|| 362.0 | \( 1.3 \times 10^{10} \) || 423.0 | \( 1.7 \times 10^{10} \) |Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy \( E_a \) for this reaction.Round your answer to 2 significant digits.#### Arrhenius EquationThe Arrhenius equation is expressed as:\[ k = A \exp \left( \frac{-E_a}{RT} \right) \]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.#### Calculating Activation EnergyTo find the activation energy, use 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 for \( E_a \):\[ E_a = - \frac{R \ln \left( \frac{k_2}{k_1} \right)}{\left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \]First, convert the temperatures from Celsius to Kelvin:\[ T_1 = 362.0 + 273.15 = 635.15 \, \text{K} \]\[ T_2 = 423.0 + 273.15 = 696.15 \, \text{K} \]Let's plug in the values:\[ \ln \left( \frac{1.7 \times 10^{10}}{1.3 \times 10^{10}} \right) = \ln \left( \frac{1.7}{1

Name: ### Example Problem on Calculating Activation EnergyThe rate constant
Uploaded: 2022-11-30T19:17:17.000Z
Channel: Bartleby
Description: ### Example Problem on Calculating Activation EnergyThe rate constant \( k \) for a certain reaction is measured at two different temperatures:| Temperature (\( \degree C \)) | \( k \) (s\(^{-1}\)) ||-------------------------------|---------