If a temperature increase from 10.0 °C to 20.0 °C doubles the rate constant for a reaction, what is the value of the activation energy for the reaction?

Verified step by step guidance

1. Identify the given information: initial temperature (T1) = 10.0 °C, final temperature (T2) = 20.0 °C, and the rate constant doubles, so k2 = 2k1.
2. Convert the temperatures from Celsius to Kelvin: T1 = 10.0 + 273.15 K and T2 = 20.0 + 273.15 K.
3. Use the Arrhenius equation in the form: \( \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \), where \( E_a \) is the activation energy and \( R \) is the gas constant (8.314 J/mol·K).
4. Substitute the known values into the equation: \( \ln(2) = \frac{-E_a}{8.314} \left( \frac{1}{293.15} - \frac{1}{283.15} \right) \).
5. Solve for the activation energy \( E_a \) by isolating it on one side of the equation.