If a temperature increase from 20.0 °C to 35.0 °C triples 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) = 20.0 °C, final temperature (T2) = 35.0 °C, and the rate constant triples, so k2 = 3k1.
2. Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature: T1 = 293.15 K and T2 = 308.15 K.
3. Use the Arrhenius equation in its ratio form: \( \frac{k_2}{k_1} = \frac{A e^{-E_a/(RT_2)}}{A e^{-E_a/(RT_1)}} = e^{-E_a/(RT_2) + E_a/(RT_1)} \), which simplifies to \( \frac{k_2}{k_1} = e^{E_a/R (1/T_1 - 1/T_2)} \).
4. Substitute the known values into the equation: \( 3 = e^{E_a/R (1/293.15 - 1/308.15)} \).
5. Solve for the activation energy \( E_a \) by taking the natural logarithm of both sides and rearranging the equation: \( E_a = R \cdot \ln(3) / (1/293.15 - 1/308.15) \), where R is the gas constant (8.314 J/mol·K).