Calculate the activation energy for the reaction that has a rate constant of 0.124 s^–1 at 125.0 °C and a rate constant of 0.345 at 200.0 °C.

• A. –2.84 kJ
• B. –21.4 kJ
• C. +2.57 kJ
• D. +2.84 kJ
• E. +21.4 kJ

Answer: (E)

Verified step by step guidance

1. Identify the given values: rate constant k1 = 0.124 s^-1 at T1 = 125.0 °C and rate constant k2 = 0.345 s^-1 at T2 = 200.0 °C.
2. Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature: T1 = 125.0 + 273.15 K and T2 = 200.0 + 273.15 K.
3. Use the Arrhenius equation in its logarithmic form: ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2), where Ea is the activation energy and R is the gas constant (8.314 J/mol·K).
4. Rearrange the equation to solve for the activation energy Ea: Ea = R * ln(k2/k1) / (1/T1 - 1/T2).
5. Substitute the known values into the equation and solve for Ea, ensuring that the units are consistent throughout the calculation.