Final answer:
To determine the activation energy for a reaction where a temperature rise doubles the rate constant, one can use the Arrhenius equation and solve for Ea, given the specific temperatures and the fact that the rate constant doubles.
Step-by-step explanation:
The question asks to determine the activation energy of a reaction given that a temperature increase from 11.0 °C to 22.0 °C doubles the rate constant. This scenario can be addressed using the Arrhenius equation, which relates the rate constant of a reaction to the temperature and activation energy. The Arrhenius equation in logarithmic form is given by ln(k) = ln(A) - (Ea/(R*T)), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the ideal gas constant (8.314 J/mol*K), and T is the temperature in Kelvin.
Considering the doubling of rate constant for a 10 °C rise in temperature, we can set up two equations for the two temperatures (11 °C or 284.15 K and 22 °C or 295.15 K) and solve for Ea. The fact that the rate constant doubles implies that ln(k2/k1) = ln(2), where k2 is the rate constant at the higher temperature and k1 is the rate constant at the lower temperature. Solving for Ea using these equations we can figure out which option (a, b, c, or d) corresponds to the calculated activation energy.