The activation energy of a certain reaction is 45.8 kJ/mol. At 20 degrees celsius, the rate constant is 0.0130 s^(-1). At what temperature would this reaction go twice as fast? Arrhenius Equation:k = Ae^(-E_a/RT) ln(k_2 / k_1) = (E_a /R)(1/T_1 - 1/T_2) Okay, so I'm not quite sure how to approach this problem. I think you're supposed to solve for k_2 and then sub it in to find the activation energy and then find the temperature. But I don't know how to solve for k_2 if I'm not given a second temperature. Any help is appreciated!