The activation energy of a certain reaction is 41.8 kJ/mol . At 29∘C , the rate constant is 0.0180s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Solve the following equation for T2: ln ⁡(k2/k1) = Ea/R (1/T1−1/T2) The temperatures must be in kelvins and the activation energy must be in J/mol. Use the following value for the gas constant: R=8.314 J/mol⋅K.

For a certain reaction, the activation energy is found to be (1.77x10^1) kJ/mol. If the rate constant is (3.6x10^-5) at (3.78x10^2) K, what is the rate constant at (6.290x10^2) K?   R = 8.314 J/Kmol;     T(K) = T(°C) + 273 Report your answer in the same units as k1, with 2 sig figs.

A researcher raises the temperature from 47.4 to 64.7 °C and finds that the rate of the reaction doubles. What was the activation energy (in Joules) for this reaction? (R = 8.3145 J/molk)

The following data were collected for the reaction between hydrogen and nitric oxide at 700°C. 2 H2(g) + 2 NO(g) - 2 H20(g) + N2(g) Experiment [H2]/M [NO]/M Initial rate/M-s-1 0.012 0.031 4.5x10-6 0.0060 0.031 2.2x10-6 3 0.012 0.0155 1.1x10-6 (a) What is the rate law for the reaction? (Rate expressions take the general form: rate = k. [A]ª . [B]b.) chemPad Help Greek - (b) Calculate the rate constant for the reaction. 4.0 (c) Suggest a plausible reaction mechanism that is consistent with the rate law. (Hint: Assume that the oxygen atom is the intermediate. Omit states-of-matter from your answer.) First step: chemPad Help Greek - Second step: chemPad Help Greek