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.

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Q: The activation energy, Ea, for a particular reaction is 19.4 kJ/mol. If the rate constant at 80 °C…

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