Explain what temperature would the reaction go twice as fast

Reference no: EM13264733

The activation energy of a certain reaction is 32.3kJ/mol . At 20 degrees C, the rate constant is 0.0130s^-1.At what temperature would this reaction go twice as fast? The equation that is given is ln(K2/K1)=Ea/R((1/T1)-(1/T2)) a. what is R in this equation? What would the answer be

Reference no: EM13264733

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