I understand how to set up the equation but I cannot seem to find K2 algebreiaca

Question

I understand how to set up the equation but I cannot seem to find K2 algebreiacaly... Please help

The activation energy of a certain reaction is 47.7kJ/mol. At 30 degree C . the rate constant is 0.0130s-1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units Given that the initial rate constant is 0.0130s-1 at an initial temperature of 30 degree C , what would the rate constant be at a temperature of 190 degree C for the same reaction described in Part A? Express your answer with the appropriate units.

Explanation / Answer

k=Ae^-Ea/RT

HENCE k1/(e^-Ea/RT1) = k2/(e^-Ea/RT2)

solving for k2 we get k2 = 9.026/s

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