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By what factor does the fraction of collisions with energy equal to or greater than activation...

By what factor does the fraction of collisions with energy equal to or greater than activation energy at 100 kJ/mol. change if the temperature increases from 35 degrees celsius to 50 degrees celsius. Give your answer in scientific notation

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Answer #1

   Ea   = 100KJ/mole

Ea     = 100000J/mole

T1 = 35+273 = 308K

T2   = 50 + 273 = 323K

logK2/K1    =   Ea/2.303R [1/T1 -1/T2]

logK2/K1     = 100000/2.303*8.314[1/308 -1/323]

logK2/K1    = 5222.7(0.00325-0.00309)

logK2/K1    = 0.8356

K2/K1    = 10^0.8356

K2/K1   = 6.85

K2/K1   = 6.85

K2     = 6.85 K1

K2 is increases 6.85 times greater than K1

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