Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles...

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Answer 1

Answer #1

temperature T1 = 63 oC = 336 K

T2 = 73 oC = 346 K

rate doubles . so k2 = 2 k1

ln (k2 / k1) = Ea / R [1 / T1 - 1 / T2]

ln (2 k1 / k1) = Ea / 8.314 x 10^-3 [1/336 - 1/346]

ln 2 = Ea / 8.314 x 10^-3 [1/336 - 1/346]

Ea = 67.0 kJ/mol

activation energy = 67 kJ/mol

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Answer 2

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