If a temperature increase from 10.0 ∘C to 21.0 ∘C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction

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08-03-2018
Chemistry

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If a temperature increase from 10.0 ∘C to 21.0 ∘C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction

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superman1987 superman1987 · Answer 1 · 2018-03-19T17:39:41+01:00

superman1987 superman1987

19-03-2018

According to this formula:
㏑(K2/K1) = Ea/R(1/T1 - 1/T2)
when K is the rate constant
Ea is the activation energy
R is the universal gas constant
and T is the temperature K
when K is doubled so K2: K1 = 2:1 & R = 8.314 J.K^-1.mol^-1
and T1 = 10 +273 = 283 k & T2 = 21 + 273 = 294 k
So by substitution:
㏑2 =( Ea / 8.314) (1/283 - 1/294 )
∴ Ea = 43588.9 J/mol = 43.6 KJ/mol

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