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

Question

vfiekz [6]

3 years ago

12

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

Chemistry

1 answer:

Ann [662] · Answer 1 · 2022-07-25T02:09:22+03:00

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