Question

A first order reaction has rate constants of `4.6xx"10"^(-2)" s"^(-1)"` and `8.1xx"10"^(-2)" s"^(-1)"` at 0 degrees Celsius and 20 degrees Celsius, respectively. What is the value of the activation energy?

Answer 1

18.81 kJ/mole

Explanation:

We have Arrhenius equation,
`k= A*e^(E/(RT))`
where `k` is rate constant
`A` is Arrhenius constant
`E` is Activation Energy and
`T` is Temperature

We have T(1) = 273 K and T(2) = 293 K
`k(1) = 4.6 *10^-2 s^-1`
`k(2) = 8.1 *10^-2 s^-1`

`(4.6 *10^-2)/(8.1*10^-2)=e^(-E/R(1/(273)-1/293))`

with `R=8.314 (kJ)/(mol. K)`
we get `E = 18.81 (kJ)/(mol)`