An organic reaction is carried out at 500 K. If the same reaction carried out in the presence of a catalyst at the same rate, the temperature required is 400 K. What is the activation energy of the uncatalysed reaction if catalyst lowers the Ea by 20 kJ/mol?

Answer

`E_(a1) = 100 kJ mol^(-1)`
`E_(a2) = 80 kJ mol^(-1)`

Explanation

`k = Ae^(-E_(a)//RT)`
or `log k = log A = - (E_(a))/(2.303 RT)`
`:. log k_(500) = log A - (E_(a))/(2.303 RT_(1))`
and `log k_(400) = log A - (E_(a_(2)))/(2.303 RT_(2))`
Given that `k_(500) = k_(400)`,
So, `(E_(a_(1)))/(T_(1)) = (E_(a_(2)))/(T_(2))`
or `(E_(a_(1)))/(500) = (E_(a_(2)))/(400)`
or `(E_(a_(1)))/(E_(a_(2))) = (5)/(4)`
It is given that
`E_(a_(1)) = E_(a_(2)) = 20`
Comparing, we get
`E_(a_(1)) = 100 kJ mol^(-1)`
`E_(a_(2)) = 80 kJ mol^(-1)`