The rate constant for the first order decompoistion of a certain reaction is described by the equation
`log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)`
(a) What is the energy of activation for the reaction?
(b) At what temperature will its half-life periof be `256 min`?

Answer

`T = 669 K`

Explanation

(a) `k = Ae^(-E_(a)//RT)`
or `log k = log A - (E_(a))/(2.303 RT)`
`= 14.34 - (1.25 xx 10^(4))/(T)K`
On careful comparison, we get
`(E_(a))/(2.303 R) = 1.25 xx 10^(4)`
or `(E_(a))/(2.303 xx 8.314) = 1.25 xx 10^(4)`
or `E_(a) = 2.3933 xx 10^(5)`
(b) `k = (0.693)/("Half time") = (0.693)/(256 xx 60) = 4.51 xx 10^(-5)s^(-1)`
`k = Ae^(-E_(a)//RT)`
or `log k = log A - (E_(a))/(2.303 RT) = 14.34 - (1.25 xx 10^(4))/(T)`
or `log (4.51 xx 10^(-5)) = 14.34 - (1.25 xx 10^(4))/(T)K`
or `T = 669 K`