Two reactions of the same order have equal exponential factors but their activation energies differ by `24.9 kJ mol^(-1)`. Calculate the ratio between the rate constants of these reactions at `27^(@)`C (Gas constant `R= 8.3 JK^(-1)mol^(-1))`

Explanation

The Arrehenius equation `k=Ae^(-Ea//Rt)`
Taking log on both sides
`log k = log A - (E_(a))/(2.303 RT)`
`log k_(1) = log A - (E_(a1))/(2.303RT)`.(i)
`log k_(2) = log A - (E_(a2))/(2.303 RT)` (ii)
Subtract equn (i) from eq. (ii),
`log k_(1)/k_(2) = (E_(a1) -E_(a2))/(2.303 RT) = (24.9 xx (1000J))/(2.303 xx (8.3JK^(-1) mol^(-1)) xx (300 K))`
`=4.342`
`k_(1)/k_(2)`= Antilog `4.342` = `2.198 xx 10^(4)`