Two reactions of the same order have equal pre exponential factors but their activation energies differ by 24.9kJ `"mol"^(-1)`. Calculate the ratio between the rate constants of these reactions at `27^(@)C`. (Gas constant R = 8.314 J `K^(-1) "mol"^(-1)`)

Explanation

The Arrhenius equation : `k = Ae^(-Ea//RT)`
Taking log on both sides : log k `="log A"-(E_(a))/(2.303RT)`
For reaction (i) log `k_(1) = "log A"- (E_(a)(1))/(2.303 RT)`
For reaction (ii) log `k_(2) = "log A" -(E_(a)(2))/(2.303 RT)`
Subtracting (i) from (ii)
`"log" (k_(1))/(k_(2)) = (E_(a)(1)-E_(a)(2))/(2.303 RT)`
`="log"(k_(1))/(k_(2)) = (24.9 xx 1000)/(2.303 xx 8.3 xx 300) = 4.342`
`(k_(1))/(k_(2)) = "anti log"(4.342) = 2.198 xx 10^(4)`