The rate constant of a reactant is `1.5 xx 10^(-3)` at `25^(@)C` and `2.1 xx 10^(-2)` at `60^(@)C`. The activation energy is

Options

A. `(35)/(333)R log_(e).(2.1 xx 10^(-2))/(1.5 xx 10^(-2))`
B. `(298 xx 333)/(35)R log_(e). (21)/(1.5)`
C. `(298 xx 333)/(35)R log_(e) 2.1`
D. `(298 xx 333)/(35) R log_(e).(2.1)/(1.5)`

Answer

B

Explanation

`T_(1) = 273 + 25 = 298 K`
`T_(2) = 273 + 60 = 333 K`
`log.(k_(2))/(k_(1)) = (E_(a))/(2.3 R)((T_(2)-T_(1))/(T_(1)T_(2)))`
or `log_(e).(k_(2))/(k_(1)) = (E_(a))/(R )((T_(2)-T_(1))/(T_(1)T_(2)))`
`log.(2.1 xx 10^(-2))/(1.5 xx 10^(-3)) = (E_(a))/(R ) xx ((35)/(33 xx 298))`
`:. E_(a) = (298 xx 333)/(35) xx R xx log_(e).(21)/(1.5)`