If the rate constants of a reaction at 500K and 700K are `0.002 s^(-1) and 0.06 s^(-1)` respectively, the value of `K^(-1)` activation energy is `(R = 8.314 J mol^(-1) K^(-1), log 3 = 0.477)`

Options

A. `49.49" kJ mol"^(-1)`
B. `98.98" kJ mol"^(-1)`
C. `24.75" kJ mol"^(-1)`
D. `12.37" kJ mol"^(-1)`

Answer

A

Explanation

For a reaction working at two different temperature we have
`log""(k_(2))/(k_(1))=(E_(a))/(2.303R)[(DeltaT)/(T_(1)T_(2))]`
Substituting `k_(1)=0.002s^(-1),k_(2)=0.06s^(-1)`
`T_(1)=500K,T_(2)=700K`
`R=8.314" J mol"^(-1)K^(-1)`
`log""(0.002)/(0.06)=(E_(a))/(2.303xx8.314)[(700-500)/(700xx500)]`
`log 0.030=(E_(a))/(19.147)[(200)/(350000)]`
`log(3.0xx10^(-2))=(E_(a))/(19.147)[(2)/(3500)]`
`E_(a)=log(3xx10^(-2))xx(19.147xx3500)/(2)`
`E_(a)=49.49" kJ mol"^(-1)`