If the rate constant of a reaction is 3.0 mol `L^(-1)s^(-1)` at 700 K and 30 mol `L^(-1)s^(-1)` at 800 K, what is the energy of activation for this reaction? `R = 8.314 JK^(-1) mol^(-1)`

Explanation

According to the available data:
`k_(1)=3.0 mol L^(-1)s^(-1), k_(2)=30 mol L^(-1)s^(-1), T_(1)=700 K, T_(2)=800K`.
R=`8.314 J K^(-1) mol^(-1)`.
`logk_(2)/k_(1)=E_(a)/(2.303R) [1/T_(1)-1/T_(2)]`
`log (30 mol L^(-1)s^(-1))/(3.0 mol L^(-1)s^(-1)) = (E_(a))/(2.303 xx 8.314 J K^(-1)mol^(-1)) [1/700K - 1/800K]`
`log 10 = (E_(a))/(2.303 xx 8.314 J mol^(-1))[(800-700)/(700 xx 800)]`
`1=E_(a)/(2.303 xx (8.314 J K^(-1)mol^(-1))[1/700K -1/800K]`
`log10 = (E_(a))/(2.303 xx 8.314 J mol^(-1))[(800-700)/(700xx800)]`
`1=(E_(a))/(2.303 xx 8.314 J mol^(-1)) xx (100)/(700 xx 800)`
`E_(a)= (2.303 xx 8.314 xx 700 xx 800)/(100)J mol^(-1)`
`=107223.99 J mol^(-1)=107.224 kJ mol^(-1)`