Question

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K.
Calculate the activation energy of the reaction . (Given : log 2 = `0.3010` , log 4 = `0.6021` , R = `8.314 J K^(-1) mol^(-1)`) .

Answer 1

`t_(1//2)` for first order reaction :
`t_(1//2) = (0.693)/(K)`
For 300 K , `" " K_(1) = (0.693)/(t_(1//2)) = (0.693)/(40) S^(-1)`
For 320 K , `" " K_(2) = (0.693)/(t_(1//2)) = (0.693)/(20) S^(-1)`
`log ((K_(2))/(K_(1))) = (E_(a))/(2.303R) ((1)/(T_(1)) - (1)/(T_(2)))`
log `(((cancel0.693)/(20))/((cancel0.693)/(40))) = (E_(a))/(2.303 xx 8.314) [(1)/(300) - (1)/(200)]`
`log(2) = (E_(a))/(2.303 xx 8.314)xx ((320 - 300)/(300 xx 320))`
`0.3010 = (E_(a))/(2.303 xx 8.314) xx (cancel20)/(cancel200_(150) xx cancel320)`
`E_(a) = 0.3010 xx 2.303 xx 8.314 xx 150 xx 32`
`E_(a) = 27663, 8 J//"mol".