A certain reactions is `50%` complete in 20 minutes at 300 K and the same reaction is again 50% completely in 5 minutes at 350 K. Calculate the activation energy if the reactions is of first order.

Explanation

`t_(1//2) = 0.693/k` or `k =0.693/t_(1//2)`
`k_(1) = 0.693/20 min= 0.03465 min^(-1)` and `k_(2) = 0.693/(5"min") = 0.1386 min^(-1)`
According to Arrhenius equation,
`logk_(2)/k_(1) = (E_(a))/(2.303R) [1/T_(1)-1/T_(2)]`
Hence, `k_(1)= 0.03465 min^(-1)` , `k_(2)=0.1386 min^(-1), T_(1) = 300K, T_(2)=350K`
`log 4= (E_(a) xx 50)/(2.303 xx (8.314 Jmol^(-1) K^(-1))xx300 xx 350)`
`0.6020 = E_(a)/(2.303 xx (8.314 J mol^(-1)) xx (2100))`
`E_(a) = 0.6020 xx 2.303 xx (8.314 J mol^(-1)) xx (2100)`
`=24205.8 J mol^(-1) = 24.2 kJmol^(-1)`