At 300K, a certain reaction is `50%` complete in 20 minutes. At 350K, the same reaction is `50%` complete in 5 minutes. Calcualte the activation energy for the reaction.

Explanation

`k_(1) = 0.693/(20 mm) = 0.03465 mm^(-1), T_(1)=300K`
`k_(2) = 0.693/(5 mm) = 0.1386 mm^(-1), T_(2)=350K`
`log k_(2)/k_(1) = E_(a)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2)]`
`log(0.1386 min^(-1))/(0.03465 min^(-1)) = (E_(a)xx[350K -300K])/(2.303 xx 8.314 JK^(-1)mol^(-1) xx 350K xx 300K)`
`log4=(E_(a) xx 50)/(2.303 xx 8.314 xx 350 xx 300(Jmol^(-1))`
`E_(a)=(0.6021 xx 2.303 xx 8.314 xx 350 xx 300)/(50) Jmol^(-1)`
`24210 J mol^(-1)=24.21 kj mol^(-1)`.