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)`) .

Explanation

`K_2=0.693//20`,
`k_1=0.693//40`
`log k_2/k_1=E_2/2.303 R [1/T_1-1/T_2]`
`k_2//k_1=2`
`log 2 =E_8/(2.303 times 1.314) [(320-300)/(320 times 300)],Ea=27.66 KJ Mol^-1`