A first order reaction is 50% completed in 20 minutes at `27^(@)`C. The energy of activation of the reaction is-

Options

A. 43.58 KJ
B. 55.14 KJ
C. 11.97 KJ
D. 6.65 KJ

Answer

B

Explanation

`"Log"((K_(2))/(K_(1)))=(Ea)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
`Log4=(Ea)/(2.303xx8)((1)/(300)-(1)/(320))`
`{K_(1)=(1n2)/(20),K_(2)=(n2)/(5)}`