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 JK^(-1)"mol"^(-1)`]

Explanation

For a first order reaction
`k=(2.303)/(t)"log"([R]_(0))/([R])`
At 300K
`k_(1)=(2.303)/(40)"log"([R]_(0))/([R]_(0)//2)=(2.303)/(40)"log 2"=(2.303)/(40)xx0.3010`
AT 320 K
`k_(2)=(2.303)/(20)"log"([R]_(0))/([R]_(0)//2)=(2.303)/(20)log2=(2.303)/(20) xx 0.3010`
`(k_(2))/(k_(1))=(2.303xx0.3010xx40)/(20xx2.303xx0.3010) =2`
Applying Arrhenius equation, we have
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Substituting the values, we have
`log 2=(E_(a))/(2.303xx8.314JK^(-1)"mol"^(-1))[(320K-300K)/(320K xx 300K)]`
or `0.3010=(E_(a))/(19.1471)[(20)/(96000)]`
or `E_(a)=(0.3010xx19.1471xx96000)/(20)=27663.7J`