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)]

A first order reaction is 50% completed in 40 minutes at 300 K and in

1.	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)]
Answer» Solution :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`