InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The rate constants of a reaction at 500 K and 700 K are `0.02s^(-1)` and `0.07s^(-1)` respectively. Calculate the values of `E_(a)` and A. |
|
Answer» According to Arrhenius equation, `logk_(2)/k_(1)= E_(a)/(2.303R)[(T_(2)-T_(1))]/(T_(1)T_(2))` `k_(1)= 0.02s^(-1),k_(2)=0.07s^(-1), T_(1)= 500K, T_(2)= 700K` `log 0.07/0.02 = (E_(a))/((2.303 xx 8.314 JK^(-1)mol^(-1)))[(700K-500K)/(700K xx 500K)]` `log 3.5 = (E_(a))/(2.303 xx 8.314 JK^(-1)mol^(-1))xx((200K)/(700K xx 500K))` `E_(a) = (0.5441 xx 2.303 xx 8.314 xx 700 xx 500)/(200) J mol^(-1)= 18231.4 J mol^(-1)= 18.23 kJmol^(-1)` log k = `(-E_(a))/(2.303R) + log A or log A=log K + (E_(a))/(2.303RT)` log A `= log 0.02 + (18231.4 J mol^(-1))/(2.303 xx 8.314JKmol^(-1) xx 500K)` `=-log 50 + 1.904 = -1.699 + 1.904 = 0.205` A= Antilog 0.205 = 1.603 |
|