InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The reaction `C_(2)H_(5)I to C_(2)H_(4) + HI` is of first order and its rate constants are `3.20 xx 10^(-4) s^(-1)` at 600 K and `1.60 xx 10^(-2)s^(-1)` at 1200 K. Calculate the energy of activation for the reaction. (Given `R= 8.314 JK^(-1)mol^(-1))` |
|
Answer» `(log) k_(2)/k_(1) = E_(a)/(2.303R) [1/T_(1)-1/T_(2)]` `k_(1) = 3.20 xx 10^(-4)s^(-1), k^(2) = 1.60 xx 10^(-2)s^(-1), T_(1) = 600 K, T_(2)=1200K` `k_(1) = 3.20 xx 10^(-4)s^(-1), k_(2) = 1.60 xx 10^(-2)s^(-1), T_(1)= 600K, T_(2)=1200K` `(log) (1.60 xx 106(-2)s^(-1))/(3.20 xx 10^(-4)s^(-1)) = (E_(a))/(2.303 xx (8.314JK^(-1) mol^(-1))) xx [(1200K - 600K)/(600K) xx 1200K]` `log50 = (E_(a))/(2.303 xx (8.314 JK^(-1)mol^(-1))) xx (600)/(600 xx 1200K)` `1.6900 = (E_(a))/(2.303 xx 8.314 xx 1200(J mol^(-1)))` `E_(a) = 1.6990 xx 2.303 xx 8.314 xx 1200(Jmol^(-1))=39037 J mol^(-1)` `=39.037 kJ mol^(-1)` |
|