InterviewSolution
Saved Bookmarks
| 1. |
The first order rate constant for the decompoistion of `C_(2)H_(5)I` by the reaction. `C_(2)H_(5)I(g)rarrC_(2)H_(4)(g)+HI(g)` at `600 K is 1.60xx10^(-5)s^(-1)`. Its energy of activation is `209 kJ mol^(-1)`. Calculate the rate constant at `700 K` |
|
Answer» `log k_(2)/k_(1) = E_(a)/(2.303)[(T_(2)-T_(1))/(T_(1)T_(2))]` `E_(a) = 209 kJmol^(-1) = 209 xx 10^(3)Jmol^(-1), T_(1)=600K, T_(2)=700K`, `log k_(2)/k_(1)=(209 xx 10^(3)Jmol^(-1) xx 100K)/(2.303 xx (8.314 JK^(-1)mol^(-1)(600K) xx (700K)) = 2.599` `k_(2)/k_(1) = "Antilog" 2.599 =397.2` `k_(2)=397.2 xx k_(1) =397.2 xx (1.6 xx 10^(-5)s^(-1)) =6.36 xx 10^(-3)s^(-1)` |
|