InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The rate constant of decomposition of methyl nitrite `(K_(1))` and ethyl nitrite `(K_(2))` follow the equations. `K_(1)(s^(-1))=10^(13)e^([(-152.3xx10^(3))/(RT)Jmol^(-1)])` and `K_(2)(s^(-1))=10^(14)e^([(-152.3xx10^(3))/(RT)Jmol^(-1)])` Calculate the temperature at which both have same rate of decomposition if `0.1M` of each is taken and both show `I` order kinetics. |
|
Answer» Rate= `K "[Reactant"]^(1)` Since [Reactant]=`0.1 M` and rates are same, thus at same temperature `K_(1)=K_(2)` `10^(13)e^((-152.3xx10^(3))/(RT))=10^(14) e^((-157.7xx10^(3))/(RT)` `e^(((+5.4xx10^(3))/(RT)))=10` `(5.4xx10^(3))/(RT)=2.303` `:. T=(5.4xx10^(3))/(8.314xx2.303)=282 K` |
|