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A first order reaction is 50% completed ...

A first order reaction is `50%` completed in `30 min` at `27^(@)C` and in `10 min` at `47^(@)C`. Calculate the reaction rate constants at `27^(@)C` and the energy of activation of the reaction in `kJ mol^(-1)`.

Text Solution

Verified by Experts

`k=0.0231" min"^(-1), 43.848" kJ mol"^(-1)`
`k=(0.693)/(t_(1//2)), (k_(2))/(k_(1)) =(E_(a))/(2.303) [(T_(2)-T_(1))/(T_(1) cdot T_(2))]`
Substituting the value at the two given conditions `k_(1)=k_(300) =(0.693)/(30) =0.0231" m in"^(-1) rArr k_(2)=k_(320) =(0.693)/(10) =0.0693" m in"^(-1)`
We also know that log `(k_(2))/(k_(1)) =(Ea)/(2.303 R) xx(T_(2)-T_(1))/(T_(1)xxT_(2))`
or `E_(a) =(2.303 R xx T_(1) xx T_(2))/(T_(2)-T_(1))" log "(k_(2))/(k_(1)) =(2.303xx8.314xx10^(-3) xx300xx320)/(320-300)xx log (0.0693)/(0.0231) =43.848" kJ mol"^(-1)`
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