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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

For firs order reaction `k=(0.693)/(t_(1//2))`
`At 27^(@)C, k_(27^(@)C)=(0.693)/(30)=0.0231"min"^(-1)`
`At 47^(@)C, k_(47^(@)C)=(0.693)/(10)=0.0693"min"^(-1)`
Now applying the following equation :
`log_(10).(k_(1))/(k_(2))=(-E)/(2.303xxR)*((T_(2)-T_(1))/(T_(2)*T_(1)))`
or `log_(10).(0.0231)/(0.0693)=(-E)/(2.3030xx8.314)*((320-300)/(320xx300))`
`-log_(10)0.3333=(E_(a))/(19.1471)xx(20)/(9600)`
`E_(a)=-(19.171xx9600)/(20)xxlog0.3333`
`=-91906xx(-0.4772)`
`=43857J"mol"^(-1)=43.857kJ"mol"^(-1)`
`=43857J"mol"^(-1)=43.857kJ"mol"^(-1)`
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