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A 1st order reaction is 50% complete in ...

A 1st order reaction is 50% complete in 30 minute at `27^(@)C` and in `10` minute at `47^(@)C`. Calculate the:
(a) Rate constant for reaction at `27^(@)C` and `47^(@)C`.
(b) Energy of activation for the reaction.
(c ) Energy of activation for the reverse reaction if heat of reaction is `-50 kJ mol^(-1)`.

Text Solution

Verified by Experts

For `I` order reaction, `K=0.693/t_(1//2)`
(a) `:. At 27^(@)C`,
`K_(1)=0.693/30=2.31xx10^(-2) "minute"^(-1)`
At `47^(@)C, K_(2)=0.693/10=6.93xx10^(-2) "minute"^(-1)`
(b) Now `2.303"log"_(10)K_(2)/K_(1)=E_(a)/R([T_(2)-T_(1)])/(T_(1)T_(2))`
`:. 2.303"log"_(10)(6.93xx10^(-2))/(2.31xx10^(-2))=E_(a)/8.314xx([320-300])/(320xx300)`
`:. E_(a)=43850 J mol^(-1)=43.85 kJ mol^(-1)`
(c ) For a reaction `E_(a_(F.R.))-E_(a_(B.R.))=DeltaH`
`43.85-E_(a_(B.R.))=-50.0`
or `E_(a_(B.R.))=43.85+50.0=93.85 kJ mol^(-1)`
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