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The time required for 10% completion of ...

The time required for `10%` completion of first order reaction at `298K` is equal to that required for its `75%` completion at `308K`. If the pre-expontential factor for the reaction is `3.56xx10^(9)s^(-1)`, calculate its rate constant at `318K` and also the energy of activation.

Text Solution

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For first order reactions,
`t=(2.303)/(k)log.(N_(0))/(N_(1))`
At `298K`, `t=(2.303)/(k_(298))log.(100)/(90)`
At `309k, t=(2.303)/(k_(308))log.(100)/(25)`
Since time is same
`(2.303)/(k_(298))log.(100)/(90)=(2.303)/(k_(308))log.(100)/(25)`
or `(0.0458)/(k_(298))=(0.1249)/(k_(308))1`
or `(k_(308))/(k_(298))=(0.1249)/(0.0458)=2.73`
According to Arrhenius equation
`2.303log.(k_(308))/(k_(298))=(E_(a))/(8.314)[(1)/(298)-(1)/(208)]`
or `2.303log2.73=(E_(a))/(8.314)[(10)/(298xx308)]`
`E_(a)=76.65kJ`
and `K_(318)=9.306xx10^(-4)s^(-1)`
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