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a) The rate of a particular reaction dou...

a) The rate of a particular reaction doubles when the temperature changes from 300 K to 310 K. Calculate the energy of activation of the reaction. [Given : `R="8.314 JK"^(-1)" mol"^(-1)`].

Text Solution

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`log.(K_(2))/(K_(1)) = (E_(a))/(2.303 R) [(T_(2)-T_(1))/(T_(1)T_(2))]`
`E_(a) = (2.303 xx 8.314 xx 310 xx 300)/(10) log.(2)/(1)`
`E_(a) = 53598.59 J` or `53.598 K J`
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