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The rate of reaction is doubled when the...

The rate of reaction is doubled when the temperature changes from `27^@C` to `37^@C`. Calculate the energy of activation.

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We know that` "log" K_2/K_1=E_a/(2.303R)[1/T_1-1/T_2]`Substituting the values,
`"log2"=E_a/(2.303xx8.314)[1/300-1/310]`
`E_a=(0.3010xx2.303xx8.314xx300xx310)/10=535986.5=53.5kj=535986.5=53.5kj`
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