The rate of most reactions become double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.
(Given, `R = 8.314" J mol"^(-1)` )

Explanation

From Arrhenius equation ,we know
`"log"K_1/K_2=E_a/(2.303R)[1/T_1-1/T_2]`
Here `K_2/K_1=2,T_1=298K,T_2=308K`.Substituting the values,we get
`"log"2=E_a/(2.303xx8.314JK^-1mol_^-1)[(T_2-T_1)/(T_1T_2)]`
`E_a=`52897.7mol^-1=52.89kj` mol.