1.

The rate constant of a reaction is doubled when the temperature increased from400K to 410K. Calculate the activation energy (Ea) [R = 8.314 JK^(-1) " mol"^(-1)] .

Answer»

SOLUTION :`log""(k_2)/(k_1) = (E_(a))/(2.303R) [(T_2 - T_1)/(T_1T_2)]`
` log 2 = (E_a)/(2.303 xx 8.314 ) [(410-400)/(400 xx 410)]`
` 0.310 = (E_a)/(19.147)[(10)/(164000)]`
`E_a = (0.301 xx 19.147 xx 164000)/(10)`
`= 94517 "Jmol"^(-1) OR 94.517 "kJmol"^(-1)`


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