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  4. What is the energy of activation of a reaction if its rate doubles when the temperature is raised from 290 K to 300 K ?

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Q. What is the energy of activation of a reaction if its rate doubles when the temperature is raised from $290 K $ to $300 K $ ?

Chemical Kinetics

Solution:

$ log \frac{k_2}{k_1} =\frac{E_a}{2.303R} \frac{[T_2 - T-1}{[T_1T_2} $
$ log2 =\frac{Ea}{2,303\times 2} \frac {[300-290]}{[290 \times 300]} $
$ (k_2 =2K_2 ,R=2 cal/ K mol) $
$ Ea=12062\, calories \approx 12 \, kcal$