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(C) The Arrhenius equation is given by: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Given: $k_2 = 2k_1$,$T_1 =

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$52.9$

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(C) The Arrhenius equation is given by: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given: $k_2 = 2k_1$,$T_1 = 298 \, K$,$T_2 = 308 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$.Substituting the values: $\log(2) = \frac{E_a}{2.303 	imes 8.314} \left[ \frac{308 - 298}{308 	imes 298} ight]$$0.3010 = \frac{E_a}{19.147} \left[ \frac{10}{91784} ight]$$E_a = \frac{0.3010 	imes 19.147 	imes 91784}{10} \approx 52897 \, J \, mol^{-1} \approx 52.9 \, kJ \, mol^{-1}$.

If the rate of a reaction doubles when the temperature increases from $298 \, K$ to $308 \, K$,the activation energy of the reaction is ........... $kJ \, mol^{-1}$.

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