The rate of a certain reaction doubles when t | vedclass

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(C) Given: $T_1 = 27\,^oC = 300\,K$,$k_1 = k$$T_2 = 37\,^oC = 310\,K$,$k_2 = 2k$Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303

Vedclass · Accepted Answer

$53.6$

Vedclass · Answer

(C) Given: $T_1 = 27\,^oC = 300\,K$,$k_1 = k$$T_2 = 37\,^oC = 310\,K$,$k_2 = 2k$Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} ight)$Substituting the values:$\log \frac{2k}{k} = \frac{E_a}{2.303 	imes 8.314} 	imes \frac{310 - 300}{300 	imes 310}$$\log 2 = \frac{E_a}{2.303 	imes 8.314} 	imes \frac{10}{93000}$$0.3010 = \frac{E_a}{19.147} 	imes 0.0001075$$E_a = \frac{0.3010 	imes 19.147}{0.0001075} \approx 53598.6\,J\,mol^{-1}$$E_a \approx 53.6\,kJ\,mol^{-1}$

The rate of a certain reaction doubles when the temperature changes from $27\,^oC$ to $37\,^oC$. The activation energy for the reaction is ........... $kJ\,mol^{-1}$.

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