The rate constant is doubled when temperature | vedclass

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(C) Given: $T_1 = 27^{\circ} C = 300 \ K$,$T_2 = 37^{\circ} C = 310 \ K$. The rate constant doubles,so $k_2 = 2k_1$. Using the Arrhenius equation: $\l

Vedclass · Accepted Answer

$53.6$

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(C) Given: $T_1 = 27^{\circ} C = 300 \ K$,$T_2 = 37^{\circ} C = 310 \ K$. The rate constant doubles,so $k_2 = 2k_1$. Using the Arrhenius equation: $\log(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} 	imes (\frac{T_2 - T_1}{T_1 T_2})$. Substituting the values: $\log(2) = \frac{E_a}{2.303 	imes 8.314} 	imes (\frac{310 - 300}{300 	imes 310})$. $0.3010 = \frac{E_a}{19.147} 	imes (\frac{10}{93000})$. $E_a = \frac{0.3010 	imes 19.147 	imes 93000}{10} \approx 53598 \ J \ mol^{-1} \approx 53.6 \ kJ \ mol^{-1}$.

The rate constant is doubled when temperature increases from $27^{\circ} C$ to $37^{\circ} C$. What is the activation energy in $kJ \ mol^{-1}$?

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