The rate of reaction becomes double when the | vedclass

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(A) Given: $T_1 = 27 + 273 = 300\,K$,$T_2 = 57 + 273 = 330\,K$. Rate becomes double,so $k_2 / k_1 = 2$. Using the Arrhenius equation: $\log(k_2 / k_1)

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$4.57$

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(A) Given: $T_1 = 27 + 273 = 300\,K$,$T_2 = 57 + 273 = 330\,K$. Rate becomes double,so $k_2 / k_1 = 2$. Using the Arrhenius equation: $\log(k_2 / k_1) = \frac{E_a}{2.303 	imes R} 	imes \frac{T_2 - T_1}{T_1 	imes T_2}$. Here $R = 1.987 	imes 10^{-3}\,k\,cal\,K^{-1}\,mol^{-1} \approx 2 	imes 10^{-3}\,k\,cal\,K^{-1}\,mol^{-1}$. $\log(2) = \frac{E_a}{2.303 	imes 2 	imes 10^{-3}} 	imes \frac{330 - 300}{300 	imes 330}$. $0.3010 = \frac{E_a}{4.606 	imes 10^{-3}} 	imes \frac{30}{99000}$. $0.3010 = \frac{E_a}{4.606 	imes 10^{-3}} 	imes \frac{1}{3300}$. $E_a = 0.3010 	imes 4.606 	imes 10^{-3} 	imes 3300$. $E_a \approx 4.57\,k\,cal\,mol^{-1}$.

The rate of reaction becomes double when the temperature of the reaction is increased from $27\,^{\circ}C$ to $57\,^{\circ}C$. The activation energy of the reaction will be ....... $k\,cal$.

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