For a particular reaction,the rate constant b | vedclass

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(B) We know that the Arrhenius equation is: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} 	imes \left( \frac{T_2 - T_1}{T_1 T_2} ight)$ Given: $k_2 =

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$12.89$

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(B) We know that the Arrhenius equation is: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} 	imes \left( \frac{T_2 - T_1}{T_1 T_2} ight)$ Given: $k_2 = 2k_1$,$T_1 = 300 \ K$,$T_2 = 310 \ K$,$R = 2 \ cal \ mol^{-1} \ K^{-1}$. Substituting the values: $\log 2 = \frac{E_a}{2.303 	imes 2} 	imes \left( \frac{310 - 300}{300 	imes 310} ight)$ $0.3010 = \frac{E_a}{4.606} 	imes \left( \frac{10}{93000} ight)$ $0.3010 = \frac{E_a}{4.606} 	imes \frac{1}{9300}$ $E_a = 0.3010 	imes 4.606 	imes 9300 \approx 12890 \ cal \ mol^{-1}$. Converting to $kcal \ mol^{-1}$: $E_a = \frac{12890}{1000} = 12.89 \ kcal \ mol^{-1}$.

For a particular reaction,the rate constant becomes double on increasing temperature from $27^{\circ} C$ to $37^{\circ} C$. Calculate the approximate activation energy (in $kcal \ mol^{-1}$,$R=2 \ cal \ mol^{-1} \ K^{-1}$).

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