What is the activation energy for a reaction | vedclass

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(A) The Arrhenius equation is given by: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \,R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$Given: $k

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$34.7$

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(A) The Arrhenius equation is given by: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \,R} \left( \frac{1}{T_{1}} - \frac{1}{T_{2}} ight)$Given: $k_{2} = 2 \,k_{1}$,$T_{1} = 20 + 273 = 293 \,K$,$T_{2} = 35 + 273 = 308 \,K$,and $R = 8.314 \,J \,mol^{-1} \,K^{-1}$.Substituting the values:$\log(2) = \frac{E_{a}}{2.303 	imes 8.314} \left( \frac{1}{293} - \frac{1}{308} ight)$$0.3010 = \frac{E_{a}}{19.147} 	imes \left( \frac{308 - 293}{293 	imes 308} ight)$$0.3010 = \frac{E_{a}}{19.147} 	imes \frac{15}{90244}$$E_{a} = \frac{0.3010 	imes 19.147 	imes 90244}{15} \approx 34673 \,J \,mol^{-1}$Converting to $kJ \,mol^{-1}$: $E_{a} = 34.673 \,kJ \,mol^{-1} \approx 34.7 \,kJ \,mol^{-1}$.

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20 \,^{\circ}C$ to $35 \,^{\circ}C$ in $kJ \,mol^{-1}$? $(R = 8.314 \,J \,mol^{-1} \,K^{-1})$

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