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(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.Given: $T_1 = 25 + 273 =

Vedclass · Accepted Answer

$\frac{2.303 	imes R 	imes 298 	imes 308}{10} \log 2$

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(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} ight)$.Given: $T_1 = 25 + 273 = 298 \, K$,$T_2 = 35 + 273 = 308 \, K$,and $K_2 = 2K_1$.Substituting these values: $\log \frac{2K_1}{K_1} = \frac{E_a}{2.303R} \left( \frac{308 - 298}{298 	imes 308} ight)$.$\log 2 = \frac{E_a}{2.303R} \left( \frac{10}{298 	imes 308} ight)$.Rearranging for $E_a$: $E_a = \frac{2.303 	imes R 	imes 298 	imes 308 	imes \log 2}{10}$.

For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

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