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(C) The rate of reaction $k$ is inversely proportional to the time taken $t$,so $k \propto 1/t$.Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_

Vedclass · Accepted Answer

$0.693 R \frac{T_1 T_2}{T_2-T_1}$

Vedclass · Answer

(C) The rate of reaction $k$ is inversely proportional to the time taken $t$,so $k \propto 1/t$.Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}ight) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.Given $t_1 = 4.0 	ext{ min}$ at $T_1$ and $t_2 = 8.0 	ext{ min}$ at $T_2$,we have $k_1 = 1/4$ and $k_2 = 1/8$.Substituting these values: $\ln \left(\frac{1/8}{1/4}ight) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.$\ln(0.5) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.Since $\ln(0.5) = -\ln(2) \approx -0.693$,we get $-0.693 = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.Rearranging for $E_a$: $E_a = -0.693 R \left[\frac{T_1 T_2}{T_1 - T_2}ight] = 0.693 R \left[\frac{T_1 T_2}{T_2 - T_1}ight]$.Note: The provided options suggest a specific sign convention; based on the derivation,the correct expression is $0.693 R \frac{T_1 T_2}{T_2 - T_1}$.

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