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(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.Since the time $t$ is the same for both temp

Vedclass · Accepted Answer

$76.75$

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(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.Since the time $t$ is the same for both temperatures,we have $\frac{2.303}{k_1} \log \frac{100}{90} = \frac{2.303}{k_2} \log \frac{100}{75}$.This simplifies to $\frac{\log(1.111)}{k_1} = \frac{\log(1.333)}{k_2}$,which gives $\frac{k_2}{k_1} = \frac{\log(1.333)}{\log(1.111)} \approx \frac{0.1248}{0.0457} \approx 2.73$.Using the Arrhenius equation $\ln \frac{k_2}{k_1} = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$,we get $\ln(2.73) = \frac{E_a}{8.314} [\frac{308 - 298}{308 	imes 298}]$.$1.004 = \frac{E_a}{8.314} 	imes \frac{10}{91784}$.$E_a = \frac{1.004 	imes 8.314 	imes 91784}{10} \approx 76600 \, J/mol = 76.6 \, kJ/mol$.The closest option is $76.75 \, kJ/mol$.

For a first-order reaction,the time taken to complete $10\%$ at $298\, K$ is equal to the time taken to complete $25\%$ at $308\, K$. The activation energy of the reaction is ................ $kJ/mol$.

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