A sample of milk splits after 60 ; min. at 30 | vedclass

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(D) The rate constant $k$ is inversely proportional to the time $t$ required for a specific change (doubling of population),so $k \propto 1/t$.Using t

Vedclass · Accepted Answer

$3.98$

Vedclass · Answer

(D) The rate constant $k$ is inversely proportional to the time $t$ required for a specific change (doubling of population),so $k \propto 1/t$.Using the Arrhenius equation in the form $\ln(k_2 / k_1) = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]$,we substitute $k_2/k_1 = t_1/t_2$:$\ln(t_1 / t_2) = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]$$\ln(60 / 40) = \frac{E_a}{8.3} [\frac{1}{300} - \frac{1}{400}]$$\ln(1.5) = \frac{E_a}{8.3} [\frac{400 - 300}{120000}]$$0.405 = \frac{E_a}{8.3} 	imes \frac{100}{120000}$$0.405 = \frac{E_a}{8.3} 	imes \frac{1}{1200}$$E_a = 0.405 	imes 8.3 	imes 1200 = 4033.8 \; J / mol \approx 4.03 \; kJ / mol$.Comparing with the given options,the closest value is $3.98 \; kJ / mol$.

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