At what temperature does the rate become doub | vedclass

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(B) Given the equation $\ln k = 10 - \frac{69 \, kJ}{RT} \cdots (i)$.Comparing this with the Arrhenius equation $\ln k = \ln A - \frac{E_a}{RT}$,we ge

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$307.7$

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(B) Given the equation $\ln k = 10 - \frac{69 \, kJ}{RT} \cdots (i)$.Comparing this with the Arrhenius equation $\ln k = \ln A - \frac{E_a}{RT}$,we get $E_a = 69 \, kJ/mol = 69000 \, J/mol$.For two temperatures $T_1 = 300 \, K$ and $T_2$,the rate constant $k_2 = 2k_1$.Using the Arrhenius equation: $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$.Substituting the values: $\ln 2 = \frac{69000}{8.314} \left( \frac{1}{300} - \frac{1}{T_2} ight)$.$0.693 = 8300 \left( \frac{T_2 - 300}{300 T_2} ight)$.Solving for $T_2$: $\frac{0.693}{8300} = \frac{T_2 - 300}{300 T_2}$.$8.349 	imes 10^{-5} = \frac{T_2 - 300}{300 T_2}$.$0.02505 T_2 = T_2 - 300$.$0.97495 T_2 = 300$.$T_2 \approx 307.7 \, K$.

At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

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