At 300 K,the activation energy and rate cons | vedclass

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(B) $1$. The reaction $2N_2O_5 	o 4NO_2 + O_2$ is a first-order reaction.$2$. The half-life formula is $t_{1/2} = \frac{0.693}{k}$. Given $t_{1/2} =

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$458$

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(B) $1$. The reaction $2N_2O_5 	o 4NO_2 + O_2$ is a first-order reaction.$2$. The half-life formula is $t_{1/2} = \frac{0.693}{k}$. Given $t_{1/2} = 2 \ hr = 7200 \ s$,we find $k_2 = \frac{0.693}{7200} \approx 9.625 	imes 10^{-5} \ s^{-1}$.$3$. Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.$4$. Given $E_a = 10000 \ J \ mol^{-1}$,$T_1 = 300 \ K$,$k_1 = 2.4 	imes 10^{-5} \ s^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.$5$. $\ln(\frac{9.625 	imes 10^{-5}}{2.4 	imes 10^{-5}}) = \frac{10000}{8.314} (\frac{1}{300} - \frac{1}{T_2})$.$6$. $\ln(4.01) = 1202.79 (0.00333 - \frac{1}{T_2}) \implies 1.389 = 4.005 - \frac{1202.79}{T_2}$.$7$. $\frac{1202.79}{T_2} = 2.616 \implies T_2 \approx 459.7 \ K$. The closest option is $458 \ K$.

At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
$2N_2O_5 \to 4NO_2 + O_2$

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At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?$2N_2O_5 \to 4NO_2 + O_2$