The rate of a reaction decreased by 3.555 tim | vedclass

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(D) Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}\right]$Given: $T_1 = 313 \, K$

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(D) Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}ight) = \frac{E_a}{R} \left[\frac{1}{T_1} - \frac{1}{T_2}ight]$Given: $T_1 = 313 \, K$ $(40^{\circ} C)$,$T_2 = 303 \, K$ $(30^{\circ} C)$.The rate decreases by $3.555$ times,so $\frac{k_1}{k_2} = 3.555$,which means $\frac{k_2}{k_1} = \frac{1}{3.555}$.$\ln \left(\frac{1}{3.555}ight) = \frac{E_a}{8.314} \left[\frac{1}{313} - \frac{1}{303}ight]$$-1.268 = \frac{E_a}{8.314} \left[\frac{303 - 313}{313 	imes 303}ight]$$-1.268 = \frac{E_a}{8.314} \left[\frac{-10}{94839}ight]$$E_a = \frac{1.268 	imes 8.314 	imes 94839}{10} \approx 99980 \, J \, mol^{-1} = 99.98 \, kJ \, mol^{-1} \approx 100 \, kJ \, mol^{-1}$.

$Step$	$Rate \ constant$	$Activation \ energy$
$Step-1$	$k_1$	$E_{a1} = 180 \ kJ/mol$
$Step-2$	$k_2$	$E_{a2} = 80 \ kJ/mol$
$Step-3$	$k_3$	$E_{a3} = 50 \ kJ/mol$

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.
[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]

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The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]