The rate constant of a first-order reaction i | vedclass

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(C) Given: $k_1 = 1.5 	imes 10^7 	ext{ s}^{-1}$, $T_1 = 300 	ext{ K}$, $k_2 = 3.0 	imes 10^7 	ext{ s}^{-1}$, $T_2 = 330 	ext{ K}$.Using the Arrh

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$19.02 	ext{ kJ mol}^{-1}$

Vedclass · Answer

(C) Given: $k_1 = 1.5 	imes 10^7 	ext{ s}^{-1}$, $T_1 = 300 	ext{ K}$, $k_2 = 3.0 	imes 10^7 	ext{ s}^{-1}$, $T_2 = 330 	ext{ K}$.Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} ight)$.$\log \left( \frac{3.0 	imes 10^7}{1.5 	imes 10^7} ight) = \frac{E_a}{19.15} \left( \frac{330 - 300}{300 	imes 330} ight)$.$\log(2) = \frac{E_a}{19.15} \left( \frac{30}{99000} ight)$.$0.3010 = \frac{E_a}{19.15} 	imes \frac{1}{3300}$.$E_a = 0.3010 	imes 19.15 	imes 3300 	ext{ J mol}^{-1}$.$E_a = 19023.225 	ext{ J mol}^{-1} \approx 19.02 	ext{ kJ mol}^{-1}$.

$Case$	$E_a \ (kcal/mol)$	$Temp. \ Change \ (K)$
$I$	$40$	$200 - 210$
$II$	$80$	$200 - 210$
$III$	$40$	$300 - 310$
$IV$	$80$	$300 - 310$

The rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$

Similar Questions

$A$ first order reaction is $50\%$ completed in $20\, \text{minutes}$ at $27\, ^oC$ and in $10\, \text{minutes}$ at $47\, ^oC$. The energy of activation of the reaction is ....... $Kcal$.

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

In which of the following cases is the percentage increase in rate constant maximum?
$Case$ $E_a \ (kcal/mol)$ $Temp. \ Change \ (K)$
$I$ $40$ $200 - 210$
$II$ $80$ $200 - 210$
$III$ $40$ $300 - 310$
$IV$ $80$ $300 - 310$

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Explain: How is the value of activation energy determined based on the Arrhenius equation?

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