Activation energy (E_a) and rate constants (k | vedclass

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(D) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ a

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

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(D) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:$\ln k_1 = \ln A - \frac{E_a}{RT_1}$$\ln k_2 = \ln A - \frac{E_a}{RT_2}$Subtracting the first equation from the second:$\ln k_2 - \ln k_1 = \left( \ln A - \frac{E_a}{RT_2} \right) - \left( \ln A - \frac{E_a}{RT_1} \right)$$\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$Alternatively,this can be written as:$\ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$Thus,both expressions $(b)$ and $(c)$ are correct.

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

Activation energy $(E_a)$ and rate constants $(k_1)$ and $(k_2)$ of a chemical reaction at two different temperatures $(T_1)$ and $(T_2)$ are related by

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