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Question

(A) The Arrhenius equation is $k = A e^{-E_a / RT}$.Given $k = (\frac{k_1 k_2}{k_3})^{2/3}$,we substitute $k_i = A_i e^{-E_{a_i} / RT}$:$A e^{-E_a / R

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(A) The Arrhenius equation is $k = A e^{-E_a / RT}$.Given $k = (\frac{k_1 k_2}{k_3})^{2/3}$,we substitute $k_i = A_i e^{-E_{a_i} / RT}$:$A e^{-E_a / RT} = [\frac{A_1 e^{-E_{a_1} / RT} 	imes A_2 e^{-E_{a_2} / RT}}{A_3 e^{-E_{a_3} / RT}}]^{2/3}$$A e^{-E_a / RT} = [\frac{A_1 A_2}{A_3}]^{2/3} 	imes e^{(-E_{a_1} - E_{a_2} + E_{a_3}) 	imes \frac{2}{3RT}}$Comparing the exponents of $e$:$-\frac{E_a}{RT} = \frac{2}{3} 	imes \frac{(-E_{a_1} - E_{a_2} + E_{a_3})}{RT}$$E_a = \frac{2}{3} [E_{a_1} + E_{a_2} - E_{a_3}]$Substituting the values:$E_a = \frac{2}{3} [180 + 80 - 50]$$E_a = \frac{2}{3} [210] = 140\ kJ/mol$.

$Step$	$Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$	$k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$	$k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$	$k_3, E_{a_3} = 50\ kJ/mol$

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