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(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$.

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

Similar Questions

For a first order chemical reaction,

The rate constant of a reaction at $25\,^oC$ is $1 \times 10^{-3}\,s^{-1}$. If the rate of the reaction doubles when the temperature is increased to $35\,^oC$,the activation energy of the reaction is .......... $kJ\, mol^{-1}$.

The rate of a reaction quadruples when the temperature changes from $293 \ K$ to $313 \ K$. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

In an exothermic reaction $A \to B$,the evolved heat is $280 \ kJ \ mol^{-1}$ and the activation energy is $200 \ kJ \ mol^{-1}$. The activation energy of the reverse reaction $B \to A$ is $.......... \ kJ \ mol^{-1}$.

At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

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