Reactant A shows two reactions:A xrightarrow{ | vedclass

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(C) According to the Arrhenius equation,$K = A \, e^{-Ea/RT}$.Assuming the pre-exponential factor $A$ is the same for both reactions:$K_1 = A \, e^{-E

Vedclass · Accepted Answer

$K_1 = K_2 \, e^{2Ea_1/RT}$

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(C) According to the Arrhenius equation,$K = A \, e^{-Ea/RT}$.Assuming the pre-exponential factor $A$ is the same for both reactions:$K_1 = A \, e^{-Ea_1/RT}$$K_2 = A \, e^{-Ea_2/RT}$Given $Ea_1 = \frac{Ea_2}{3}$,we have $Ea_2 = 3Ea_1$.Substituting $Ea_2$ into the expression for $K_2$:$K_2 = A \, e^{-3Ea_1/RT}$Now,divide $K_1$ by $K_2$:$\frac{K_1}{K_2} = \frac{A \, e^{-Ea_1/RT}}{A \, e^{-3Ea_1/RT}} = e^{(-Ea_1 + 3Ea_1)/RT} = e^{2Ea_1/RT}$Therefore,$K_1 = K_2 \, e^{2Ea_1/RT}$.

Reactant $A$ shows two reactions:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:

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Reactant $A$ shows two reactions:$A \xrightarrow{K_1} B$,activation energy $= Ea_1$$A \xrightarrow{K_2} C$,activation energy $= Ea_2$If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is: