The activation energy of a reaction at a give | vedclass

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(B) The Arrhenius equation is given by $k = A e^{-E_{a} / RT}$.Taking the ratio of rate constant $k$ to the Arrhenius factor $A$,we get $k/A = e^{-E_{

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$0.1$

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(B) The Arrhenius equation is given by $k = A e^{-E_{a} / RT}$.Taking the ratio of rate constant $k$ to the Arrhenius factor $A$,we get $k/A = e^{-E_{a} / RT}$.Given the activation energy $E_{a} = 2.303 \ RT \ J \ mol^{-1}$.Substituting $E_{a}$ into the equation: $k/A = e^{-(2.303 \ RT) / RT} = e^{-2.303}$.Taking the natural logarithm on both sides: $\ln(k/A) = -2.303$.Since $\ln(x) = 2.303 \log_{10}(x)$,we have $2.303 \log_{10}(k/A) = -2.303$.Dividing by $2.303$,we get $\log_{10}(k/A) = -1$.Therefore,$k/A = 10^{-1} = 0.1$.

$A \rightarrow B$	$slow ; \Delta H=+ve$
$B \rightarrow C$	$fast ; \Delta H=-ve$
$C \rightarrow D$	$fast ; \Delta H=-ve$

The activation energy of a reaction at a given temperature is found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of rate constant to the Arrhenius factor is

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