Catalyst A reduces the activation energy for | vedclass

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(C) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.For the catalysed reaction,$k_{	ext{cat}} = A e^{-E_{a, 	ext{cat}} / RT}$.For the unca

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.For the catalysed reaction,$k_{	ext{cat}} = A e^{-E_{a, 	ext{cat}} / RT}$.For the uncatalysed reaction,$k_{	ext{uncat}} = A e^{-E_{a, 	ext{uncat}} / RT}$.The ratio is $\frac{k_{	ext{cat}}}{k_{	ext{uncat}}} = e^{(E_{a, 	ext{uncat}} - E_{a, 	ext{cat}}) / RT}$.Given $\Delta E_a = E_{a, 	ext{uncat}} - E_{a, 	ext{cat}} = 10 \ kJ \ mol^{-1} = 10000 \ J \ mol^{-1}$.Substituting the values: $\frac{k_{	ext{cat}}}{k_{	ext{uncat}}} = e^{10000 / (8.31 	imes 300)} = e^{10000 / 2493} = e^{4.011}$.Comparing with $e^x$,we get $x \approx 4$.

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