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Calorimetry

For a given r...

Question

For a given reaction, energy of activation for forward reaction $(E_{a f}) i s 80 k J . m o l^{- 1} . Δ H = - 40 k J . m o l^{- 1}$ for the reaction. A catalyst lowers $E_{a f} t o 20 k J . m o l^{- 1}$ . Find out the ratio of energy of activation for reverse reaction before and after addition of catalyst.

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Solution

$Δ H = E_{f} - E_{b} - 40 = 80 - E_{b}, E_{b} = 120 k J / m o l e$

Catalyst lower the $E_{a f}$ to $20 k J / m o l e$ for forward Rxn then $E_{f}^{'} = 20 k J / m o l$

We know catalyst decreases the activation energy equal amount in both directions:

${E_{b}}^{^{'}} = (120 - 60) = 60 k j / m o l$

$\frac{E_{b}}{E_{b}^{'}} = \frac{120}{60} = 2.0$

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