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Rate of Reaction

For a given r...

Question

For a given reaction, energy of activation for forward reaction $(E_{a f})$ is $80 k J / m o l$ . $Δ H = - 40 k J / m o l$ for the reaction. A catalyst lowers $E_{a f}$ to $20 k J / m o l$ . The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

1.0

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0.5

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1.2

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2.0

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Solution

The correct option is B $2.0$
Initially
$E_{a b} = E_{a f} + | | △ H | | = 120 K J m o l^{- 1}$
Due to catalyst
$E_{a b}^{'} = E_{a} f^{'} + | | △ H | | = 60 K J m o l^{- 1}$
$∴ E_{a b} : E_{a b}^{'} = 2 : 1$

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