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Activation energy of an uncatalysed reaction at 1000 K is 5 Kcal. In presence of catalyst the activation energy of same reaction at same temperature will be (Ratio of rate constants of catalyzed reaction to uncatalysed reaction is 10:1)

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394 cal

4606 cal

2303 cal

931 cal

answer is A.

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Detailed Solution

$K_{catalyst} = {Ae}^{- \frac{{E_{a}}_{1}}{RT}}$

$K_{un catalyst} = {Ae}^{- \frac{{E_{a}}_{2}}{RT}}$

$\frac{K_{C}}{K_{UC}} = e^{\frac{E_{a_{(2)}} - E_{a_{(1)}}}{R T}}$

$\log \frac{K_{c a t a l y s t}}{K_{u n c a t a l y s t}} = \frac{E_{a_{(2)}} - E_{a_{(1)}}}{2.303 R T}$

$E_{a_{(2)}} - E_{a_{(1)}} = 4606 cal$

Activation energy of catalysed reaction E_a1 = 5000 - 4606 = 394 cal

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