The rate of a catalyzed reaction at 27° C is e 2 times the rate of un catalyzed reaction at 727° C. if the catalyst decreased the activation energy by 6 Kcal/ mol, the activation energy of uncatalyzed reaction is (consider the same value of pre-exponential factors in both cases) :

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$\frac{48}{7} Kcal / mol$

$\frac{6}{7} Kcal / mol$

$6 Kcal / mol$

$\frac{90}{7} Kcal / mol$

answer is A.

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Detailed Solution

We have

$\begin{aligned} T_{1} = 27^{\circ} C = 300 K \\ T_{2} = 727^{\circ} = 1000 K \end{aligned}$

Now

$K = A e^{- F A R T}$

$\frac{k_{C}}{k_{u C}} = e^{- [\frac{E - b}{300} - \frac{E}{1000}] \times \frac{10^{3}}{2}}$

$\Rightarrow 4 = E - \frac{(E - 6)}{0.3}$

$\Rightarrow E = \frac{48}{7} Kcal / mol$

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