The activation energy of the reaction is $75 k J / m o l$ in absence of a catalyst and it lowers to $50.14 k J / m o l$ with a catalyst. How many times will the rate of reaction grow in presence of a catalyst if the reaction proceeds at $25^{\circ} C$ ?

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Solution

We know that Arrhenius for calculation of activation energy is given by:-
$K = A$ $e^{- E a / R T}$ $- (i)$
If $K_{1}$ and $K_{2}$ are two rate constants at activation energy $E a_{1}$ and $E a_{2}$ . Then,
$K_{1} = A$ $e^{- E a / R T}$ $- (i i)$
$K_{2} = A$ $e^{- E a / R T}$ $- (i i i)$
Now, $(i i i) / (i i)$
$\Rightarrow \frac{K_{2}}{K_{1}} = \frac{e^{- E a_{2} / R T}}{e^{- E a_{1} / R T}}$
$\Rightarrow K_{2} = K_{1}$ $e^{+ \frac{1}{R T} (E a_{1} - E a_{2})}$
Given, $T = 25^{\circ} C = 25 + 273 = 298 K$
$R = 8.314$ $J K^{- 1} m o l^{- 1}$
$E a_{1} = 75 K J / m o l$ $E a_{2} = 50.14 K J / m o l$
$\Rightarrow \frac{K_{2}}{K_{1}} = e^{+ (75 - 50.14) \times 10^{3} / 8.314 \times 298}$
$= e^{+ 10.034}$
$\Rightarrow \frac{K_{2}}{K_{1}} = 222789$

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The activation energy of the reaction is 75kJ/mol in absence of a catalyst and it lowers to 50.14kJ/mol with a catalyst. How many times will the rate of reaction grow in presence of a catalyst if the reaction proceeds at 25∘C?

The activation energy of the reaction is $75 k J / m o l$ in absence of a catalyst and it lowers to $50.14 k J / m o l$ with a catalyst. How many times will the rate of reaction grow in presence of a catalyst if the reaction proceeds at $25^{\circ} C$ ?