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QJEEM22Ch180c: JEE Main 2022 Chemistry - Class 12 JEEM22Ch180c Solution: Catalyst A reduces the activation energy for a reaction by 10…

Catalyst A reduces the activation energy for a reaction by 10 kJ mol −- 1 at 300 K. The ratio of rate constants, kT, CatalysedkT, Uncatalysed{{{}^kT,\,Catalysed} \over {{}^kT,\,Uncatalysed}} is ex. The value of x is ___________. [nearest integer]

[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K −- 1 mol −- 1]

Step-by-Step Verified Solution

Activation Energy : It is the minimum amount of energy required to activate the molecules/ atoms so that they can undergo the chemical reaction.

A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.

In the Question,

Given : Ecat −Euncat =10 kJ/mol\mathrm{E}_{\text {cat }}-\mathrm{E}_{\text {uncat }}=10 \mathrm{~kJ} / \mathrm{mol}

T=300 K\mathrm{T}=300 \mathrm{~K}

According to Arrhenius Equation,

amp;K=Ae−Ea/RTamp;Ecat Euncat =eEa−Ea1RT\begin{aligned}& \mathrm{K}=A \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \\\\& \frac{\mathrm{E}_{\text {cat }}}{\mathrm{E}_{\text {uncat }}}=e^{\frac{E a-E_a^1}{R T}}\end{aligned}

amp;=e10×1038.21×300amp;=e4amp;=4\begin{aligned}& =e^{\frac{10 \times 10^3}{8.21 \times 300}} \\\\& =e^4 \\\\& =4\end{aligned}

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