JEE MAIN - Chemistry (2022 - 26th June Evening Shift - No. 22)
Catalyst A reduces the activation energy for a reaction by 10 kJ mol$$-$$1 at 300 K. The ratio of rate constants, $${{{}^kT,\,Catalysed} \over {{}^kT,\,Uncatalysed}}$$ 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]
Answer
4
Explanation
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 : $\mathrm{E}_{\text {cat }}-\mathrm{E}_{\text {uncat }}=10 \mathrm{~kJ} / \mathrm{mol}$
$$ \mathrm{T}=300 \mathrm{~K} $$
According to Arrhenius Equation,
$$ \begin{aligned} & \mathrm{K}=A \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \\\\ & \frac{\mathrm{E}_{\text {cat }}}{\mathrm{E}_{\text {uncat }}}=e^{\frac{E a-E_a^1}{R T}} \end{aligned} $$
$$ \begin{aligned} & =e^{\frac{10 \times 10^3}{8.21 \times 300}} \\\\ & =e^4 \\\\ & =4 \end{aligned} $$
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 : $\mathrm{E}_{\text {cat }}-\mathrm{E}_{\text {uncat }}=10 \mathrm{~kJ} / \mathrm{mol}$
$$ \mathrm{T}=300 \mathrm{~K} $$
According to Arrhenius Equation,
$$ \begin{aligned} & \mathrm{K}=A \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \\\\ & \frac{\mathrm{E}_{\text {cat }}}{\mathrm{E}_{\text {uncat }}}=e^{\frac{E a-E_a^1}{R T}} \end{aligned} $$
$$ \begin{aligned} & =e^{\frac{10 \times 10^3}{8.21 \times 300}} \\\\ & =e^4 \\\\ & =4 \end{aligned} $$
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