JEE Main 2022: Catalyst Rate Ratio Problem

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Daily Challenge JEE Main 2022 (26 Jun, Shift 2)

Catalyst Effect on Rate Constant Ratio

Apply the Arrhenius equation to find the exponential relationship between catalysed and uncatalysed rates.

Question:

Catalyst $A$ reduces the activation energy for a reaction by $10 \text{ kJ mol}^{-1}$ at $300 \text{ K}$.

The ratio of rate constants, $\frac{k_{\text{Catalysed}}}{k_{\text{Uncatalysed}}}$ is $e^x$.

The value of $x$ is ________. [nearest integer]

[Assume that the pre-exponential factor is same in both the cases. Given $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$]

Enter your integer answer for x:

Detailed Solution

Step 1: Set up the Arrhenius Equations

The Arrhenius equation relates the rate constant ($k$) to the activation energy ($E_a$) and temperature ($T$): $k = A e^{-\frac{E_a}{RT}}$

We can write this for both the uncatalysed and catalysed reactions. We are given that the pre-exponential factor ($A$) is the same for both.

• Uncatalysed: $\quad k_{uncat} = A e^{-\frac{E_{a, uncat}}{RT}}$
• Catalysed: $\quad\;\;\; k_{cat} = A e^{-\frac{E_{a, cat}}{RT}}$

Step 2: Find the Ratio

Divide the catalysed rate constant by the uncatalysed rate constant:

$$ \frac{k_{cat}}{k_{uncat}} = \frac{A e^{-\frac{E_{a, cat}}{RT}}}{A e^{-\frac{E_{a, uncat}}{RT}}} $$

The '$A$' factors cancel out. Using exponent rules ($\frac{e^a}{e^b} = e^{a-b}$):

$$ \frac{k_{cat}}{k_{uncat}} = e^{\frac{-E_{a, cat}}{RT} - \left(\frac{-E_{a, uncat}}{RT}\right)} $$ $$ \frac{k_{cat}}{k_{uncat}} = e^{\frac{E_{a, uncat} - E_{a, cat}}{RT}} $$

We are given that this ratio equals $e^x$. Therefore:

$$ x = \frac{E_{a, uncat} - E_{a, cat}}{RT} $$

The Number One Exam Trap!

The problem states the catalyst reduces the activation energy by $10 \text{ kJ mol}^{-1}$. This means the difference $(E_{a, uncat} - E_{a, cat}) = 10 \text{ kJ mol}^{-1}$.

Stop! Look at the units of $R$!
$R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$. Because $R$ is in Joules, you MUST convert the energy difference into Joules before plugging it into the equation.
$\Delta E_a = 10 \text{ kJ mol}^{-1} = \mathbf{10000 \text{ J mol}^{-1}}$

Step 3: Calculate the final value

Now, substitute the values into our equation for $x$:

• $E_{a, uncat} - E_{a, cat} = 10000 \text{ J mol}^{-1}$
• $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$
• $T = 300 \text{ K}$

$$ x = \frac{10000}{8.31 \times 300} $$ $$ x = \frac{10000}{2493} $$ $$ x \approx 4.011 $$

The question asks for the nearest integer.

Final Answer: 4

Review Catalyst Effects

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