Q53 / 180 · COMEDK 2024 (Evening Shift) · Chemistry
In the presence of a catalyst at a given temperature of $27^{\circ} \mathrm{C}$, the Activation energy of a specific reaction is reduced by $100 \mathrm{~J} / \mathrm{mol}$. What is the ratio between the rate constants for the catalysed $(\mathrm{k}_2)$ and uncatalysed $(\mathrm{k}_1)$ reactions?
- A1.97
- B$1.04 \times 10^{-2}$
- Correct answerC1.04
- D$2.32 \times 10^{-2}$
Solution
The Arrhenius equation for the rate constant is given by $k = A e^{-E_a / RT}$.
For the uncatalysed reaction, $k_1 = A e^{-E_{a1} / RT}$.
For the catalysed reaction, $k_2 = A e^{-E_{a2} / RT}$.
The ratio of the rate constants is $\dfrac{k_2}{k_1} = \dfrac{A e^{-E_{a2} / RT}}{A e^{-E_{a1} / RT}} = e^{(E_{a1} - E_{a2}) / RT}$.
Given that the activation energy is reduced by $100 \text{ J/mol}$, we have $E_{a1} - E_{a2} = 100 \text{ J/mol}$.
The temperature $T = 27^{\circ} \text{C} = 27 + 273 = 300 \text{ K}$.
The gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.
Substituting these values, $\dfrac{k_2}{k_1} = e^{100 / (8.314 \times 300)} = e^{100 / 2494.2} = e^{0.04009}$.
Calculating the value, $e^{0.04009} \approx 1.0409$.
Rounding to two decimal places, the ratio is $1.04$.
Answer: 1.04
For the uncatalysed reaction, $k_1 = A e^{-E_{a1} / RT}$.
For the catalysed reaction, $k_2 = A e^{-E_{a2} / RT}$.
The ratio of the rate constants is $\dfrac{k_2}{k_1} = \dfrac{A e^{-E_{a2} / RT}}{A e^{-E_{a1} / RT}} = e^{(E_{a1} - E_{a2}) / RT}$.
Given that the activation energy is reduced by $100 \text{ J/mol}$, we have $E_{a1} - E_{a2} = 100 \text{ J/mol}$.
The temperature $T = 27^{\circ} \text{C} = 27 + 273 = 300 \text{ K}$.
The gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.
Substituting these values, $\dfrac{k_2}{k_1} = e^{100 / (8.314 \times 300)} = e^{100 / 2494.2} = e^{0.04009}$.
Calculating the value, $e^{0.04009} \approx 1.0409$.
Rounding to two decimal places, the ratio is $1.04$.
Answer: 1.04
Hence, the correct answer is option: 'C'.