nius equation is given by:

$$ k = A e^{-\frac{E_a}{RT}} $$

where:

$$k$$ is the rate constant.
$$A$$ is the pre-exponential factor.
$$E_a$$ is the activation energy.
$$R$$ is the gas constant (approximately $$8.314 \, J/(mol \cdot K)$$).
$$T$$ is the temperature in Kelvin.

Given:

The reduction in Activation Energy is $$100 \mathrm{~J} / \mathrm{mol}$$.
Temperature is $$27^{\circ} \mathrm{C}$$, which is $$300 \, \mathrm{K}$$.

Let's denote the activation energy of the uncatalysed reaction as $$E_{a1}$$ and the activation energy of the catalysed reaction as $$E_{a2}$$. Since the activation energy is reduced by $$100 \mathrm{~J} / \mathrm{mol}$$ due to the catalyst:

$$ E_{a2} = E_{a1} - 100 $$

We want to find the ratio $$\frac{k_2}{k_1}$$. Using the Arrhenius equation for both the catalysed and uncatalysed reactions, we get:

$$ k_1 = A e^{-\frac{E_{a1}}{RT}} $$

$$ k_2 = A e^{-\frac{E_{a2}}{RT}} $$

Combining these equations gives:

$$ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a2}}{RT}}}{A e^{-\frac{E_{a1}}{RT}}} = e^{-\frac{E_{a2}}{RT}} e^{\frac{E_{a1}}{RT}} = e^{\frac{E_{a1} - E_{a2}}{RT}} $$

Substituting $$E_{a2} = E_{a1} - 100$$, we get:

$$ \frac{k_2}{k_1} = e^{\frac{100}{RT}} $$

Now, substitute $$R = 8.314 \, J/(mol \cdot K)$$ and $$T = 300 \, K$$:

$$ \frac{k_2}{k_1} = e^{\frac{100}{8.314 \times 300}} $$

Calculating the exponent:

$$ \frac{100}{8.314 \times 300} \approx 0.0401 $$

So:

$$ \frac{k_2}{k_1} = e^{0.0401} \approx 1.041 $$

Hence, the ratio between the rate constants for the catalysed and uncatalysed reactions is approximately $$1.04$$. The correct answer is:

Option D: 1.04

Chemical Kinetics

Continue this paper

Similar Chemical Kinetics PYQs

More practice from this question