JEE Chemistry 2023 Question with Solution

Given: Rate constants are $K_{200} = 0.03 \, \text{min}^{-1}$ at $200 \, \text{K}$ and $K_{300} = 0.05 \, \text{min}^{-1}$ at $300 \, \text{K}$.

Find: Activation energy $E_a$.

Use the Arrhenius relation:

$$\log \left( \frac{K_{300}}{K_{200}} \right) = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$$

Substituting the given values:

$$\log \left( \frac{0.05}{0.03} \right) = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{200} - \frac{1}{300} \right)$$

Now,

$$\log \left( \frac{0.05}{0.03} \right) = \log 5 - \log 3 = 0.70 - 0.48 = 0.22$$

So,

$$0.22 = \frac{E_a}{2.3 \times 8.314} \times \left( \frac{1}{200} - \frac{1}{300} \right)$$

And,

$$\frac{1}{200} - \frac{1}{300} = \frac{1}{60000}$$

Hence,

$$0.22 = \frac{E_a}{2.3 \times 8.314} \times \frac{1}{60000}$$

Solving for $E_a$:

$$E_a = 2519.88 \, \text{J}$$

Therefore, the activation energy is $2520 \, \text{J}$ to the nearest integer.

Given: The required relation is between the ratio of rate constants and activation energy.

Find: Activation energy $E_a$.

A quick route is to first evaluate the logarithmic term directly:

$$\log \left( \frac{0.05}{0.03} \right) = \log 5 - \log 3 = 0.22$$

Then use the temperature difference factor:

$$\frac{1}{200} - \frac{1}{300} = \frac{1}{60000}$$

Substitute these two simplified values into the Arrhenius form and solve for $E_a$.

This works because the equation depends only on the ratio of the two rate constants, not on their individual absolute values separately.

Therefore, the activation energy comes out to $2520 \, \text{J}$ approximately.