JEE Chemistry 2026 Question with Solution

Given: Pre-exponential factors are identical, so the Arrhenius equations for the two reactions have the same $A$. Also, $E_{a1} - E_{a2} = 20 \, \text{kJ/mol} = 20000 \, \text{J/mol}$, $T = 300 \, \text{K}$, and $R = 8.3 \, \text{J K}^{-1} \text{ mol}^{-1}$.

Find: $\ln\left(\frac{k_2}{k_1}\right)$.

Using the Arrhenius equation:

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

and

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

Taking the ratio,

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

Taking natural logarithm,

$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_{a1}-E_{a2}}{RT}$$

Substitute the given values:

$$\ln\left(\frac{k_2}{k_1}\right) = \frac{20000}{8.3 \times 300}$$ $$= \frac{20000}{2490} \approx 8.032$$

Therefore, the nearest integer value of $\ln\left(\frac{k_2}{k_1}\right)$ is $8$.

Given: The difference in activation energies is $20 \, \text{kJ mol}^{-1}$.

Find: Why conversion to joules is necessary before substitution.

The gas constant is given as

$$R = 8.3 \, \text{J K}^{-1} \text{ mol}^{-1}$$

so activation energy must be in joules per mole, not kilojoules per mole. Hence,

$$20 \, \text{kJ mol}^{-1} = 20000 \, \text{J mol}^{-1}$$

Using kilojoules directly with $R$ in joules would make the ratio numerically incorrect by a factor of $1000$. After proper conversion, the logarithmic expression evaluates to approximately $8.032$, so the nearest integer is $8$.