JEE Chemistry 2025 Question with Solution

Given: The reaction follows first-order kinetics at $1000 \, \text{K}$ with activation energy $E_a = 191.48 \times 10^3 \, \text{J mol}^{-1}$, frequency factor $A = 10^{20}$, and $R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}$.

Find: The time required for $50\%$ of A to convert into B, that is, the half-life.

For a first-order reaction, the half-life is

$$t_{1/2} = \frac{0.693}{K}$$

where the rate constant is given by the Arrhenius equation

$$K = Ae^{-\frac{E_a}{RT}}$$

Substituting the given values,

$$K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}}$$ $$= 10^{20} \times e^{-23.031}$$ $$= 10^{20} \times e^{-\ln(10^{10})}$$ $$= 10^{20} \times 10^{-10}$$ $$= 10^{10} \, \text{sec}^{-1}$$

Thus,

$$t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{sec}$$ $$= 69.3 \times 10^{-12} \, \text{sec}$$

Therefore, the time required is $69$ picoseconds, so the correct option is A.

Given: First-order isomerisation, $T = 1000 \, \text{K}$, $E_a = 191.48 \, \text{kJ mol}^{-1}$, and $A = 10^{20}$.

Find: Half-life in picoseconds.

Since $50\%$ molecules of A become B, the required time is the half-life of a first-order reaction.

Use

$$t_{1/2} = \frac{0.693}{K}$$

and

$$K = Ae^{-\frac{E_a}{RT}}$$

Now evaluate the exponent first:

$$\frac{E_a}{RT} = \frac{191.48 \times 10^3}{8.314 \times 1000} = 23.031$$

Hence,

$$K = 10^{20} e^{-23.031}$$

Using the relation from the provided working,

$$e^{-23.031} = e^{-\ln(10^{10})} = 10^{-10}$$

So,

$$K = 10^{20} \times 10^{-10} = 10^{10} \, \text{sec}^{-1}$$

Now calculate the half-life:

$$t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{sec}$$

Convert seconds into picoseconds:

$$1 \, \text{ps} = 10^{-12} \, \text{sec}$$

Therefore,

$$6.93 \times 10^{-11} \, \text{sec} = 69.3 \times 10^{-12} \, \text{sec} = 69.3 \, \text{ps}$$

Nearest integer:

$$69$$

Therefore, the correct option is A.