JEE Chemistry 2025 Question with Solution

Given: A first order decomposition reaction reaches $\frac{1}{4}$ of its initial concentration in time $t_1$ and $\frac{1}{8}$ of its initial concentration in time $t_2$.

Find: The ratio $\frac{t_1}{t_2}$.

For a first-order reaction,

$$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$$

For $t_1$, the remaining concentration is $\frac{1}{4}$ of the initial concentration, so

$$t_1 = \frac{2.303}{k} \log \frac{1}{1/4} = \frac{2.303}{k} \log 4$$

For $t_2$, the remaining concentration is $\frac{1}{8}$ of the initial concentration, so

$$t_2 = \frac{2.303}{k} \log \frac{1}{1/8} = \frac{2.303}{k} \log 8$$

Therefore,

$$\frac{t_1}{t_2} = \frac{\log 4}{\log 8}$$

Using

$$\log 4 = 2 \log 2$$

and

$$\log 8 = 3 \log 2$$

we get

$$\frac{t_1}{t_2} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3}$$

Therefore, the correct option is C.

Given: The reactant follows first-order kinetics.

Find: The ratio $\frac{t_1}{t_2}$ when the concentration becomes $\frac{1}{4}$ and $\frac{1}{8}$ of the initial value.

In a first-order reaction, equal half-lives are observed.

• To reach $\frac{1}{4}$ of the initial concentration, the reactant undergoes 2 half-lives.
• To reach $\frac{1}{8}$ of the initial concentration, the reactant undergoes 3 half-lives.

So,

$$t_1 = 2t_{1/2}, \qquad t_2 = 3t_{1/2}$$

Hence,

$$\frac{t_1}{t_2} = \frac{2t_{1/2}}{3t_{1/2}} = \frac{2}{3}$$

Therefore, the correct option is C.