JEE Chemistry 2025 Question with Solution

Given: The reaction is zero-order. Half-life is $1 \, \text{hour}$ and initial concentration is $[A]_0 = 2.0 \, \text{mol L}^{-1}$.

Find: The time required to decrease concentration of $A$ from $0.50 \, \text{mol L}^{-1}$ to $0.25 \, \text{mol L}^{-1}$.

For a zero-order reaction, the half-life is given by

$$t_{1/2} = \frac{[A]_0}{2k}$$

Substituting the given values,

$$1 = \frac{2.0}{2k}$$

So,

$$k = 1.0 \, \text{mol L}^{-1} \, \text{h}^{-1}$$

Now, for a zero-order reaction, the time for concentration to change from $[A]_0$ to $[A]$ is

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

Using $[A]_0 = 0.50 \, \text{mol L}^{-1}$ and $[A] = 0.25 \, \text{mol L}^{-1}$,

$$t = \frac{0.50 - 0.25}{1.0} = 0.25 \, \text{hours}$$

Converting to minutes,

$$0.25 \times 60 = 15 \, \text{minutes}$$

Therefore, the time required is $15 \, \text{min}$. The correct option is C.

Given: Zero-order reaction with half-life $1 \, \text{hour} = 60 \, \text{min}$ and initial concentration $A_0 = 2.0 \, \text{mol L}^{-1}$.

Find: Time for concentration to fall from $0.50$ to $0.25 \, \text{mol L}^{-1}$.

For zero-order reaction,

$$t_{1/2} = \frac{A_0}{2k}$$

Hence,

$$60 = \frac{2}{2k}$$

So,

$$k = \frac{1}{60} \, \text{mol L}^{-1} \, \text{min}^{-1}$$

Now use

$$A_t = A_0 - kt$$

which gives

$$t = \frac{A_0 - A_t}{k}$$

Substituting,

$$t = \frac{0.5 - 0.25}{\frac{1}{60}} = 0.25 \times 60 = 15 \, \text{min}$$

Therefore, the required time is $15 \, \text{minutes}$, so the correct option is C.