In the first order reaction the concentration of the reactant is reduced to \(\frac{1}{4}^{th}\) in \(60\) minutes, what will be its half-life?

30 minutes

The correct answer is option (3) → 30 minutes

A first-order reaction has a constant half-life ($t_{1/2}$), meaning the time it takes for the concentration to drop by half is always the same.

1. Initial State: $100\%$ (or $1$)
2. After 1st Half-life ($t_{1/2}$): Concentration becomes $\frac{1}{2}$
3. After 2nd Half-life ($2 \times t_{1/2}$): Concentration becomes $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Since the concentration reached $\frac{1}{4}$ in 60 minutes, we know that two half-lives have passed:

$2 \times t_{1/2} = 60 \text{ minutes}$

$t_{1/2} = \frac{60}{2} = \mathbf{30 \text{ minutes}}$