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

30 minutes

The correct answer is option 3.30 minutes.

For a first-order reaction, the relationship between the half-life (\(t_{1/2}\)) and the rate constant (\(k\)) is given by the formula:
\( t_{1/2} = \frac{\ln(2)}{k} \)
Given that the concentration of the reactant is reduced to \(\frac{1}{4}\) in \(60\) minutes, we can use the first-order integrated rate law to find the rate constant (\(k\)):
\( \ln\left(\frac{[A]}{[A]_0}\right) = -kt \)
where:
\([A]\) is the concentration at time \(t\),
\([A]_0\) is the initial concentration,
\(k\) is the rate constant,
\(t\) is time.
Since the concentration is reduced to \(\frac{1}{4}\) in \(60\) minutes, we can express this as:
\( \ln\left(\frac{[A]}{[A]_0}\right) = -k(60) \)
Solving for \(k\):
\( k = \frac{\ln(4)}{60} \)
Now, substitute this value into the formula for half-life:
\( t_{1/2} = \frac{\ln(2)}{k} \)
\( t_{1/2} = \frac{\ln(2)}{\frac{\ln(4)}{60}} \)
\( t_{1/2} = \frac{60 \cdot \ln(2)}{\ln(4)} \)
\( t_{1/2} = \frac{60 \cdot \ln(2)}{2 \cdot \ln(2)} \)
\( t_{1/2} = 30 \ \text{minutes} \)
Therefore, the correct answer is (3) 30 minutes.