If half-life period for a first order reaction in A is 2 minutes, how long will it take [A] to reach 10 % of its initial concentration?

6.65 min

The correct answer is option 4. 6.65 min.

To solve the problem, we need to determine the time it takes for the concentration of reactant A to reach 10% of its initial concentration in a first-order reaction, given that the half-life of the reaction is 2 minutes.

The half-life \((t_{1/2})\) for a first-order reaction is given by the formula:

\(t_{1/2} = \frac{0.693}{k}\)

Given that \(t_{1/2} = 2\) minutes, we can rearrange the formula to find k:

 \(k = \frac{0.693}{t_{1/2}}\)

\(⇒ k = \frac{0.693}{2} = 0.3465\, \ min^{-1}\)

When the concentration reaches 10% of the initial concentration, 90% of the initial concentration has reacted. Using the same first-order rate equation:

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

Here, \([A]_0 = 100\%\) and \([A] = 10\%\). Thus,

\(t = \frac{2.303}{0.3465} log{100}{10}\)

\(⇒ t = \frac{2.303}{0.3465}log (10)\)

\(⇒ t = \frac{2.303}{0.3465}\)

\(⇒ t \approx 6.65\, \ minutes\)