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?
Answer & explanation
Correct answer: option 4
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,
\(t = \frac{2.303}{0.3465} log{100}{10}\)
\(⇒ t = \frac{2.303}{0.3465}log (10)\)
\(⇒ t = \frac{2.303}{0.3465}\)
\(⇒ t \approx 6.65\, \ minutes\)