75% of a first order reaction was completed in 16 minutes. When was it half completed?

(Given: $\log(0.75) = -0.125, \log(0.25) = -0.602$)

8 min

The correct answer is Option (1) → 8 min

First-Order Rate Law (Integrated): $k=\frac{2.303}{t}\log\left(\frac{[A]_0}{[A]}\right)$

Where: $k$ = rate constant, $t$ = time, $[A]_0$ = initial concentration, $[A]$ = concentration at time $t$

Given: 75% completed in 16 minutes.

If $[A]_0=100$, then $[A]=100-75=25$

$k=\frac{2.303}{16}\log\left(\frac{100}{25}\right)=\frac{2.303}{16}\log(4) $

Since $\log(4)=2\log(2)\approx2\times0.301=0.602 $

$ k=\frac{2.303\times0.602}{16} $

For Half-Life $(t_{1/2})$:

Half-life is the time when the reaction is 50% completed.

So, if $[A]_0=100$, then $[A]=50 $

$ k=\frac{2.303}{t_{1/2}}\log\left(\frac{100}{50}\right)=\frac{2.303}{t_{1/2}}\log(2) $

$ t_{1/2}=\frac{2.303\log(2)}{k}=\frac{0.693}{k}\quad(\text{since }2.303\times\log(2)\approx0.693) $

Relate 75% Completion to Half-Life:

For a first-order reaction:

After one half-life, 50% remains. After two half-lives, 25% remains.

If 25% remains, then 75% is completed.

So, 16 minutes represents two half-lives: $2\times t_{1/2}=16 $

$ t_{1/2}=\frac{16}{2}=8\,\text{minutes.} $