Question

The half-life period of a first order process is $1.6$min. It will be $90\%$ complete in :

• A. $5.3$ min
• B. $10.6$ min
• C. $43.3$ min
• D. $99.7$ min

Answer 1

The correct option is A $5.3$ min

For a first order reaction,
$k=\frac{2.303}{t}log\frac{a}{a-x}$

At half-time,
$x=\frac{a}{2}$

$\therefore k=\frac{2.303}{1.6}\times log\frac{a}{a-\frac{a}{2}}$

$=\frac{2.303}{1.6}\times log2$

$=\frac{2.303}{1.6}\times 0.3010$

$=0.4332$

For $90\%$ completion,
$x=0.9$a

$\therefore k=\frac{2.303}{t}log\frac{a}{a-0.9a}$

$0.4332=\frac{2.303}{t}log\frac{a}{0.1a}$

$=\frac{2.303}{t}log10$

$=\frac{2.303}{t}\times 1$

$t=\frac{2.303}{0.4332}$

$=5.3$min.