Question

The half-life period of a first-order chemical reaction is $6.93$ minutes. The time required for the completion of $99\%$ of the chemical reaction will be: $[\log 2=0.301]$

• A. 230.3 minutes
• B. 23.03 minutes
• C. 46.06 minutes
• D. 460.6 minutes

Answer 1

Answer: (C)

46.06 minutes

$K=\frac{0.693}{t_{\frac{1}{2}}}$

$K=\frac{0.693}{6.93}={10}^{-1}mi{n}^{-1}$
${10}^{-1}=\frac{2.303}{t}\log \frac{100}{1}$
$t=\frac{2.303\times 2}{{10}^{-1}}=46.06min$