Question

Half life of a first order reaction is 10 min. What % of reaction will be completed in 100 min ?
${10}^{3.009}\approx 1020.9$

• A. 25%
• B. 99.9%
• C. 75%
• D. 80%

Answer 1

The correct option is B 99.9%

Given, $n=1,t_{1/2}=10\text{min},t=100min$
Let initial amout be a=100
Applying first order formula
$\frac{0.693}{t_{1/2}}=\frac{2.303}{t}lo{g}_{10}\left(\frac{a}{a-x}\right)$
$\frac{0.693}{10}=\frac{2.303}{100}log\left(\frac{100}{100-x}\right)$
$log\left(\frac{100}{a-x}\right)=3.009$
$\frac{100}{100-x}={10}^{3.009}=1020.9$

x = 99.9%