Question

The half-life period of a first-order reaction is $10min$. What percentage of the reaction will be completed in $100min$?

• A. $25$%
• B. $50$%
• C. $99.9$%
• D. $75$%

Answer 1

The correct option is C $99.9$%

$\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$

$\frac{0.693}{10}=\frac{2.303}{100}\log \left(\frac{100}{100-x}\right)\Rightarrow x=99.9$%.