Question

The half-life period of a $1^{st}$ order reaction is $60$ minutes. What percentage will be left over after $240$ minutes?

• A. $6.25\%$
• B. $4.25\%$
• C. $5\%$
• D. $6\%$

Answer 1

The correct option is A $6.25\%$

For a first order reaction: $t_{1/2}=\frac{ln2}{k}$

$k=\frac{0.693}{t_{1/2}}$

$=\frac{ln2}{60}mi{n}^{-1}$

$\frac{\left[A_t\right]}{\left[A_o\right]}$

$=e^{-kt}$

$=e^{-4ln2}$

$=2^{-4}$

$=0.0625$

Hence $6.25\%$ of reactant will be left after $240$ minutes