Question

A first order reaction takes $69.3$ minutes for $50\%$ completion. How much time will be needed for $80\%$ completion?

Answer 1

$t_{1/2}=69.3min=\frac{ln2}{K}$

$K=\frac{ln2}{69.3}mi{n}^{-1}$

For $80$% conversion, if we assume initial concentration
to be $a_o$, concentration left would be $\frac{a_o}{5}$

$t\times \frac{ln2}{69.3}=ln\left(\frac{a_o}{a_o/5}\right)$

$t=\frac{69.3ln5}{ln2}=161mi{n}^{-1}$