Question

A first-order reaction is $40\%$ complete in $50$ min. How long(minutes) will it take to $80\%$ completion?

Answer 1

Since we know $k=\frac{2.303}{50}\log \left(\frac{a}{a-2a/5}\right)$
$⟹k=\frac{2.303}{50}\log \frac{5}{3}$

Now , $t=\frac{2.303\times 50}{2.303}\frac{log\left(5\right)}{log\left(\frac{5}{3}\right)}$
$⟹t=50\times \frac{0.699}{0.222}$
$⟹t=50\times 3.15$
$⟹t=157.57min$