Question

Half life of a 1st order reaction is 30 sec. Calculate the time required for 2/3 completion of the reaction.

Answer 1

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

$\Rightarrow 30=\frac{0.693}{K}$

$\Rightarrow k=0.0231$

Let $A_o$ be initial concentration. At $2/3$ completetion of reaction, remaining concentration

$\Rightarrow A=A_o-\frac{2}{3}{A}_o=/cfrac13A$

$\therefore t_{2/3}=\frac{2.303}{K}log\left[\frac{A_o}{A}\right]$

$=\frac{2.303}{0.0231}log\left[\frac{A_o}{\frac{1}{3}{A}_o}\right]$

$t_{2/3}=47.56sec$