Question

For a first order reaction how much half lives are nearly required for 90 completion of reaction.

Answer 1

$t_{90\%}=\frac{2.303}{k}log\frac{a}{(a-0.9a)}$

$t_{\frac{1}{2}}=\frac{2.303}{k}log\frac{a}{a-\frac{a}{2}}=\frac{2.303}{k}log2$

$t_{90\%}=\frac{2.303}{k}\times 0.3$

For a first order reaction,

$k=\frac{2.303}{t}log\frac{a}{a-x}$

$t=\frac{2.303\times t_{\frac{1}{2}}}{0.693}log\frac{100}{100-90}[\because k=\frac{t_{\frac{1}{2}}}{0.693}]$

$=\frac{2.303\times t_{\frac{1}{2}}}{0.693}log10$

$=3.3t_{\frac{1}{2}}$

= 3.3 times that of half-life.