[અહી આપેલ $\left.\log _{10} 2=0.3010\right]$
$t_{1 / 2}=\frac{0.693}{2.303 \times 10^{-3}}=301 s$
The time required for $40 \mathrm{g}$ of reactant to reduce to $10 \mathrm{g}$ $\mathrm{t}_{75 \%}=2 \times \mathrm{t}_{1 / 2}$
$\mathrm{t}_{75\%} =2 \times 301=602\; \mathrm{s}$
(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ) $[$ ઉપયોગ કરો : $\left. R =8.31 \,J \,K ^{-1} \,mol ^{-1}\right]$
( $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$ આપેલ છે.)