[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 \(10 \mathrm{g}\) \(\mathrm{t}_{75 \%}=2 \times \mathrm{t}_{1 / 2}\)
\(\mathrm{t}_{75\%} =2 \times 301=602\; \mathrm{s}\)
(આપેલ : $\log 2=0.30, \log 3=0.48$ )
[લો; $R =8.314 \,J\, mol ^{-1}\, K ^{-1}$ In $3.555=1.268$]