Question

$60$% of the first-order reaction was completed in $60$ min. The time taken for reactants to decompose to half of their original amount will be:

• A. $\approx 30min$
• B. $\approx 60min$
• C. $\approx 90min$
• D. $\approx 45min$

Answer 1

The correct option is D $\approx 45min$

$t_{1/2}=\frac{0.69}{k}=\frac{0.3\times 2.3}{k}$

$t_{60\%}=\frac{2.3}{k}log\frac{100}{100-60}=\frac{2.3}{k}log\frac{10}{4}$

$k=\frac{2.3}{60}log\frac{10}{4}$

$\frac{0.3\times 2.3}{t_{1/2}}=\frac{2.3}{60}log\frac{10}{4}=\frac{1}{60}[1-2log2]=\frac{1}{60}[1-0.6]$
$\frac{0.3}{t_{1/2}}=\frac{0.4}{60}$
$\therefore t_{1/2}=\frac{60\times 0.3}{0.4}=45min$
Use direct relation
$t_{1/2}=0.3,t_{x\%}=\left(log\frac{100}{100-x}\right)$
$t_{60\%}=log\frac{10}{4}=\left(0.4\right)$
$\frac{t_{1/2}}{t_{60\%}}=\frac{0.3}{0.4}$
$\therefore t_{1/2}=t_{60\%}\times \frac{0.3}{0.4}=\frac{60\times 3}{4}=45min$