Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the rate constant is given by $k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$

For $60\%$ completion in $20$ minutes:

$k=\frac{2.303}{20}\log \left(\frac{100}{100-60}\right)=\frac{2.303}{20}\log (2.5)$

For $84\%$ completion in time $t$:

$k=\frac{2.303}{t}\log \left(\frac{100}{100-84}\right)=\frac{2.303}{t}\log (6.25)$

Equating the two expressions for $k$:

$\frac{1}{20}\log (2.5)=\frac{1}{t}\log (2.5^2)$

$\frac{1}{20}\log (2.5)=\frac{2}{t}\log (2.5)$

$t=40$ min

Answer: $40$ min