Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the rate constant $k$ is given by:

$k=\frac{2.303}{t}{\log }_{10}\left(\frac{[A{]}_0}{[A{]}_t}\right)$

Given:
$[A{]}_0=20$ milli mol dm${}^{-3}$
$[A{]}_t=8$ milli mol dm${}^{-3}$
$t=40$ minute

Substituting the values:

$k=\frac{2.303}{40}{\log }_{10}\left(\frac{20}{8}\right)$

$k=\frac{2.303}{40}{\log }_{10}(2.5)$

$k=\frac{2.303}{40}\times 0.3979$

$k=\frac{0.9164}{40}$

$k=0.02291$ minute${}^{-1}$

$k\approx 0.023$ minute${}^{-1}$

Answer: 0.023 minute ${}^{-1}$