Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the time required is given by the integrated rate law:

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

Given:
$k=1.15\times 10^{-3}{\text{ s}}^{-1}$
$[A{]}_0=5\text{ g}$
$[A{]}_t=3\text{ g}$

Substituting the values into the formula:

$t=\frac{2.303}{1.15\times 10^{-3}}{\log }_{10}\left(\frac{5}{3}\right)$

$t=\frac{2.303}{1.15\times 10^{-3}}({\log }_{10}5-{\log }_{10}3)$

$t=\frac{2.303}{1.15\times 10^{-3}}(0.6990-0.4771)$

$t=\frac{2.303}{1.15\times 10^{-3}}\times 0.2219$

$t=2.0026\times 10^3\times 0.2219$

$t\approx 444\text{ s}$

Answer: 444 second