Chemistry - Chemical Kinetics Question with Solution | TestHub

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

$k=\frac{0.693}{t_{1/2}}$

Substituting the given half-life $t_{1/2}=850$ s:

$k=\frac{0.693}{850}=8.15\times 10^{-4}{\text{ s}}^{-1}$

The integrated rate law for a first order reaction is:

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

Substituting the values $t=1200$ s and $[A{]}_0=0.06{\text{ mol dm}}^{-3}$:

$8.15\times 10^{-4}=\frac{2.303}{1200}\log \left(\frac{0.06}{[A{]}_t}\right)$

$\log \left(\frac{0.06}{[A{]}_t}\right)=\frac{8.15\times 10^{-4}\times 1200}{2.303}\approx 0.425$

Taking the antilog on both sides:

$\frac{0.06}{[A{]}_t}=10^{0.425}\approx 2.66$

$[A{]}_t=\frac{0.06}{2.66}\approx 0.0226{\text{ mol dm}}^{-3}$

Rounding off, the concentration remaining is approximately $0.023{\text{ mol dm}}^{-3}$.

Answer: $0.023$ mol ${\text{dm}}^{-3}$