Question

The rate constant of a first order reaction is $6.8\times {10}^{-4}{s}^{-1}.$ If the initial concentration of the reactant is $0.04M,$ what is its molarity after $20$ minutes?

Answer 1

Given, $K=6.8\times {10}^{-4}$ $s^{-1}$; $\left[A_o\right]=0.04$ $M$;

$t=20$ $min=20\times 60=1200$ $sec$

$K=\frac{2.303}{t}\log \frac{[A{]}_o}{[A_t{]}_1}$

$⟹6.8\times {10}^{-4}=\frac{2.303}{1200}{\log }_{10}\left(\frac{0.04}{\left[A_t\right]}\right)$

$⟹Antiog\left(0.3545\right)=\frac{0.04}{\left[A_t\right]}$

$⟹2.261=\frac{0.04}{\left[A_t\right]}$

$⟹\left[A_t\right]=0.0177$ $M$