Chemistry - Chemical Kinetics Question with Solution | TestHub

I.$\log (\mathrm{a} – \mathrm{x})$, where $(\mathrm{a} – \mathrm{x})$ is remaining concentration, decreases with passage of time.

$\log \left(\mathrm{a}-\mathrm{x}\right)=\log a-\frac{\mathrm{kt}}{2.303}$

$\mathrm{y}=\log \mathrm{a}-\frac{\mathrm{k}}{2.3030}\mathrm{x}$

$\mathrm{slope}=-\mathrm{ve}$
II. $\mathrm{k}=\frac{2.303}{\mathrm{t}}{\log }_{10}\left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)$
$\log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)=\frac{\mathrm{k}}{2.303}\times \mathrm{t}=\mathrm{y}=\frac{\mathrm{k}}{2.303}\mathrm{x}$
$\mathrm{y}=\mathrm{m}\times \mathrm{x}$
Plot of $\log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)$ against 't' gives straight line through origin.
III. ${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}};$ Half-life of first order reaction does not depend on initial concentration. Thus, graph of ${\mathrm{t}}_{1/2}$ against 'a' gives horizontal straight line.