Chemistry - Chemical Kinetics Question with Solution | TestHub

As$\mathrm{k}=\frac{0.6932}{{\mathrm{t}}_{1/2}}$
$=\frac{0.6932}{\mathrm{6.93 }}\mathrm{m}\mathrm{i}{\mathrm{n}}^{-1}$
Also$\mathrm{t}=\frac{2.303}{\mathrm{\lambda }}\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[{\mathrm{A}}_{\mathrm{o}}\right]}{\left[\mathrm{A}\right]}$
$\left[{\mathrm{A}}_{\mathrm{n}}\right]$= initial concentration (amount)
[A] = final concentration (amount)
So$\mathrm{t}=\frac{2.303\times 6.93}{0.6932}\mathrm{l}\mathrm{o}\mathrm{g}\frac{100}{1}$
$=\mathrm{46.06 }\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}$