Chemistry - Chemical Kinetics Question with Solution | TestHub

${\text{C}}_{\text{t}}\text{=}\frac{{\text{C}}_{\text{0}}}{{\left(\text{2}\right)}^{\text{n}}}$
${\left(\text{2}\right)}^{\text{n}}\text{=}\frac{\text{800}}{\text{50}}\text{=16}$
${\left(\text{2}\right)}^{\text{n}}\text{=}{\left(\text{2}\right)}^{\text{4}}$
So,$\text{n}\text{=}\text{4}$
${\text{4T}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\text{200}\text{sec}$
${\text{T}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}\text{=}\text{50}\text{sec}$
$\text{K}\text{=}\frac{\text{0}\text{.693}}{{\text{T}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}}\text{=}\frac{\text{0}\text{.693}}{\text{50}}\text{=}\text{1}\text{.386}\text{×}{\text{10}}^{-\text{2}}{\text{sec}}^{-\text{1}}$