Chemistry - Chemical Kinetics Question with Solution | TestHub

Integrated rate law for first order reaction is
$\mathrm{k}=\frac{1}{\mathrm{t}}1\mathrm{n}\left(\frac{{\mathrm{C}}_{\mathrm{o}}}{{\mathrm{C}}_{\mathrm{t}}}\right)$
[Where${\mathrm{C}}_{\mathrm{t}}\to$concentration of reactant at time$\mathrm{t},{\mathrm{C}}_{\mathrm{o}}\to$initial concentration of reactant]
$1\mathrm{n}\left(\frac{{\mathrm{C}}_{\mathrm{o}}}{{\mathrm{C}}_{\mathrm{t}}}\right)=1\mathrm{n}{\mathrm{C}}_{\mathrm{o}}-1\mathrm{n}{\mathrm{C}}_{\mathrm{t}}=\mathrm{k}\mathrm{t}$
1n$\left({\mathrm{C}}_{\mathrm{t}}\right)=-\mathrm{k}\mathrm{t}+1\mathrm{n}\left({\mathrm{C}}_{\mathrm{o}}\right)$
If$\left({\mathrm{C}}_{\mathrm{t}}\right)=\left[\mathrm{R}\right]=$concentration of reactant
$1\mathrm{n}\left[\mathrm{R}\right]=-\mathrm{k}\mathrm{t}+1\mathrm{n}\left({\mathrm{C}}_{\mathrm{o}}\right)$
On comparing with$\mathrm{y}=\text{mx}+\mathrm{C}$
$\Rightarrow$Slope$\mathrm{m}=-\mathrm{k}$for graph between 1n (R) and t with intercept 1n$\left({\mathrm{C}}_{\mathrm{o}}\right)$
$\Rightarrow$First graph is of first order
Integrated rate law for zero order reactions is$\left[{\mathrm{C}}_{\mathrm{t}}\right]=\left[{\mathrm{C}}_{\mathrm{o}}\right]-\mathrm{k}\mathrm{t}$or$\left[{\mathrm{C}}_{\mathrm{t}}\right]=-\mathrm{k}\mathrm{t}+\left[{\mathrm{C}}_{\mathrm{o}}\right]$or$\left[\mathrm{R}\right]=-\mathrm{k}\mathrm{t}+\left[{\mathrm{C}}_{\mathrm{o}}\right]({\mathrm{C}}_{\mathrm{t}}=\mathrm{R}$given)
On comparing with$\mathrm{y}=\text{mx}+\mathrm{C}$
$\Rightarrow$slope$\mathrm{m}=-\mathrm{k}$for graph between (R) and t with intercept${\mathrm{C}}_{\mathrm{o}}$
$\Rightarrow$Second graph is of zero order reactions.