Chemistry - Chemical Kinetics Question with Solution | TestHub

Intergrated 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$\mathrm{y}=\text{mx}+\mathrm{C}$
slope$\mathrm{m}=-\mathrm{k}$for graph between (R) and t with intercept${\mathrm{C}}_{\mathrm{o}}$
first graph is of zero order reactions.
Integrated rate law for first order reaction is
$\mathcal{l}\mathrm{n}\left[\mathrm{R}\right]=-\mathrm{k}\mathrm{t}+\mathcal{l}\mathrm{n}\left({\mathrm{C}}_{\mathrm{o}}\right)$
On comparing with$\mathrm{y}=\text{mx}+\mathrm{C}$
Slope$\mathrm{m}=-\mathrm{k}$for graph between 1n (R) and t with intercept 1n$\left({\mathrm{C}}_{\mathrm{o}}\right)$
Second graph is of first order