Chemistry - Chemical Kinetics Question with Solution | TestHub

(A) Zero order reaction →

$[\mathrm{P}]{\mathrm{t}}_{\text{comp. }}=\frac{[\mathrm{A}{]}_0}{\mathrm{K}}$

complete in finite time.

[Q] $\frac{1}{[\mathrm{At}{]}^{\mathrm{n}-1}}\mathrm{v}/\mathrm{st}$ for zero order " $\mathrm{n}=0$ "

At $\mathrm{v}/\mathrm{st}$ is straight line between " $[\mathrm{At}]\propto \mathrm{t}$ "

(B) ${\mathrm{I}}^{\text{st }}$ order reaction ⟶

[S] rate of reaction change directly with change in conc. of reactant

[T] It may be elementry on complex reaction.

(C) ${\mathrm{II}}^{\text{nd }}$ order reaction →

$[\mathrm{Q}]\left[{\mathrm{A}}_{\mathrm{t}}\right]\propto$ so for $\mathrm{n}=2$ graph will show straight line.

$[\mathrm{R}]{\mathrm{t}}_{1/2}=\frac{2}{\mathrm{K}[\mathrm{A}{]}_0}$ so ${\mathrm{t}}_{1/2}$ increases on decreases $\left[{\mathrm{A}}_0\right]$

$[\mathrm{S}]$ rate of reaction decreases during reaction.

(D) Half order reaction →

$[\mathrm{P}]$ Reaction completes infinite time.

[Q] At $\propto \mathrm{t}$ for half order so graph between $\frac{1}{{\left({\mathrm{A}}_{\mathrm{t}}\right)}^{\mathrm{t}-1}}$ vs t will be straight line

[S] rar decreases during reaction.