Chemistry - Chemical Kinetics Question with Solution | TestHub

Match the coloumn :-

(P) Constant boling mixture → Azeotrope

(Q) Cryscopic constant is property of solvent.

$(\mathrm{R})\mathrm{A}\to \mathrm{B}$ first order reaction

$$\text{ Half life }\begin{array}{rlrl}{t}_{1/2}& =\frac{1}{K}\ln \frac{\left[A_0\right]}{\left[A_0\right]1/2}& & \end{array}$$

$$t_{1/2}=\frac{{\ln }^2}{K}$$

$$t_{3/4}=\frac{1}{K_1}\ln \frac{\left[A_0\right]}{\left[A_0\right]/4}=\frac{{\ln }^2}{K}=2t_{1/2}$$

(S) 2A → second order reaction.

${\mathrm{t}}_{1/2}=\frac{1}{\mathrm{K}\left[{\mathrm{A}}_0\right]}$ So First ${\mathrm{t}}_{1/2}=\frac{1}{\mathrm{K}\left[{\mathrm{A}}_0\right]}$

After first ${\mathrm{t}}_{1/2}:-[\mathrm{A}]=\frac{\left[{\mathrm{A}}_{\mathrm{o}}\right]}{2}$

$$\text{ So, }{2}^{\text{nd }}{\mathrm{t}}_{1/2}=\frac{2}{\mathrm{K}[\mathrm{A}.]}$$

After $2^{\text{nd }}{\mathrm{t}}_{1/2}:-[\mathrm{A}]=\frac{\left[{\mathrm{A}}_{\mathrm{o}}\right]}{2}$ So $3^{\text{rd }}{\mathrm{t}}_{1/2}=\frac{4}{\mathrm{K}\left[{\mathrm{A}}_{\mathrm{o}}\right]}$

So ${\mathrm{I}}^{\mathrm{st}}{\mathrm{t}}_{1/2}:2^{\mathrm{nd}}{\mathrm{t}}^{1/2}:3^{\mathrm{rd}}{\mathrm{t}}_{1/2}=1:2:4$