A mixture of $2$ gases $-\mathrm{A}$ (reactive) and $\mathrm{X}$ (inert) is kept in a closed flask when A decomposes as per the following reaction at $500\mathrm{K}$:
$\mathrm{A}\left(\mathrm{g}\right)⟶2 \mathrm{B}\left(\mathrm{g}\right)+3\mathrm{C}\left(\mathrm{l}\right)+4\mathrm{D}\left(\mathrm{s}\right)$

$\mathrm{P}°=210 \mathrm{mmHg}$

${\mathrm{P}}_{\mathrm{t}}\left(10\mathrm{mins}\right)=330\mathrm{mmHg}$

${\mathrm{P}}_{\mathrm{x}}=430\mathrm{mmHg}$

Vapour pressure of $\mathrm{C}\left(\mathrm{l}\right)$ at $500\mathrm{K}$ is $20\mathrm{mmHg}$.
Calculate the half life for $\mathrm{A}$:

Which of the following is/are correct for the first order reaction?

(I) 

(II) 

(III)  

Consider the following reaction occurring in gas phase:

${\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathrm{CHCl}}_3\left(\mathrm{g}\right)⟶\mathrm{HCl}\left(\mathrm{g}\right)+{\mathrm{CCl}}_4\left(\mathrm{g}\right)$

The reaction is experimentally found to obey the following rate law:

$\mathrm{R}=\mathrm{k}\left[{\mathrm{CHCl}}_3\right]{\left[{\mathrm{Cl}}_2\right]}^{\frac{1}{2}}$

Step 1: ${\mathrm{Cl}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{Cl}\left(\mathrm{g}\right)$

Step 2: $\mathrm{Cl}\left(\mathrm{g}\right)+{\mathrm{CHCl}}_3\left(\mathrm{g}\right)⟶\mathrm{HCl}\left(\mathrm{g}\right)+{\mathrm{CCl}}_3\left(\mathrm{g}\right)$

Step 3: $\mathrm{Cl}\left(\mathrm{g}\right)+{\mathrm{CCl}}_3\left(\mathrm{g}\right)⟶{\mathrm{CCl}}_4\left(\mathrm{g}\right)$

Based on this information, what conclusion can be drawn about this proposed mechanism:

Consider this gas phase reaction

${\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathrm{CHCl}}_3\left(\mathrm{g}\right)⟶\mathrm{HCl}\left(\mathrm{g}\right)+{\mathrm{CCl}}_4\left(\mathrm{g}\right)$

The reaction is found experimentally to follow this rate law.

rate $=\mathrm{k}\left[{\mathrm{CHCl}}_3\right] {\left[{\mathrm{Cl}}_2\right]}^{1/2}$

Based on this information, what conclusions can be drawn about this proposed mechanism?

Step $1$. ${\mathrm{Cl}}_2\left(\mathrm{g}\right)⟶2\mathrm{Cl}\left(\mathrm{g}\right)$

Step $2$. $\mathrm{Cl}\left(\mathrm{g}\right)+{\mathrm{CHCl}}_3\left(\mathrm{g}\right)⟶\mathrm{HCl}\left(\mathrm{g}\right)+{\mathrm{CCl}}_3\left(\mathrm{g}\right)$

Step $3$. $\mathrm{Cl}\left(\mathrm{g}\right)+{\mathrm{CCl}}_3\left(\mathrm{g}\right)⟶{\mathrm{CCl}}_4\left(\mathrm{g}\right)$

For a gas reaction at $\mathrm{T} \left(\mathrm{K}\right)$, the rate is given by $-\frac{{\mathrm{dP}}_{\mathrm{A}}}{\mathrm{dt}}={\mathrm{k}}^{\text{'}}{\mathrm{P}}_{\mathrm{A}}^2 \mathrm{atm} {\mathrm{hr}}^{-1}$. If the rate equation is expressed as $-{\mathrm{r}}_{\mathrm{A}}=-\frac{1}{\mathrm{V}}\frac{{\mathrm{dn}}_{\mathrm{A}}}{\mathrm{dt}}={\mathrm{kC}}_{\mathrm{A}}^2$,$\mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{hr}}^{-1}$, the rate constant $\mathrm{k}$ is given by

(where, $\mathrm{R}=$ ideal gas law constant, $\mathrm{cal} {\mathrm{g}}^{-1} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$)