Chemical Kinetics and Nuclear ChemistryHard

Question

The suggested mechanism for the reaction CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) is

$${Cl_{2}\overset{\quad K_{1}K_{2}\quad}{\leftleftarrows}2\overset{\bullet}{Cl}\ \left( \text{fast} \right) }{CHCl_{3} + \overset{\bullet}{Cl}\overset{\quad K_{3}\quad}{\rightarrow}HCl + \overset{\bullet}{C}Cl_{3}\left( \text{slow} \right) }$$$\overset{\bullet}{C}Cl_{3} + \overset{\bullet}{Cl}\overset{\quad K_{4}\quad}{\rightarrow}CCl_{4}\left( \text{fast} \right)$

The experimental rate law consistent with the mechanism is

Options

A.rate = $K_{3}\left\lbrack CHCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack$
B.rate = $K_{4}\left\lbrack CCl_{3} \right\rbrack\lbrack Cl\rbrack$
C.rate = $K_{eq}\left\lbrack CHCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack$
D.rate = $K_{3}K_{eq}^{1/2}\left\lbrack CHCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack^{1/2}$

Solution

$r = K_{3}\left\lbrack CHCl_{3} \right\rbrack\lbrack Cl\rbrack\lbrack Cl\rbrack\text{ and }\frac{K_{1}}{K_{2}} = \frac{\lbrack Cl\rbrack^{2}}{\left\lbrack Cl_{2} \right\rbrack}$

$\therefore r = \sqrt{\frac{K_{1}}{K_{2}}}.K_{3}\left\lbrack CHCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack^{1/2}$

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