Question

he following mechanism has been proposed for the gas-phase reaction of chloroform (CHCl3) and chlorine. Cl2 ⇌ 2Cl (fast, reversible) Cl + CHCl3 → HCl + CCl3 (slow) Cl + CCl3 → CCl4 (fast) What rate law does this mechanism predict? (Choose from the list below and enter your answers in alphabetical order, e.g. ABC ). A)k G) [CCl3]1/2 M) [HCl]2 B) [Cl] H) [HCl]1/2 N) [Cl2]2 C) [CHCl3] I) [Cl2]1/2 O) [Cl]2 D) [CCl3] J) [Cl]1/2 P) [CHCl3]2 E) [HCl] K) [CHCl3]1/2 F) [Cl2] L) [CCl3]2 Tries 0/99 Decide which of the following reactive intermediates could exist for the reaction above. Cl Cl2 CHCl3 CCl3 CCl4 Tries 0/99

Answer 1

Answer: The rate law predicted by the given mechanism is ACI

Explanation:

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The intermediate reaction of the mechanism follows:

Step 1:  $Cl_2\rightleftharpoons 2Cl;\text{ (fast)}$

Step 2:  $Cl+CHCl_3\rightarrow HCl+CCl_3;\text{(slow)}$

Step 3:  $Cl+CCl_3\rightarrow CCl_4;\text{(fast)}$

Overall reaction:  $Cl_2+CHCl_3\rightarrow HCl+CCl_4$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=K_2[Cl][CHCl_3]$      ......(1)

As, [Cl] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for [Cl] from step 1 and step 3, we get:

$K_1=\frac{[Cl]^2}{[Cl_2]}$

$[Cl]=\sqrt{K_1[Cl_2]}$

$K_3=\frac{[CCl_4]}{[CHCl_3][Cl]}$

$[Cl]=\frac{[CCl_4]}{K_3\times [CCl_3]}$

Putting the value of [Cl] in equation 1, we get:

$\text{Rate}=K_2(\sqrt{K_1[Cl_2]}\times (\frac{[CCl_4]}{K_3[CCl_3]})[CHCl_3]\\\\\text{Rate}=k[Cl]^{1/2}[CCl_4][CCl_3]^{-1}[CHCl_3]$

Hence, the rate law predicted by the given mechanism is ACI