Question

1. The following mechanism has been proposed for the pas-phase reaction of chloroform and chlorine Ch> 20 (fast) CI + CHCI, - HCI +CCI, (slow) Cl + CCI - CCI (fast) a) Identify any intermediate(s): b) Identify any catalyst(s): b) Write the balanced equation for the overall reaction: c) Molecularity of each step: step 1. step 2. step 3 d) rds (circle): 1 2 3 d) What is the observable rate law predicted by this mechanism? e) What is the overall order for the reaction? f) What are the units for the rate constant for the predicted rate law? B) Formulate the reaction rate laws for each component in the reaction: h) Based on the rate determining step (i.e., using ka <<<k, k., ks) what simplification/s can be made to the set of formulated equations? Show on the above formulated equations if you wish.

Answer 1

Based on the proposed mechanism, the overall chemical reaction can be written by adding all the three partial equations of the mechanism. The method of addition is shown below:

Step 1) Cl₂$\rightarrow$2Cl.

Step 2) Cl. + CHCl₃$\rightarrow$ HCl + .CCl₃

Step 3) $\cdot$ CCl₃ + $\cdot$ Cl $\rightarrow$ CCl₄

The free-radicals (shown in bold and italics) which are produced in one step of mechanism and are consumed in another step get cancelled out and will not appear in the overall reaction. For example, in the first step, 2 Cl. are produced but get used up in step 2 and 3. So they cancel out and will not appear in the overall reaction. Similar is the fate of .CCl₃ (produced in step 2 and used up in step 3). Such chemical species are commonly referred to as intermediates.

$\therefore$1a) Identification of intermediates: Cl. and .CCl3 free-radicals

1b) No catalysts. This is because there are no substances which appear to speed up the reaction but do not get used in it.

1b) The overall balanced chemical reaction (derived by the addition of steps 1 to 3 of mechanism).

Cl₂ + CHCl₃$\rightarrow$ CCl₄ + HCl

<Note: Each step of the mechanism should be separately balanced and intermediates should cancel out i.e. their numbers produced must be equal to their numbers used>

1c) Molecularity: refers to how many molecules react to form products in each step of the mechanism

Molecularity of step 1 = 1

Molecularity of step 2 = 2

Molecularity of step 3 = 2

1d) Rate law for the overall reaction: Rate = k [Cl₂]^1/2. [CHCl₃]; k is rate constant and [ ] indicate molar concentrations

Note that the exponents or powers of the molar concentrations are experimentally dependent numbers and the rate law given above is a representative of the class of reactions called free-radical halogenation of alkanes.

1e) Overall order of reaction

The overall order of reaction is got by adding the exponents of all the molar concentration in the rate law or rate equation.

In this case, the overall order of reaction = 1/2 +1 = 1.5

1f) Units of the rate constant

The units of rate are mol L^-1 s^-1 and the units of molar concentration are molL^-1. Doing dimensional analysis from the rate law written in 1d

molL^-1s^-1 = k. (molL^-1)^1/2. (molL^-1)

=> k = (molL^-1s^-1)/(molL^-1)^1/2. (molL^-1) = mol^-1/2.L^1/2 s^-1

If molL^-1 is written as M, then units of k = {1/(molL^-1)^1/2}s^-1 = M^-1/2.s^-1

1g) Reaction rate laws for each of the mechanism steps (can be written from their molecularity i.e the sum of exponents is equal to the molecularity)

For the first step: Rate₁ = k₁.[Cl₂]

For the second step: Rate₂ = k₂ [Cl.][CHCl₃]

For the third step: Rate₃ = k₃ [Cl.][.CCl₃]

1h) Since, k₁ and k₃ are much greater than k₂. The elementary or mechanism steps written above can be simplified so that the first step occurs only once and leads to step 2 and 3 happening repeatedly. This can be achieved as follows:

Step 1) Cl₂$\rightarrow$ 2 Cl.

Step 2) Cl. + CHCl₃$\rightarrow$ HCl + .CCl₃

Step 3) .CCl₃ + Cl₂$\rightarrow$ CCl₄ + .Cl

Now, Steps 2 and 3 can be called chain propagating steps and step 1 can be called chain initiation step