His mechanism has been proposed for the reaction between chloroform and chlorine. Step 1: Cl2(g) 2Cl(g) fast Step 2: CHCl3(g) Cl(g) CCl3(g) HCl(g) slow Step 3: CCl3(g) Cl(g) CCl4(g) fast a. Write the stoichiometric equation for the overall reaction. b. Identify any reaction intermediates in this mechanism. c. Write the rate equation for the rate determining step. d. Show how the rate equation in c. can be used to obtain the rate law for the overall reaction. e. If the concentrations of the reactants are doubled, by what ratio does the reaction rate change

a) Balanced Overall Stoichiometric Equation

CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)

b) The reaction intermediates include Cl(g) and CCl₃(g)

c) The rate of reaction of the rate determining step is:

Rate = k [CHCl₃] [Cl]

d) The rate of overall reaction is given as

Rate = K [CHCl₃] √[Cl₂]

e) The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.

Explanation:

Step 1: Cl₂(g) → 2Cl(g) fast

Step 2: CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow

Step 3: CCl₃(g) + Cl(g) → CCl₄(g) fast

a) The overall reaction is a reaction between Chloroform (CHCl₃) and Chlorine (Cl₂). It can be obtained by summing all the elementary equations and eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations).

Cl₂(g) + CHCl₃(g) + Cl(g) + CCl₃(g) + Cl(g) → 2Cl(g) + CCl₃(g) + HCl(g) + CCl₄(g)

Now, eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations), we are left with

Cl₂(g) + CHCl₃(g) → HCl(g) + CCl₄(g)

CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)

b) Like I mentioned in (a), the reaction intermediates are the species that appear on both sides upon summing up all the elementary equations. They include:

Cl(g) and CCl₃(g)

c) The rate determining step is usually the slowest step among the elementary equations.

The rate of reaction expression is usually written from the slowest step.

The slowest step is step 2.

CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow

The rate of reaction is then given as

Rate = k [CHCl₃] [Cl]

d) The rate of reaction for the overall reaction is obtained by substituting for any intermediates that appear in the rate of reaction for the rate determining step

The rate of reaction of the rate determining step is:

Rate = k [CHCl₃] [Cl]

But we can substitute for [Cl] by obtaining an expression for it from the step 1.

Step 1: Cl₂(g) → 2Cl(g)

K₁ = [Cl]²/[Cl₂]

[Cl]² = K₁ [Cl₂]

[Cl] = √{K₁ [Cl₂]}

We then substitute this into the rate determining step

Rate = k [CHCl₃] [Cl]

Rate = k [CHCl₃] √{K₁ [Cl₂]}

Rate = (k)(√K₁) [CHCl₃] √[Cl₂]

Let (k)(√K₁) = K (the overall rate constant)

Rate = K [CHCl₃] √[Cl₂]

e) If the concentrations of the reactants are doubled, by what ratio does the reaction rate change

Old Rate = K [CHCl₃] √[Cl₂]

If the concentrations of the reactants are doubled, the new rate would be

New Rate = K × 2[CHCl₃] × √{2 × [Cl₂]}

New Rate = K × 2[CHCl₃] × √2 × √[Cl₂]

New Rate = 2√2 K [CHCl₃] √[Cl₂]

Old Rate = K [CHCl₃] √[Cl₂]

New Rate = 2√2 × (Old Rate)

The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.

Hope this Helps