Answer:
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.
Step-by-step 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!!!!