Propose the mechanisms for the reactions given below. Prove that your mechanisms are consistent with — Physical Chemistry — Kinetics Chemistry Question
Kinetic puzzles
Propose the mechanisms for the reactions given below. Prove that your mechanisms are consistent with the experimentally observed rate laws. Use proper approximations if necessary.
Oxidation of bromide ion by permanganate in acidic solution
2 MnO4 – + 10 Br– + 16 H+ = 2 Mn2+ + 5 Br2 + 8 H2O
a) at low concentrations of Br– and H+
r = k с(MnO4 –) с2(Br–) с3(H+)
b) at high concentrations of Br– and H+
r = k с(MnO4 –) с(Br–) с(H+)
where с are the total concentrations of reactants. In both cases с(MnO4 –) << с(Br–), с(H+).
Model Answer
The schematic mechanism is:
2 H+ + Br– + MnO4 – ↔ H2MnO4Br K fast
H2MnO4Br + H+ + Br– → H3MnO4 + Br2 k limiting
H3MnO4 → products fast
At low concentrations of proton and bromide the equilibrium of the first reaction is shifted to the left, hence the concentration of the complex H2MnO4Br is
[H2MnO4Br] = K [MnO4 –] [Br–] [H+]2 ≈ K с(MnO4 –) с(Br–) с2(H+)
At high concentrations of proton and bromide the equilibrium of the first reaction is shifted to the right, hence the concentration of complex H2MnO4Br equals the total concentration of permanganate:
[H2MnO4Br] ≈ с(MnO4 –)
The rate of the reaction
2 MnO4 – + 10 Br– + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O
is half of that of the rate-determining step:
r = ½ k[H2MnO4Br][H+][Br–]
in the case (a)
r = ½ k [H2MnO4Br] [H+] [Br–] ≈ keff с(MnO4 –) с2(Br–) с3(H+)
where keff = ½ k K.
In the case (b)
r = ½ k [H2MnO4Br] [H+] [Br–] ≈ keff с(MnO4 –) с(Br–) с(H+)
where keff = ½ k.
Oxidation of benzamide by peroxydisulfate in the presence of Ag+ ions in water-acetic acid solution
2 C6H5CONH2 + 2 H2O + 3 S2O8 2– = 2 C6H5COOH + 6 SO4 2– + N2 + 6 H+
r = k [Ag+] [S2O8 2–]
Model Answer
The catalytic effect of silver is due to formation of silver(II) ions and sulfate ion radicals upon reaction of Ag+ with persulfate. The mechanism is:
Ag+ + S2O8 2 – → ⋅SO4 – + SO4 2– + Ag2+ slow, rate-determining
⋅SO4 – + PhCONH2 → products fast
Ag2+ + PhCONH2 → products fast
The first reaction is the rate-determining step; therefore the overall oxidation reaction has the same order as the rate-determining step:
r = k [Ag+] [S2O8 2–]
Oxidation of formate ion by peroxydisulfate in water solution
HCOO– + S2O8 2– = CO2 + 2 SO4 2– + H+
r = k [HCOO–]1/2 [S2O8 2–]
Model Answer
The minimal mechanism includes the following steps:
S2O8 2– → 2 ⋅SO4 – k1 very slow
HCOO– + ⋅SO4 – → H+ + ⋅CO2 – + SO4 2– k2 fast
⋅CO2 – + S2O8 2– → ⋅SO4 – + CO2 k3 fast
⋅CO2 – + ⋅SO4 – → SO4 2– + CO2 k4 fast
The second and the third reaction make a chain process involving consumption of peroxydisulfate and formate. The first reaction is very slow, so most of peroxydisulfate is consumed in the third reaction. Applying the steady-state approximation to ⋅SO4 – and ⋅CO2 – we get:
2 r1 – r2 + r3 – r4 = 0
r2 – r3 – r4 = 0
Hence
r1 = r4
r2 – r3 = r1
Since the rate of the first reaction is very low, then
r1 = r4
r2 = r3
Applying the rate laws we get:
k1 [S2O8 2–] = k4 [⋅CO2 –] [⋅SO4 –]
k2 [HCOO–] [⋅SO4 –] = k3 [⋅CO2 –] [S2O8 2–]
Hence
[⋅CO2 –] = (k1k2)1/2 (k3k4)–1/2 [HCOO–]1/2
[⋅SO4 –] = (k1k3)1/2 (k2k4)–1/2 [S2O8 2–] [HCOO–]–1/2
The rate of the reaction is equal to the rate of formate consumption:
r = r2 = k2 [HCOO–] [⋅SO4 –] = (k1k2k3)1/2 k4–1/2 [HCOO–]1/2 [S2O8 2–] =
= keff [HCOO–]1/2 [S2O8 2–]
A more complex mechanism includes the formation of OH radicals and several chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.
Oxidation of azide ion by iodine in carbon disulfide solution
I2 + 2 N3 – = 3 N2 + 2 I–
r = k [N3 –]
Model Answer
The rate-determining step is the addition of azide ion to the solvent, carbon disulfide:
The oxidation of this ion by iodine is a series of fast reactions. The overall rate of the reaction
I2 + 2 N3 – → 3 N2 + 2 I–
is half of that of the azide-CS2 reaction:
r = ½ k [N3 –] [CS2].
Introducing the effective constant keff = ½ k [CS2] we get:
r = keff [N3 –].
Condensation of aldehydes with acryl esters in the presence of the base – 1,4-diazabicyclo[2.2.2]octane (DABCO) in tetrahydrofurane solution
r = k 2 [DABCO]
Model Answer
The reaction mechanism includes several steps. The first step is the reversible addition of DABCO to ether:
The next two steps are the reversible additions of two molecules of aldehyde to the zwitter ion formed in the previous step:
The rate-determining step is an intramolecular proton transfer followed by the elimination of DABCO:
After that, the product rapidly eliminates one molecule of aldehyde. Applying quasi-equilibrium conditions to the first three steps, we get:
r = kRDS K1 K2 K3 2 [DABCO] = keff 2 [DABCO]
It is worth mentioning that in protic solvents the rate-determining step is the solvent-assisted proton transfer in DABCO-ether-aldehyde adduct, hence the reaction order is one with respect to either aldehyde, or ether or base.
Decomposition of peroxyacids in water solution
2 RCO3H = 2 RCO2H + O2
where c(RCO3H) is the total concentration of acid. Consider the following: when the mixture of normal RCO–O–O–H and isotopically labeled RCO–18O–18O–H peroxyacid is used as a reactant, the main species of evolving oxygen are 16O2 and 18O2.
Model Answer
The first step of the reaction is the reversible addition of peroxyacid anion to the carboxylic group of peroxyacid:
The next, rate-determining step is a decomposition of the addition product:
Applying a quasi-equilibrium approximation, we get:
r = keff [RCO3H] [RCO3 –]
The concentrations of peroxyacid and its anion are related to the total concentration of peroxy compound с(RCO3H) and proton concentration [H+] as follows:
where Ka is the acidity constant of peroxyacid. Substituting these concentrations to the rate law we obtain:
Note that at given c(RCO3H) the reaction rate is maximum if [RCO3H] = [RCO3 –] (and [H+] = Ka).