Equilibrium constants are often included into the rate equations for complex chemical reactions. For — Physical Chemistry — Kinetics Chemistry Question
Quasi-equilibrium model
Equilibrium constants are often included into the rate equations for complex chemical reactions. For some rapid reversible steps the ratio of concentration of products to concentration of reactants is assumed to be equal to equilibrium constant, though reaction as a whole still proceeds and chemical equilibrium is not attained. This is quasi-equilibrium approximation. The concept of quasi-equilibrium makes rate equations much simpler which is vitally important for complex reactions.
Consider kinetics of a complex reaction:
A + B → C + D (keff)
The following mechanism was proposed:
A + B ↔ AB (k1, k-1)
AB → C + D (k2)
The rates of the forward and reverse reaction of the first step are almost equal, r1 ≈ r-1, i.e., quasi-equilibrium is reached.
Metallic platinum loses its mass interacting with the flow of atomic fluorine at T = 900 K. Partial pressure of F in the incident flow near the surface is 10–5 bar, see Fig.1.
Fig. 1. Gasification of Pt by a flow of atomic fluorine
No solid products of interaction of Pt with F were found on the surface. Gaseous species PtF4 and PtF2 were detected in the flow desorbed from the surface. The ratio p(PtF2)^2 / p(PtF4) was equal to 1 ⋅10–4 bar and did not vary with the change of the incident atomic fluorine flow.
Use the data in the Table to answer the following questions:
Table
Reaction: 2 F(g) = F2(g) | Kр (900 К), bar–1: 1.7 ⋅10^3 | Gaseous species: PtF2 | р(900 K), bar: 2 ⋅10–6
Reaction: Pt(s) + 2F(g) = PtF2(g) | Kр (900 К), bar–1: 5 ⋅10^8 | Gaseous species: PtF4 | р(900 K), bar: 4 ⋅10–8
Calculate keff, if k1 / k–1 = 10 mol-1 dm3, k2 = 20 s–1.
Model Answer
In this case, the quasi-equilibrium step precedes the rate-limiting one, r1 ≈ r–1, k1[A][B] >> k–1[AB] and k–1[AB] >> k2[A][B]. Using the stationary state condition d[AB]/dt ≈ 0 one gets r = k2[AB] = (k1k2/k–1)[A][B] = keff[A][B]. keff = k2(k1/k–1) = 20 * 10 = 200 mol-1 dm3 s-1.
Find the maximum partial pressure of molecular fluorine near the platinum surface under the given experimental conditions. Assume first that gasification does not proceed.
Model Answer
Maximum partial pressure of F2 will be attained if the equilibrium is reached in the reaction 2 F(g) = F2(g) (2 a).
Kp = p(F2) / p(F)^2 = 1.7 * 10^3 bar^-1; p(F2) = Kp * p(F)^2 = 1.7 * 10^3 bar^-1 * (10^-5 bar)^2 = 1.7 * 10^-7 bar.
Partial pressure of molecular fluorine near the surface is negligible.
Why is the ratio p(PtF2)^2 / p(PtF4) = 10–4 bar constant near the surface?
Model Answer
It is safe to assume that quasi-equilibrium is achieved in the reaction Pt(s) + PtF4(g) = 2 PtF2(g) (2 b). The measured ratio p(PtF2)^2 / p(PtF4) is equal to the equilibrium constant of this reaction.
Make the necessary assumptions and estimate the partial pressure of atomic fluorine in the desorbed flow.
Model Answer
One may assume that quasi-equilibrium is also achieved in the reaction Pt(s) + 2 F(g) = PtF2(g) (2 c) within the desorbed flow. Then p(F) = (p(PtF2) / K_eq.2c)^1/2 = (2 * 10^-6 / 5 * 10^8)^1/2 = (4 * 10^-15)^1/2 = 6.3 * 10^-8 bar.
Put forward the quasi-equilibrium model to account for the rate of gasification of Pt with atomic fluorine,
r_Pt = dn_Pt / dt = {mol of Pt / Pt surface area / time}
Make use of the dimensionless equilibration probability, α, which is equal to the fraction of incident fluorine flow involved in gasification. Consider other steps of gasification as quasi-equilibrium. The flow ρi of each gaseous species i is related to its partial pressure pi as ρi = c(pi / mi^1/2) where mi is a molecular mass, c is constant.
Model Answer
For the gasification of platinum, the following mechanism could be proposed:
F(Inc) → αF(s) → F(Des)
2 F + Pt = PtF2, 4 F + Pt = PtF4
Here the rate-limiting step is “equilibration” of atomic fluorine on the surface. It precedes the quasi-equilibrium steps. “Equilibrated” fluorine takes part in the quasi-equilibrium gasification of Pt. Interaction products do not accumulate on Pt surface. Hence ρ_F(Inc) * α = ρ_F(Des) + 2 ρ_PtF2 + 4 ρ_PtF4, or α * c * p_F(Inc)/(m_F)^1/2 = c * p_F(Des)/(m_F)^1/2 + 2 * c * p_PtF2/(m_PtF2)^1/2 + 4 * c * p_PtF4/(m_PtF4)^1/2.
Under the experimental conditions: α * p_F(Inc)/(m_F)^1/2 >> 2 * p_PtF2/(m_PtF2)^1/2.
The rate of gasification is r_Pt = dn_Pt / dt = ρ_PtF2 + ρ_PtF4 >> ρ_PtF2 >> (α / 2) ρ_F(Inc).
Estimate the equilibration probability α under the experimental conditions, described in the Table.
Model Answer
As it was shown in 2.4, α * p_F(Inc)/(m_F)^1/2 >> 2 * p_PtF2/(m_PtF2)^1/2.
p_F(Inc) = 10^-5 bar, p_PtF2 = 2 * 10^-6 bar.
α * (10^-5 / 19^1/2) = 2 * (2 * 10^-6 / 233^1/2)
α * (10^-5 / 4.35) = 2 * (2 * 10^-6 / 15.26)
α = 0.4 * (4.35 / 15.26) = 0.1
How many grams of Pt will be gasified from 1 cm2 of the Pt surface in 15 minutes, if the incident flow of atomic fluorine is 2⋅1018 atoms/cm2/s?
Model Answer
r_Pt = dn_Pt / dt = ρ_PtF2 + ρ_PtF4 >> ρ_PtF2 >> (α / 2) ρ_F(Inc).
r_Pt = (0.1 / 2) * ρ_F(Inc) = 0.05 * 2 * 10^18 = 10^17 atoms / cm^2 / s.
In 15 minutes: N = 10^17 * 15 * 60 = 9 * 10^19 atoms / cm^2 = 1.5 * 10^-4 mol / cm^2.
m_Pt = n * M_Pt = 1.5 * 10^-4 * 195.08 = 0.029 g / cm^2 will be gasified.