Foundations of Chemical Kinetics. Lecture 32: Heterogeneous kinetics: Gases and surfaces

Transcription

1 Foundations of Chemical Kinetics Lecture 32: Heterogeneous kinetics: Gases and surfaces Marc R. Roussel Department of Chemistry and Biochemistry

2 Gas-surface reactions

3 Adsorption Adsorption: sticking of molecules to a surface Enthalpy-driven process Surface coverage (θ): fraction of surface to which molecules are adsorbed Physisorption: adsorption based on intermolecular forces only Chemisorption: bond formation between an adsorbate and the surface Typically forms much stronger gas-surface associations than physisorption Dissociative chemisorption: chemisorption with bond dissociation in the adsorbate

4 Gas-surface rate processes Consider the process of adsorption of a single species of molecule to a surface. Suppose that the gas phase is well stirred so that in any small volume, the number of molecules per unit volume (the concentration, c) is the same. Consider a volume of thickness δ near a surface of area A. This volume contains cδa molecules of the gas. Let the direction perpendicular to the surface be z. The mean speed along the z axis is vz. Half the molecules would be moving toward the surface and the other half away from it, so the number of molecules that can collide with the surface is N = cδa/2.

5 Gas-surface rate processes (continued) For molecules in this small volume, the mean distance to the surface is δ/2. The mean time before a molecule that is moving toward the surface impacts it is therefore t = δ/2v z. The rate of collisions is therefore N/t = cavz. (Units?) The probability that the molecule sticks to the surface depends on two factors: An intrinsic probability of adsorption per collision event, Pad The probability that the molecule meets an unoccupied site on the surface, 1 θ The rate of adsorption is therefore given by v ad = cav z P ad (1 θ). The concentration can be rewritten in terms of the gas pressure using the ideal gas law: c = n/v = p/rt, so v ad = pav z P ad (1 θ)/rt

6 Defining k ad = Av z P ad/rt Gas-surface rate processes (continued) v ad = pav z P ad (1 θ)/rt we get v ad = k ad p(1 θ) (Units?) Note that the rate constant is proportional to the surface area.

7 Langmuir adsorption isotherm Now consider a molecule present in a container at pressure p which can adsorb and desorb from a surface (usually of an added solid material, but possibly also of the container itself): A (g) A (ad) v ad = k ad p A (1 θ) A (ad) A (g) v de = k de θ At equilibrium, k ad p A (1 θ) = k de θ k ad p A θ = k ad p A + k de

8 Langmuir adsorption isotherm (continued) where θ = p A p A + K 1 ad K ad = k ad /k de This equation, and closely related variations, is called the Langmuir adsorption isotherm. (Units of K ad?)

9 The Langmuir adsorption isotherm in practice In practice, we rarely measure the surface coverage. Rather, we measure the amount of gas adsorbed to the surface as an equivalent volume at the experimental temperature and a constant measurement pressure. Suppose that ρs is the areal density of adsorption sites, and that A is the total surface area. Then the total number of sites (usually in mol) is ρ S A. If the coverage is θ, then the number of moles of gas adsorbed is θρ S A. Invoking the ideal gas law, we have θρs A = pv ad /RT, or θ = pv ad ρ S ART

10 The Langmuir adsorption isotherm in practice (continued) As the pressure of the adsorbate goes to infinity, θ 1, so V = ρ S ART /p and θ = V /V The Langmuir adsorption isotherm becomes V ad = V p A p A + K 1 ad

11 The Langmuir adsorption isotherm in practice Graphical methods For the purpose of extracting the constants, we rewrite V 1 ad = V 1 + (K ad V p A ) 1 Double-reciprocal plots have terrible statistical and visual properties. Ideally, we would fit the isotherm directly, but for graphical purposes, it is convenient to have a linearized form. To get one, multiply through by V ad V and rearrange: V = V ad + K 1 ad (V ad/p A ) V ad = V K 1 ad (V ad/p A )

12 Example: Adsorption of CO on charcoal The following data were obtained for the adsorption of CO on charcoal at 273 K: p CO /torr V ad /cm The linearized graph is V ad /cm V ad p A /cm 3 torr -1 Slope = 979 ± 27 torr, intercept = ± 2.2 cm 3

13 Example: Adsorption of CO on charcoal (continued) From the intercept, we get V = ± 2.2 cm 3. Since θ = V /V, we can infer the surface coverages corresponding to each experimental point. For example, the highest coverage reached in this experiment is θ = 46.1 cm 3 /109.4 cm 3 = From the slope, we get K ad = (979 ± 27 torr) 1 = (1.021 ± 0.028) 10 3 torr 1.

14 Example: Adsorption of CO on charcoal (continued) Take a second look at the graph: V ad /cm V ad p A /cm 3 torr -1 Note the slight curvature. The derivation of the Langmuir isotherm assumes that sites are independent. Adsorbate-adsorbate interactions alter the binding constant K ad at higher coverages, i.e. we run into non-ideal behavior.

15 Surface reactions Once molecules have adsorbed onto a surface, they can diffuse and react. We often represent surface reactions much as we do reactions in other media, but we must bear in mind that the symbols have different meanings. For example, if we write A (ad) + B (ad) product with v = k[a (ad) ][B (ad) ], we must keep in mind that [A] and [B] are areal concentrations, and that v and k will typically have different units from those used in gas-phase kinetics.

16 Example: surface recombination Consider the following mechanism for surface recombination: A (g) + S A (ad) v ad = k ad [A (g) ][S] A (ad) A (g) + S v de = k de [A (ad) ] A (ad) + A (ad) A 2(g) + 2S v r = k r [A (ad) ] 2 where S represents a free surface site. Note the use of concentration symbols here rather than pressures or coverages. It is best to think hard about the units before getting too far into treating a surface reaction. Suppose we want to use units of mol m 3 for [A (g) ], and of mol m 2 for [A (ad) ] and [S]. Because of the different units of the concentrations, we would normally formulate the rates in mol s 1. What are the units of the rate constants?

17 Example: surface recombination (continued) A (g) + S A (ad) v ad = k ad [A (g) ][S] A (ad) A (g) + S v de = k de [A (ad) ] A (ad) + A (ad) A 2(g) + 2S v r = k r [A (ad) ] 2 Rate equations: d[a (g) ] dt d[a (ad) ] dt d[s] dt = 1 ( kad [A V (g) ][S] + k de [A (ad) ] ) = 1 (k ad [A A (g) ][S] k de [A (ad) ] 2k r [A (ad) ] 2) = 1 ( k ad [A A (g) ][S] + k de [A (ad) ] + 2k r [A (ad) ] 2)

18 Example: surface recombination (continued) Note that [A(ad) ] + [S] = S 0 is a constant. d[a (g)] = 1 { kad [A dt V (g) ] ( S 0 [A (ad) ] ) + k de [A (ad) ] } d[a (ad) ] = 1 {k ad [A dt A (g) ] ( S 0 [A (ad) ] ) k de [A (ad) ] 2k r [A (ad) ] 2} Suppose that adsorption/desorption is in pseudo-equilibrium, i.e. that these processes are fast compared to reaction. Then k ad [A (g) ] ( S 0 [A (ad) ] ) k de [A (ad) ] [A (ad) ] k ad[a (g) ]S 0 k ad [A (g) ] + k de v = k r [A (ad) ] 2 k r kad 2 [A (g)] 2 S0 2 ( ) 2 = k r S0 2[A (g)] 2 ( kad [A (g) ] + k de [A(g) ] + K 1 ad ) 2

19 Example: surface recombination (continued) If we define vmax = k r S0 2, we get v v max[a (g) ] 2 ( [A(g) ] + K 1 ad Note that vmax depends on the square of S 0, which in turn is proportional to the total surface area. Because kad depends on the area, as discussed earlier, the apparent equilibrium constant K ad is also proportional to A. ) 2

20 Example: surface recombination (continued) The rate of reactant depletion is d[a (g) ] 1 dt V v max [A (g) ] 2 ( [A(g) ] + K 1 ad Suppose we carry out the reaction spherical flasks of different radii, r: ) 2 d[a (g) ] dt 1 r 4 [A (g) ] 2 r 3 {[A (g) ] + (K ad (r)) 1} 2 = r [A (g) ] 2 {[A (g) ] + (K ad (r)) 1} 2

21 Example: surface recombination (continued) d[a (g) ] dt [A (g) ] 2 r {[A (g) ] + (K ad (r)) 1} 2 At higher reactant pressures where K 1 ad is less significant, the rate would tend to increase with increasing flask size. At very low pressures, the rate reduces to d[a (g) ] dt [A (g) ] 2 r (K ad (r)) 2 r [A (g)] 2 r 4 = r 5 [A (g) ] 2 and the rate increases even more rapidly with area. This leads to inconsistent rate constants depending on the glassware used, and is typical of reactions with a key step occurring on the flask surface.