Foundations of Chemical Kinetics Lecture 32: Heterogeneous kinetics: Gases and surfaces Marc R. Roussel Department of Chemistry and Biochemistry
Gas-surface reactions
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
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 c δ A molecules of the gas. ◮ Let the direction perpendicular to the surface be z . ◮ The mean speed along the z axis is v z . ◮ 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 = c δ A / 2.
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/ 2 v z . ◮ The rate of collisions is therefore N / t = cAv z . (Units?) ◮ The probability that the molecule sticks to the surface depends on two factors: ◮ An intrinsic probability of adsorption per collision event, P ad ◮ 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
Gas-surface rate processes (continued) v ad = pAv z P ad (1 − θ ) / RT ◮ Defining k ad = Av z P ad / RT we get v ad = k ad p (1 − θ ) (Units?) ◮ Note that the rate constant is proportional to the surface area.
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
Langmuir adsorption isotherm (continued) p A ∴ θ = p A + K − 1 ad where K ad = k ad / k de This equation, and closely related variations, is called the Langmuir adsorption isotherm. (Units of K ad ?)
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 ρ S 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 θρ S A = pV ad / RT , or pV ad θ = ρ S ART
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 ∞ p A V ad = p A + K − 1 ad
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 )
Example: Adsorption of CO on charcoal ◮ The following data were obtained for the adsorption of CO on charcoal at 273 K: p CO / torr 100 200 300 400 500 600 700 V ad / cm 3 10.2 18.6 25.5 31.4 36.9 41.6 46.1 ◮ The linearized graph is 50 40 30 V ad /cm 3 20 10 0 0.06 0.07 0.08 0.09 0.1 0.11 -1 /cm 3 torr -1 V ad p A Slope = − 979 ± 27 torr, intercept = 109 . 4 ± 2 . 2 cm 3
Example: Adsorption of CO on charcoal (continued) ◮ From the intercept, we get V ∞ = 109 . 4 ± 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 = 0 . 421 ◮ From the slope, we get K ad = (979 ± 27 torr) − 1 = (1 . 021 ± 0 . 028) × 10 − 3 torr − 1 .
Example: Adsorption of CO on charcoal (continued) ◮ Take a second look at the graph: 50 40 30 V ad /cm 3 20 10 0 0.06 0.07 0.08 0.09 0.1 0.11 -1 /cm 3 torr -1 V ad p A ◮ 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.
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.
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) ] v r = k r [A (ad) ] 2 A (ad) + A (ad) → A 2(g) + 2S 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?
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) ] v r = k r [A (ad) ] 2 A (ad) + A (ad) → A 2(g) + 2S ◮ Rate equations: d [A (g) ] = 1 � � − k ad [A (g) ][S] + k de [A (ad) ] dt V d [A (ad) ] = 1 � k ad [A (g) ][S] − k de [A (ad) ] − 2 k r [A (ad) ] 2 � dt A d [S] = 1 � − k ad [A (g) ][S] + k de [A (ad) ] + 2 k r [A (ad) ] 2 � dt A
Example: surface recombination (continued) ◮ Note that [A (ad) ] + [S] = S 0 is a constant. ∴ d [A (g) ] = 1 � � � � − k ad [A (g) ] S 0 − [A (ad) ] + k de [A (ad) ] dt V d [A (ad) ] = 1 � − k de [A (ad) ] − 2 k r [A (ad) ] 2 � � � k ad [A (g) ] S 0 − [A (ad) ] dt A ◮ 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) ] k ad [A (g) ] S 0 ∴ [A (ad) ] ≈ k ad [A (g) ] + k de k r k 2 ad [A (g) ] 2 S 2 k r S 2 0 [A (g) ] 2 ∴ v = k r [A (ad) ] 2 ≈ 0 � 2 = � 2 [A (g) ] + K − 1 � k ad [A (g) ] + k de � ad
Example: surface recombination (continued) ◮ If we define v max = k r S 2 0 , we get v max [A (g) ] 2 v ≈ � 2 [A (g) ] + K − 1 � ad ◮ Note that v max depends on the square of S 0 , which in turn is proportional to the total surface area. ◮ Because k ad depends on the area, as discussed earlier, the apparent equilibrium constant K ad is also proportional to A .
Example: surface recombination (continued) ◮ The rate of reactant depletion is v max [A (g) ] 2 d [A (g) ] ≈ − 1 � 2 dt V [A (g) ] + K − 1 � ad ◮ Suppose we carry out the reaction spherical flasks of different radii, r : r 4 [A (g) ] 2 [A (g) ] 2 d [A (g) ] ∼ 1 [A (g) ] + ( K ad ( r )) − 1 � 2 = r r 3 [A (g) ] + ( K ad ( r )) − 1 � 2 dt � �
Example: surface recombination (continued) [A (g) ] 2 d [A (g) ] ∼ r [A (g) ] + ( K ad ( r )) − 1 � 2 dt � ◮ 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 ( K ad ( r )) − 2 ∼ r [A (g) ] 2 [A (g) ] 2 d [A (g) ] = r 5 [A (g) ] 2 ∼ r r − 4 dt 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.
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