foundations of chemical kinetics - lecture 32...
TRANSCRIPT
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
I Enthalpy-driven process
Surface coverage (θ): fraction of surface to which molecules areadsorbed
Physisorption: adsorption based on intermolecular forces only
Chemisorption: bond formation between an adsorbate and thesurface
I Typically forms much stronger gas-surfaceassociations than physisorption
Dissociative chemisorption: chemisorption with bond dissociationin the adsorbate
Gas-surface rate processes
I Consider the process of adsorption of a single species ofmolecule to a surface.
I Suppose that the gas phase is well stirred so that in any smallvolume, the number of molecules per unit volume (theconcentration, c) is the same.
I Consider a volume of thickness δ near a surface of area A.
I This volume contains cδA molecules of the gas.
I Let the direction perpendicular to the surface be z .
I The mean speed along the z axis is vz .
I Half the molecules would be moving toward the surface andthe other half away from it, so the number of molecules thatcan collide with the surface is N = cδA/2.
Gas-surface rate processes (continued)
I For molecules in this small volume, the mean distance to thesurface is δ/2.
I The mean time before a molecule that is moving toward thesurface impacts it is therefore t = δ/2vz .
I The rate of collisions is therefore N/t = cAvz . (Units?)I The probability that the molecule sticks to the surface
depends on two factors:I An intrinsic probability of adsorption per collision event, Pad
I The probability that the molecule meets an unoccupied site onthe surface, 1− θ
I The rate of adsorption is therefore given byvad = cAvzPad(1− θ).
I The concentration can be rewritten in terms of the gaspressure using the ideal gas law: c = n/V = p/RT , so
vad = pAvzPad(1− θ)/RT
Gas-surface rate processes (continued)
vad = pAvzPad(1− θ)/RT
I Definingkad = AvzPad/RT
we getvad = kadp(1− θ)
(Units?)
I Note that the rate constant is proportional to the surface area.
Langmuir adsorption isotherm
I Now consider a molecule present in a container at pressure pwhich can adsorb and desorb from a surface (usually of anadded solid material, but possibly also of the container itself):
A(g) → A(ad) vad = kadpA(1− θ)A(ad) → A(g) vde = kdeθ
I At equilibrium,
kadpA(1− θ) = kdeθ
∴ θ =kadpA
kadpA + kde
Langmuir adsorption isotherm (continued)
∴ θ =pA
pA + K−1ad
whereKad = kad/kde
This equation, and closely related variations, is called the Langmuiradsorption isotherm.
(Units of Kad?)
The Langmuir adsorption isotherm in practice
I In practice, we rarely measure the surface coverage.
I Rather, we measure the amount of gas adsorbed to thesurface as an equivalent volume at the experimentaltemperature and a constant measurement pressure.
I Suppose that ρS is the areal density of adsorption sites, andthat A is the total surface area. Then the total number ofsites (usually in mol) is ρSA. If the coverage is θ, then thenumber of moles of gas adsorbed is θρSA.
I Invoking the ideal gas law, we have θρSA = pVad/RT , or
θ =pVad
ρSART
The Langmuir adsorption isotherm in practice(continued)
I As the pressure of the adsorbate goes to infinity, θ → 1, so
V∞ = ρSART/p
andθ = V /V∞
I The Langmuir adsorption isotherm becomes
Vad =V∞pA
pA + K−1ad
The Langmuir adsorption isotherm in practiceGraphical methods
I For the purpose of extracting the constants, we rewrite
V−1ad = V−1∞ + (KadV∞pA)−1
I Double-reciprocal plots have terrible statistical and visualproperties.Ideally, we would fit the isotherm directly, but for graphicalpurposes, it is convenient to have a linearized form.To get one, multiply through by VadV∞ and rearrange:
V∞ = Vad + K−1ad (Vad/pA)
∴ Vad = V∞ − K−1ad (Vad/pA)
Example: Adsorption of CO on charcoal
I The following data were obtained for the adsorption of CO oncharcoal at 273 K:
pCO/torr 100 200 300 400 500 600 700
Vad/cm3 10.2 18.6 25.5 31.4 36.9 41.6 46.1
I The linearized graph is
0
10
20
30
40
50
0.06 0.07 0.08 0.09 0.1 0.11
Vad
/cm
3
Vad pA-1/cm3 torr-1
Slope = −979± 27 torr, intercept = 109.4± 2.2 cm3
Example: Adsorption of CO on charcoal(continued)
I From the intercept, we get V∞ = 109.4± 2.2 cm3.I 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 cm3/109.4 cm3 = 0.421
I From the slope, we getKad = (979± 27 torr)−1 = (1.021± 0.028)× 10−3 torr−1.
Example: Adsorption of CO on charcoal(continued)
I Take a second look at the graph:
0
10
20
30
40
50
0.06 0.07 0.08 0.09 0.1 0.11
Vad
/cm
3
Vad pA-1/cm3 torr-1
I Note the slight curvature.
I The derivation of the Langmuir isotherm assumes that sitesare independent.
I Adsorbate-adsorbate interactions alter the binding constantKad at higher coverages, i.e. we run into non-ideal behavior.
Surface reactions
I Once molecules have adsorbed onto a surface, they can diffuseand react.
I We often represent surface reactions much as we do reactionsin other media, but we must bear in mind that the symbolshave different meanings.For example, if we write A(ad) + B(ad) → product withv = k[A(ad)][B(ad)], we must keep in mind that [A] and [B] areareal concentrations, and that v and k will typically havedifferent units from those used in gas-phase kinetics.
Example: surface recombination
I Consider the following mechanism for surface recombination:
A(g) + S→ A(ad) vad = kad[A(g)][S]A(ad) → A(g) + S vde = kde[A(ad)]
A(ad) + A(ad) → A2(g) + 2S vr = kr [A(ad)]2
where S represents a free surface site.I Note the use of concentration symbols here rather than
pressures or coverages.I 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 ofmol m−2 for [A(ad)] and [S].Because of the different units of the “concentrations”, wewould 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) vad = kad[A(g)][S]A(ad) → A(g) + S vde = kde[A(ad)]
A(ad) + A(ad) → A2(g) + 2S vr = kr [A(ad)]2
I Rate equations:
d [A(g)]
dt=
1
V
(−kad[A(g)][S] + kde[A(ad)]
)d [A(ad)]
dt=
1
A
(kad[A(g)][S]− kde[A(ad)]− 2kr [A(ad)]
2)
d [S]
dt=
1
A
(−kad[A(g)][S] + kde[A(ad)] + 2kr [A(ad)]
2)
Example: surface recombination (continued)
I Note that [A(ad)] + [S] = S0 is a constant.
∴d [A(g)]
dt=
1
V
{−kad[A(g)]
(S0 − [A(ad)]
)+ kde[A(ad)]
}d [A(ad)]
dt=
1
A
{kad[A(g)]
(S0 − [A(ad)]
)− kde[A(ad)]− 2kr [A(ad)]
2}
I Suppose that adsorption/desorption is in pseudo-equilibrium,i.e. that these processes are fast compared to reaction. Then
kad[A(g)](S0 − [A(ad)]
)≈ kde[A(ad)]
∴ [A(ad)] ≈kad[A(g)]S0
kad[A(g)] + kde
∴ v = kr [A(ad)]2 ≈
krk2ad[A(g)]
2S20(
kad[A(g)] + kde)2 =
krS20 [A(g)]
2([A(g)] + K−1ad
)2
Example: surface recombination (continued)
I If we define vmax = krS20 , we get
v ≈vmax[A(g)]
2([A(g)] + K−1ad
)2I Note that vmax depends on the square of S0, which in turn is
proportional to the total surface area.
I Because kad depends on the area, as discussed earlier, theapparent equilibrium constant Kad is also proportional to A.
Example: surface recombination (continued)
I The rate of reactant depletion is
d [A(g)]
dt≈ − 1
V
vmax[A(g)]2(
[A(g)] + K−1ad
)2I Suppose we carry out the reaction spherical flasks of different
radii, r :
d [A(g)]
dt∼ 1
r3r4[A(g)]
2{[A(g)] + (Kad(r))−1
}2= r
[A(g)]2{
[A(g)] + (Kad(r))−1}2
Example: surface recombination (continued)
d [A(g)]
dt∼ r
[A(g)]2{
[A(g)] + (Kad(r))−1}2
I At higher reactant pressures where K−1ad is less significant, therate would tend to increase with increasing flask size.
I At very low pressures, the rate reduces to
d [A(g)]
dt∼ r
[A(g)]2
(Kad(r))−2∼ r
[A(g)]2
r−4= r5[A(g)]
2
and the rate increases even more rapidly with area.
I This leads to inconsistent rate constants depending on theglassware used, and is typical of reactions with a key stepoccurring on the flask surface.