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.