Foundations of Chemical Kinetics Lecture 32: Heterogeneous - - PowerPoint PPT Presentation

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 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.

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/2vz.

◮ 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

vad = cAvzPad(1 − θ).

◮ The concentration can be rewritten in terms of the gas

pressure 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

◮ Defining

kad = AvzPad/RT we get vad = kadp(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) vad = kadpA(1 − θ) A(ad) → A(g) vde = kdeθ

◮ At equilibrium,

kadpA(1 − θ) = kdeθ ∴ θ = kadpA kadpA + kde

Langmuir adsorption isotherm (continued)

∴ θ = pA pA + K −1

ad

where Kad = kad/kde This equation, and closely related variations, is called the Langmuir adsorption isotherm. (Units of Kad?)

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 ρSA. If the coverage is θ, then the number of moles of gas adsorbed is θρSA.

◮ Invoking the ideal gas law, we have θρSA = pVad/RT, or

θ = pVad ρSART

The Langmuir adsorption isotherm in practice

(continued)

◮ As the pressure of the adsorbate goes to infinity, θ → 1, so

V∞ = ρSART/p and θ = V /V∞

◮ The Langmuir adsorption isotherm becomes

Vad = V∞pA pA + 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 ∞ + (KadV∞pA)−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 VadV∞ and rearrange: V∞ = Vad + K −1

ad (Vad/pA)

∴ Vad = V∞ − K −1

ad (Vad/pA)

Example: Adsorption of CO on charcoal

◮ The following data were obtained for the adsorption of CO on

charcoal 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

◮ The linearized graph is

10 20 30 40 50 0.06 0.07 0.08 0.09 0.1 0.11 Vad /cm3 Vad pA

• 1/cm3 torr-1

Slope = −979 ± 27 torr, intercept = 109.4 ± 2.2 cm3

Example: Adsorption of CO on charcoal

(continued)

◮ From the intercept, we get V∞ = 109.4 ± 2.2 cm3. ◮ 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

◮ From the slope, we get

Kad = (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:

10 20 30 40 50 0.06 0.07 0.08 0.09 0.1 0.11 Vad /cm3 Vad pA

• 1/cm3 torr-1

◮ Note the slight curvature. ◮ The derivation of the Langmuir isotherm assumes that sites

are independent.

◮ Adsorbate-adsorbate interactions alter the binding constant

Kad 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) 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.

◮ 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) 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

◮ 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)

◮ 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

◮ 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 ≈ krk2

ad[A(g)]2S2

• kad[A(g)] + kde

2 = krS2

0[A(g)]2

• [A(g)] + K −1

ad

2

Example: surface recombination (continued)

◮ If we define vmax = krS2

0, we get

v ≈ vmax[A(g)]2

• [A(g)] + K −1

ad

2

◮ Note that vmax depends on the square of S0, which in turn is

proportional to the total surface area.

◮ Because kad depends on the area, as discussed earlier, the

apparent equilibrium constant Kad is also proportional to A.

Example: surface recombination (continued)

◮ The rate of reactant depletion is

d[A(g)] dt ≈ − 1 V vmax[A(g)]2

• [A(g)] + K −1

ad

2

◮ Suppose we carry out the reaction spherical flasks of different

radii, r: d[A(g)] dt ∼ 1 r3 r4[A(g)]2

• [A(g)] + (Kad(r))−12 = r

[A(g)]2

• [A(g)] + (Kad(r))−12

Example: surface recombination (continued)

d[A(g)] dt ∼ r [A(g)]2

• [A(g)] + (Kad(r))−12

◮ 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 ∼ 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.

◮ This leads to inconsistent rate constants depending on the

glassware used, and is typical of reactions with a key step

• ccurring on the flask surface.