Reactions on surfaces

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By reactions on surfaces it is understood reactions in which at least one of the steps of the reaction mechanism is the adsorption of one or more reactants. The mechanisms for these reactions, and the rate equations are of extreme importance for heterogeneous catalysis

1 Simple decomposition
2 Bimolecular reaction
- 2.1 Langmuir-Hinshelwood mechanism
- 2.2 Eley-Rideal mechanism
3 References

[edit] Simple decomposition

If a reaction occurs through these steps:

A + S ⇌ AS → Products

Where A is the reactant and S is an adsorption site on the surface. If the rate constants for the adsorption, desorption and reaction are k₁, k_-1 and k₂ then, the global reaction rate is: $r=-\frac {dC_A}{dt}=k_2 C_{AS}=k_2 \theta C_S$

where $C A S$ is the concentration of occupied sites, $θ$ is the surface coverage and $C S$ is the total number of sites (occupied or not).

$C S$ is highly related to the total surface area of the adsorbent, the bigger surface area, the more sites and the faster the reaction, this is the reason why heterogeneous catalysts are usually sought to have great surface areas (in the order of hundred m²/gram)

If we apply the steady state approximation to AS then

$\frac {dC_{AS}}{dt}= 0 = k_1 C_A C_S (1-\theta)- k_2 \theta C_S -k_{-1}\theta C_S$ so $\theta =\frac {k_1 C_A}{k_1 C_A + k_{-1}+k_2}$ and $r=-\frac {dC_A}{dt}= \frac {k_1 k_2 C_A C_S}{k_1 C_A + k_{-1}+k_2}$ .

The result is completely equivalent to the Michaelis-Menten kinetics. The rate equation is complex, and the reaction order is not clear, in experimental work, usually two extreme cases are looked for, in order to prove the mechanism. In them, the rate-determining step can be:

Limiting step: Adsorption/Desorption

$k_2 >> \ k_1C_A, k_{-1}$ , so $r \approx k_1 C_A C_S$ . The order respect to A is 1. Examples of this mechanism are N₂O on gold and HI on platinum

Limiting Step: Reaction

$k_2 << \ k_1C_A, k_{-1}$ so $\theta =\frac {k_1 C_A}{k_1 C_A + k_{-1}}$ which is just Langmuir isotherm and $r= \frac {K_1 k_2 C_A C_S}{K_1 C_A+1}$ . Depending on the concentration of the reactant the rate changes:

Low concentrations, then $r = K 1 k 2 C A C S$ , that is to say a first order reaction.
High concentration, then $r = k 2 C S$ . It is a zeroth order reaction.

[edit] Bimolecular reaction

[edit] Langmuir-Hinshelwood mechanism

This mechanism proposes that both molecules adsorb and the adsorbed molecules experiment a bimolecular reaction:

A + S ⇌ AS

B + S ⇌ BS

AS + BS → Products

The rate constants are now $k 1$ , $k - 1$ , $k 2$ , $k - 2$ and $k$ for adsorption of A, adsorption of B, and reaction. The rate law is: $r=k_2 \theta_A \theta_B C_S^2$

Proceeding as before we get $\theta_A=\frac{k_1C_A\theta_E}{k_{-1}+kC_S\theta_B}$ , where $θ E$ is the fraction of empty sites, so $θ A + θ B + θ E = 1$ . Let us assume now that the rate limiting step is the reaction of the adsorbed molecules, which is easily understood: the probability of two adsorbed molecules colliding is low. Then $θ A = K 1 C A θ E$ , which is nothing but Langmuir isotherm for two adsorbed gases, with adsorption constants $K 1$ and $K 2$ . Calculating $θ E$ from $θ A$ and $θ B$ we finally get

$r=C_S^2 \frac{K_1K_2C_AC_B}{(1+K_1C_A+K_2C_B)^2}$ .

The rate law is complex and there is no clear order respect to any of the reactants but we can consider different values of the constants, for which it is easy to measure integer orders:

Both molecules have low adsorption

That means that $1 > > K 1 C A, K 2 C B$ , so $r=C_S^2 K_1K_2C_AC_B$ . The order is one respect to both the reactants

One molecule has very low adsorption

In this case $K 1 C A,1 > > K 2 C B$ , so $r=C_S^2 \frac{K_1K_2C_AC_B}{(1+K_1C_A)^2}$ . The reaction order is 1 respect to B. There are two extreme possibilities now:

At low concentrations of A, $r=C_S^2 K_1K_2C_AC_B$ , and the order is one respect to A.
At high concentrations, $r=C_S^2 \frac{K_2C_B}{K_1C_A}$ . The order es minus one respect to A. The higher the concentration of A, the slower the reaction goes, in this case we say that A inhibits the reaction.

One molecule has very high adsorption

One of the reactants has very high adsorption and the other one doesn's adsorb much.

$K 1 C A > > 1, K 2 C B$ , so $r=C_S^2 \frac{K_2C_B}{K_1C_A}$ . The reaction order is 1 respect to B and -1 respect to A. Reactant A inhibits the reaction at all concentrations.

The following reactions follow a Langmuir-Hinshelwood mechanism [1]:

2 CO + O₂ → 2 CO₂ on a platinum catalyst.
CO + 2H₂ → CH₃OH on a ZnO catalyst.
C₂H₄ + H₂ → C₂H₆ on a copper catalyst.
N₂O + H₂ → N₂ + H₂O on a platinum catalyst.
C₂H₄ + ½ O₂ → CH₃CHO on a palladium catalyst.

[edit] Eley-Rideal mechanism

This mechanism proposes that only one of the molecules adsorbs and the other one reacts with it directly, without adsorbing:

A + S ⇌ AS

AS + B → Products

Constants are $k 1, k - 1$ and $k$ and rate equation is $r = k C S θ A C A C B$ . Applying steady state approximation to AS and proceeding as before (considering the reaction the limiting step once more) we get $r=C_S C_B\frac{K_1C_A}{K_1C_a+1}$ . The order is one respect to B. There are two possibilities, depending on the concetration of reactant A:

At low concentrations of A, $r = C S K 1 K 2 C A C B$ , and the order is one with respect to A.

At high concentrations of A, $r = C S K 2 C B$ , and the order is zero with respect to A.

The following reactions follow a Langmuir-Hinshelwood mechanism [2]:

C₂H₄ + ½ O₂ (adsorbed) → H₂COCH₂ The dissociative adsorption of oxygen is also possible, which leads to secondary products carbon dioxide and water.
CO₂ + H₂(ads.) → H₂O + CO
2NH₃ + 1½ O₂ (ads.) → N₂ + 3H₂O on a platinum catalyst
C₂H₂ + H₂ (ads.) → C₂H₄ on nickel or iron catalysts