Reactions on surfaces

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.

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 $$C_{AS}$$ is the concentration of occupied sites, $$\theta$$ 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 the surface area, the more sites and the faster the reaction. This is the reason why heterogeneous catalysts are usually chosen 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 $$r=-\frac {dC_A}{dt}= \frac {k_1 k_2 C_A C_S}{k_1 C_A + k_{-1}+k_2}$$. Please notice that, with $$K_1=\frac{k_1}{k_{-1}}$$, the formula was divided by $$k_{-1}$$.

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:

$$k_2 >> \ k_1C_A, k_{-1}$$, 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: Adsorption/Desorption

$$k_2 << \ k_1C_A, k_{-1}$$ 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:
 * Limiting Step: Reaction
 * 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.

Langmuir-Hinshelwood mechanism
This mechanism proposes that both molecules adsorb and the adsorbed molecules undergo 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: $$r=k \theta_A \theta_B C_S^2 $$

Proceeding as before we get $$\theta_A=\frac{k_1C_A\theta_E}{k_{-1}+kC_S\theta_B}$$, where $$\theta_E$$ is the fraction of empty sites, so $$\theta_A+\theta_B+\theta_E=1$$. 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 $$\theta_A=K_1C_A\theta_E$$, with $$K_i=k_i/k_{-1}$$, which is nothing but Langmuir isotherm for two adsorbed gases, with adsorption constants $$K_1$$ and $$K_2$$. Calculating $$\theta_E$$ from $$\theta_A$$ and $$\theta_B$$ we finally get
 * $$r=k C_S^2 \frac{K_1K_2C_AC_B}{(1+K_1C_A+K_2C_B)^2}$$.

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:

That means that $$1 >> K_1C_A, K_2C_B$$, so $$r=C_S^2 K_1K_2C_AC_B$$. The order is one respect to both the reactants
 * Both molecules have low adsorption

In this case $$K_1C_A, 1>>K_2C_B$$, 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:
 * One molecule has very low adsorption
 * 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 of the reactants has very high adsorption and the other one doesn't adsorb strongly.
 * One molecule has very high adsorption

$$K_1C_A >> 1, K_2C_B$$, 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 :
 * 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.
 * CO + OH → CO2 + H+ + e- on a platinum catalyst.

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 $$r = k C_S \theta_A C_A C_B$$. 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, $$r=C_S K_1K_2C_AC_B$$, and the order is one with respect to A.


 * At high concentrations of A, $$r=C_S K_2C_B$$, and the order is zero with respect to A.

The following reactions follow a Eley-Rideal mechanism :
 * 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