Unimolecular and Bimolecular Surface reactions

Simple Langmuir isothermThe simplest situation in which the gas atoms or molecules occupy single sites on the surface and are not dissociated can be represented as follows:A A + S S

Langmuir kinetic derivation of isotherm:Let be the fraction of surface that is covered. then 1- be the fraction of surface that is bare. The rate of adsorption may be written as

ka [A ](1

The rate of desorption is

va k a

At equlibrium, the rates are equal and therefore,

ka [A](1 k a=>

ka [A] (1 ) k a

=>

(1 )

K [A]Equlibrium constant;

=> K[A] K[A] => =>

ka K k a

(1 K[A]) K[A]K [A] 1 K [A]

Unimolecular surface reactionsSurface reactions involving single adsorbed molecules and therefore terms as unimolecular and are treated by langmuir adsorption isotherm as follows: The rate is proportional to the fraction of surface that is covered and is thus

kK[ A ] ..............(1) v k 1 K [A]The dependence of rate on [A] is shown in figure below:

vkZero order kinetics)

kK[A] v 1 K [A]

v

v kK[A] (First order kinetics)[A]Figure 7.4 (a) [Laidler] Variation of rate with concentration for simple unimolecular process/ At sufficiently high concentrations of A, the rate is independent of the

concentrations, which means the kinetics are zero order.At low concentrations, when K[A] 1, the kinetics are first order.

InhibitionSometimes a substance other than the reactant is adsorbed on the surface, with the result that the effective surface area and therefore the rate are reduced.

Suppose, a substance A is undergoing a unimolecular reaction on a surface and that a nonreacting subtance, I, known as a inhibitor or a poison, is also adsorbed. As per competative adsorption, the fraction of the surface covered by A is

K A [A] (1 K A [A] K I [I] )Where KA and KI are the adsorption constants for A and I.

Original equation

K A [A] A (1 K A [A ] K B [ B] )

The rate of equation, equals to k , is thus

kKA [A ] v (1 K A [A ] K I [I] )

..............(2)

In the absence of inhibitor this equation reduces to

kKA [ A ] ..............(3) Simple langmuir equation eq. (1) v 1 K A [A]

Special case:When the surface is covered only sparsely by the reactant but is covered fairly fully by the inhibitor, KI[I] 1+ KA[A] The rate is then

kKA [ A ] v K I [ I]

..............(4)

Bimolecular surface reactionsThere are two distinctive mechasims for a surface reactions bettween two reactants A and B .

In the Langmuir-Hinshelwood mechanism, reaction occurs between A and B molecules when both are adsorbed on the surface.Alternatively, in the Langmur- Rideal mechansim, the reactions occurs between an adsorbed molecule and a molecule in the gas phase. 7.8.1: Reactions between two adsorbed molecules: Langmuir-Hinshelwood mechanism In a Langmuir-Hinshelwood mechanism the rate is proportional to the fraction of the molecules A and B that are adsorbed. These fractions are given by the following two equations (from competitive adsorption).

K A [A] ; A (1 K A [A ] K B [ B] )The rate is therefore

K B [ B] B (1 K A [ A ] K B [ B] )

v k A BkKA K B [A ][ B] ..............(1) 2 (1 K A [A ] K B [B])If [B] is held constant and [A] is varied, the rate varies in accordance with the following figure.

v

K A K B [ A ][ B] (1 K A [ A ] K B [ B] )

v

v [ A][A] (with [B] held constant)

v

1 [A]

Figure 7.4 (b) [Laidler] Variation of rate with concentration for bimolecu;ar process occurring by a Langmuir-Hinshelwood mechasins.

The rate first increases then passes through a maximum and finally decreases. Why?

The explanation of the fall off in the rate at high concentrations is that one reactant replaces the other as its concentration is increased.

The maximum rate corresponds to the existence of the maximum number of neighboring A-B pairs on the surface.

Two special cases of interest:1. Sparsely covered surface:If [A] and [B] are both sufficiently low that KA[A] and KB[B] may be neglected in comparison with unity, the rate equation becomes

kKA K B [A][ B] v (1 K A [A ] K B [B]) 2

Since KA[A] & KB[B] 1

v kKA K B [A][ B]

..............(2)

The reaction is therefore Second order, being first order in both a and B.

2. One reactant very weakly adsorbed:If reactant A is very weakly adsorbed, KA[A] in the denominator of eq.1 may be neglected, and the rate equation becomes

kKA K B [A][ B] v (1 K A [A ] K B [B]) 2

Since KA[A] 1

=>

kKA K B [ A ][ B] ..............(3) v 2 (1 K B [ B])

the rate still passes through a maxiumum as [B] increases, but as long as the condition Since KA[A] 1+ KB[B] remains satisfied the rate is proportional to [A].

If reactant B is adsorbed sufficiently strongly:then KB[B] 1, and the rate equation (3) becomes

kKA K B [ A ][ B] v (1 K B [ B]) 2=>

kKA K B [ A ][ B] v ( K B [ B]) 2 kKA [ A ] v K B [ B]..............(4)

=>

The rate is now inversely proportional to [B], and the order with respect to B is -1.