### Mechanism of Catalytic Reaction

Develop the overall rate equation for the catalytic reaction A + B → C. The effect of diffusive mass transfer may be neglected. The following steps may be considered:

1. Adsorption of A and B

For the cases

1. The product C is not adsorbed
2. The product C is adsorbed

Assume step (b) is the rate controlling step.

Calculations:

(i) Product C is not adsorbed:

A + s → A.s

B + s → B.s

Where s is the vacant site.

Rate of adsorption of A = raA = kaA CA θV

Rate of desorption of A = rdA = kdA θA

At equilibrium raA = rdA

Therefore,

kaA CAθV = kdAθA

θA = KA CAθV → 1

where KA = kaA/kdA

Similarly, for B

θB = KB CBθV → 2

θV = fraction of unoccupied sites = 1 - θA - θB

substituting for θA and θB from equn.1 and 2,

θV = 1 - KA CAθV - KB CBθV

i.e.,

θV = 1 / (1 + KA CA + KB CB)

Surface reaction:

A.s + B.s → C + 2s

-rA = k θAθB → 3

i.e., the overall rate equation is,

-rA = k KAKB CACB / (1 + KA CA + KB CB)2

A + s → A.s

B + s → B.s

A.s + B.s → C.s + s

Desorption of C:

C.s → C + s

For the above steps and some similarity to the part (i)

θA = KA CAθV

θB = KB CBθV

-rA = k θAθB → 4

and for the desorption of C:

rdC = kdC θC

raC = kaC CCθV

For equilibrium desorption of C,

rdC = raC

Therefore,

kdC θC = kaC CCθV

θC = KC CCθV

where KC = kaC/kdC

Here, θV = 1 - θA - θB - θC

Therefore,

θV = 1 / (1 + KACA + KBCB + KCCC )

and θA = KACA / (1 + KACA + KBCB + KCCC )

θB = KBCB / (1 + KACA + KBCB + KCCC )

substituting for θA and θB in equn.4,

-rA = k KAKB CACB / (1 + KA CA + KB CB + KC CC)2

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