# Mechanism of Catalytic Reaction

Develop the overall rate equation for the catalytic reaction A + B à 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 à A.s

B + s à B.s

Where s is the vacant site.

Rate of adsorption of A = raA = kaA CA qV

Rate of desorption of A = rdA = kdA qA

At equilibrium raA = rdA

Therefore,

kaA CA qV = kdA qA

qA = KA CAqV à 1

where KA = kaA/kdA

Similarly, for B

qB = KB CBqV à 2

qV = fraction of unoccupied sites = 1 - qA - qB

substituting for qA and qB from equn.1 and 2,

qV = 1 - KA CAqV - KB CBqV

i.e.,

qV = 1 / (1 + KA CA + KB CB)

Surface reaction:

A.s + B.s à C + 2s

-rA = k qAqB à 3

i.e., the overall rate equation is,

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

A + s à A.s

B + s à B.s

A.s + B.s à C.s + s

Desorption of C:

C.s à C + s

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

qA = KA CAqV

qB = KB CBqV

-rA = k qAqB à 4

and for the desorption of C:

rdC = kdC qC

raC = kaC CCqV

For equilibrium desorption of C,

rdC = raC

Therefore,

kdC qC = kaC CCqV

qC = KC CCqV

where KC = kaC/kdC

Here, qV = 1 - qA - qB - qC

Therefore,

qV = 1 / (1 + KACA + KBCB + KCCC )

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

qB = KBCB / (1 + KACA + KBCB + KCCC )

substituting for qA and qB in equn.4,

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