Unformatted text preview: ensity of the mixture C is constant (e.g. a process that occurs at constant
T and P and the gas mixture is ideal), then we can readily integrate (11) to determine the concentration
profile of xA . But we do not know what 1 and 2 are. So the problem is illposed. Unless we can pin
down the fluxes A and B we can not find a solution to (11) that will allow us to predict the concentration of a given species in the gas layer! Heterogeneous Reaction
At the catalytic surface at z=∆, Axiom II gives us a constraint on the molar rates of production Ri s of the
species participating in the surface reaction. If we know the molecular weights of the species participating in the reaction we can solve 2 b at the surface Alternatively, we can use the fact for a single reaction schema (1) we can express the Ris in terms of the stoichiometric coefficients of the reaction. Thus
RAs ΝA RBs ΝB 1
(12) n Hence the molar rates of production Ri are constrained by
n RAs RBs (13) We will show shortly that as a consequence of the species jump condition, the molar fluxes of our two
species at the catalytic surface must satisfy
B n A at z ∆ And because the molar fluxes of the two species is constant throughout the gas layer, the constraint
given by (14) holds everywhere. (14) IntroReactionDiffusion.nb 3 Suppose the catalytic reaction is first order in cA :
k cA ∆ RAs (15) where k is the rate constant for the reaction. Then from the species jump condition at the solid/gas
interface we have
cA vA ΞΒΓ w RAs 0 (16) The subscripts Β and Γ refer to the solid and gas phases respectively. The velocity of the ΒΓ interface is
w, and Ξ ΒΓ
ez is unit normal directed from the Β phase into the Γ phase. The notation for the jump
· is
cA vA ΞΒΓ w cAΒ vAΒ w cAΓ vAΓ ΞΒΓ w In our system the ΒΓ interface is stationary so that w 0, Ξ ΒΓ that there is no surface diffusion of species A so that vA
cA vA RA Β (17) ez , and at the solid surface we assume 0. Hence the jump condition gives ∆ at z (18) Thus at the catalytic surface we have the following requirement
A cA vA RAs k cA ∆ (19) The units of k are [k]= m/s as the units of RAs are RAs mol m2 s .This analysis shows that we now have two additional r...
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This document was uploaded on 03/14/2014.
 Winter '14

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