M. Li et al. / Journal of Catalysis 235 (2005) 201–208
207
which effectively transforms a 2+ charge carrier into two
mobile 1+ charged carriers. The IR results in this paper
demonstrate that the [NO]+ ions formed in this process on
BaNa/Y exhibit a fairly strong IR band, but no such band
is detectable with BaO/γ -Al2O3 or γ -Al2O3. Because the
[NO]+ ion is known to form nitrous acid (HONO) on reac-
tion with water, its presence provides a path for the forma-
tion of ammonia nitrite. This path may be less important in
the presence of a high partial pressure of ammonia, because
reduction of nitrate ions to nitrite ions provides another ef-
ficient route toward ammonia nitrite. However, for catalytic
reactions where ammonia must be formed by NOx reacting
with an organic molecule, nitrite formation through hydra-
tion of the [NO]+ ion is likely to become dominant, because
the concentration of ammonia will be low in the steady state.
In terms of point 3, the third characteristic difference
between zeolite and nonzeolite supports is the heat of ph-
ysisorption in the pores of the support. Data obtained by
calorimetry or derived from isosteric equilibria have been
reported by various authors, including Breck [42] and Stach
et al. [43]. Using silicalite analogs of MFI and FAU lattices,
Eder and Lercher [44] showed that the heat of adsorption
of n-hexane depends strongly on the pore diameter, being
>70 kJ/mol on MFI but only 45 kJ/mol on FAU. When
discussing the consequences of adsorption on the effective-
ness of SCR catalysts, one must consider that actual de-NOx
processes always occur in the presence of a partial pressure
of water exceeding that of the organic reductant or ammo-
nia. Adsorption of the reductant thus must compete with
adsorption of water on the same sites. Work focusing on
this competitive adsorption has been published by Hunger
et al. [45], who used temperature-programmed desorption
to determine the activation energy, Edes, which will be very
similar to the enthalpy of adsorption. These authors showed
that methanol displaces adsorbed water from the strongest
adsorption sites of Na/MFI. The Edes of methanol is 75–
100 kJ/mol. This finding explains why oxygenates such as
acetaldehyde or ethanol can achieve NOx reduction in the
presence of water vapor at a lower temperature than alka-
nes, which also require that a transition metal oxo-complex
be partially oxidized.
We conclude that enhanced adsorption of the reductant and
formation of NO+ ions are likely causes of the superior ac-
tivity of zeolite-based de-NOx catalysts. Because adsorption
of the reducing agent occurs in competition with water va-
por, one can understand that for nonpolar hydrocarbons, the
“hydrophobic” zeolites [46] with low Al/Si ratio are op-
timum, but with polar molecules, such as acetaldehyde or
ethanol, high N2 yields are obtained at lower temperatures
over “hydrophilic” zeolites with high Al/Si ratios.
Acknowledgments
This research was supported by the Chemical Sciences,
Geosciences and Biosciences Division, Office of Basic En-
ergy Sciences, Office of Science, US Department of Energy
(Grant No. DE-FG02-03ER15457). The authors thank the
donors of the American Chemical Society Petroleum Re-
search Fund for their partial support of this work and Pro-
fessor R. Snurr for his helpful suggestions.
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The concept of competitive adsorption of reductant and
water is illustrated by the well-known negative effect of wa-
ter vapor on the NOx reduction rate over such catalysts as
Cu/MFI and Co/MFI. It is, therefore, quite remarkable that
water vapor has a much less inhibiting effect when an oxy-
genate is used as the reductant. In the case of NOx reduction
with acetaldehyde over BaNa/Y, our data show higher NOx
conversion in the presence of water than in its absence [20].
This remarkable result has been rationalized by considering
that the presence of water will shift the equilibrium for aldol
condensation (7) to the left, thus minimizing the formation
of crotonaldehyde, which is a potential precursor of coke that
deactivates catalyst sites:
2CH3CHO = CH3–CH=CH–CHO + H2O.
(7)