催化剂研究方法与表征外教课件.doc

Surface enrichmentThe composition of the surface of most materials is different from that of the bulk and this has large consequences, not only in catalysis, but in all areas where the surface plays an important role. The reason for the surface enrichment is that atoms at the surface have a different surrounding than in the bulk and therefore thermodynamics favours some atoms to be at the surface and others to be in the bulk. We will study this with the example of alloys, but the principles are also valid for other materials, like metal oxides and sulfides. Let us take a look at the surface of a binary alloy AB. We will assume that an atom in the bulk has p direct neighbours in the plane parallel to the surface and m direct neighbours perpendicular to that plane. For an fcc metal p +m = 12 and for a bcc metal p +m = 8. An atom at the surface will have the same number (p) of neighbours in the surface plane, but only m/2 neighbours perpendicular to the plane, because the neighbours above the plane are missing. We will assume that the broken-bond model applies, which assumes that only bonds between direct neighbouring atoms contribute to the energy of the alloy, and that surface enrichment only occurs in the top surface layer. All other layers have the bulk composition, also the subsurface layer. When xs is the concentration of atom A at the surface and xb its concentration in the bulk, then the energy EAs of atom A at the surface isEAs = pxsEAA + (1- xs)EAB + m/2xbEAA + (1-xb)EAB EAs = (pxs +mxb/2) EAA + p(1- xs) + m(1-xb)/2EAB where EAA is the binding energy between two neighbouring A atoms and EAB is the binding energy between two neighbouring A and B atoms.The energy of an atom A in the bulk isEAb = (p + m)xbEAA + (1-xb)EABSimilarly, for atom B at the surface and in the bulk we get the energies EBs and Ebb, respectively:EBs = (pxs +mxb/2) EAB + p(1- xs) + m(1-xb)/2EBB and EBb = (p + m)xbEAB + (1-xb)EBBThe energy difference DE between atom A at the surface and atom B in the bulk on the one hand and atom B at the surface and atom A in the bulk on the other hand, isDE = (EAs + EBb) (EBs + EAb) = (EAs - EAb) (EBs - EBb) = = p(xs xb) mxb/2EAA 2p(xs xb) mxb + m/2EAB + p(xs xb) - mxb/2 + m/2EBBThe binding energies EAA and EBB can be experimentally determined from the heats of sublimation of the pure metals A and B, because in the broken bond model the molar heat of sublimation of metal A is equal to (p + m)EAA/2. The binding energy EAB cannot directly be determined in the same way. Normally, EAB is in between the energies EAA and EBB and we may introduce the parameter a which indicates how much EAB deviates from the arithmetic mean of EAA and EBB. EAB = (EAA + EBB)/2 + aSubstitution in DE givesDE = - a2p(xs xb) mxb + m/2 + m(EBB EAA)/4To determine the thermodynamic equilibrium composition of the alloy surface, we need to determine the surface composition at which the free energy has a minimum as a function of xs. We thus must find xs at whichd(DG)/dxs = d(DH TDS)/dxs = 0 or d(DH)/dxs = Td(DS)/dxsd(DH)/dxs = dNs IntDE dxs/dxs = DE.NsThe remaining task is to calculate d(DS)/dxs and therefore we must first calculate the entropy of a mixture. The entropy of mixing is known from statistical thermodynamics to beS0 = - kN x ln x + (1 - x) ln (1 - x)The entropy of mixing of two elements to a surface of composition xs and a bulk of composition xb thus isS1 = - kNs xs ln xs + (1 - xs) ln (1 xs) - kNbxb ln xb + (1 - xb) ln (1 - xb)Because of the surface enrichment of A in the surface, the bulk concentration xb of A is lower than xs and the overall, average concentration x of A is in between:x = xs l = xb + L, with NAb L= NAs lReplacing xb by x- L one obtainsxb ln xb = (x- L) ln (x- L) = (x- L) ln x(1 - L/x) = (x- L) ln x + ln (1 - L/x) and if L/x 1, xb ln xb = x ln x - L ln x - Land similarly (1 - xb) ln (1 - xb) = (1 - x) ln (1 - x) + L ln (1 - x) + LIn the equation for the entropy S1 this givesS1 = - kNsxs ln xs + (1 - xs) ln (1 - xs) - kNbx ln x - L ln x - L + (1-x) ln (1 - x) + L ln (1 - x) + L and with S0 = - k(Ns + Nb)x ln x + (1 - x) ln (1 - x) we getDS = S1- S0 = kNs x ln x + (1 x) ln (1 x) - xs ln xs - (1 - xs) ln (1 - xs) + kNbL ln x/(1-x) = kNs x ln x + (1 x) ln (1 x) - xs ln xs - (1 - xs) ln (1 - xs) + l ln x/(1-x)Because dl/dxs = d(xs x)/dxs = 1,d(DS)/dxs = kNs - ln xs - 1 + ln (1 - xs) + 1 + ln x/(1-x) = kNs ln x/(1-x).(1-xs)/xsFinally, from d(DH)/dxs = Td(DS)/dxs one obtainsNs- a2p(xs xb) mxb + m/2 + m(EBB EAA)/4 = kT.Ns ln x/(1-x).(1-xs)/xsor m(EBB EAA)/4 + a(2p + m)xb 2pxs - m/2 = - kT ln cwhere c is the surface enrichment factor x/(1-x).(1-xs)/xs. When a (EAA + EBB)/2, which is certainly the case for ideal solutions with EAB = (EAA + EBB)/2 and no heat of mixing, then c = exp - m(EBB EAA)/4kT. When B has a higher binding energy than A (B has the higher sublimation energy), EBB EAA 1. The surface is enriched with component A, the component with the lowest sublimation energy. The driving force for the surface enrichment is the fact that the component with the highest binding energy (in this case B) wants to have as many neighbours as possible. That can better be reached in the bulk than at the surface. As a consequence, the other element A loses out and must go to the surface. One can also say that it costs less energy to place the element with the lowest binding energy at the surface. Figure 1 shows that xAs can be much larger than xAb for relatively small differences in the sublimation energies of metals A and B. Surface enrichment thus can be strong. The surface enrichment depends also on m, the number of neighbouring atoms perpendicular to the surface. The larger m, the stronger the effect of the missing atoms above the surface plane will be felt and the stronger the enrichment will be. Rough surfaces (kinks and steps, see Chapter ) have a larger surface enrichment than regular planes. The atoms of the element with the lower sublimation energy are forced to take up the energetically unfavourable kink and step positions.Figure 1. Surface composition xAs as a function of bulk composition xAbThe surface structure of endothermic or exothermic alloys is more difficult to predict than of ideal solutions. Endothermic alloys have a negative heat of formation and AA and BB bonds are stronger than two AB bonds. That is why endothermic alloys should show phase segregation into two phases at low temperature, when the entropy does no longer stabilize one phase. The compositions and the proportions of these two phases are determined by the phase diagram (cf. Figure 2). Small metal particles, whose average composition is in the multi-phase region, will consist of a kernel and an outer shell. The outer shell will have the lower surface energy and will in addition be surface enriched (cherry model). For instance, in case of PtAu (Figure 2) the outer shell contains 18% Pt and 82% Au, while the kernel contains 97% Pt and 3% Au. According to the lever rule, a PtAu alloy with 50-50 average composition will have a (50 -18) : (97 50) = 1 : 1.5 mass ratio of kernel and outer shell. On top of that, the surface of the outer shell may be even richer in Au than 82%, because of the lower sublimation energy of Au than of Pt.Figure 2. Phase diagram of Pt-Au.AB bonds of exothermic alloys are stronger than half the sum of AA and BB bonds. Some combinations of metals from the left and the right side of the periodic table belong to this category (PdZr, Ni3Al) and also PtSn and Pt3Sn. This has been ascribed to Lewis base (Pt, Pd, Ni)-Lewis acid (Al, Zr, Sn) type interactions. Also in exothermic alloys the surface will be enriched in the component with the lowest sublimation energy. Because of the strong AB bonds this enrichment will not occur at the cost of the whole metal particle, but mainly at the cost of the second layer, the subsurface layer.Not only the difference in surface energy, but any factor that influences the free energy may cause surface enrichment. The size of the atoms will also play a role, which explains why the surface of M(3d)M(4d) and M(3d)M(5d) alloys is often enriched in the 4d or 5d metal, because the diameter of a 4d or 5d metal atom is larger than that of a 3d metal atom. For instance, the surface of Au-Cu particles is enriched in gold, although copper has a lower sublimation energy than gold. The reason is that a smaller number of relatively large gold atoms can cover the surface equally well as a larger number of small copper atoms and that, because of the smaller number of atoms, the gold enriched surface has a lower surface energy. A gas phase above the surface adds another thermodynamical factor that can influence the surface composition. This explains why NiCu surfaces in vacuum are enriched in copper and that this enrichment becomes lower when CO gas is brought in contact with the NiCu alloy, because CO binds stronger with Ni atoms than with Cu atoms and pulls the Ni atoms to the surface. For alloys in contact with air, the surface should be enriched with the less noble component. Finally, temperature plays an important role as well, because the entropy factor is proportional to the temperature. The higher the temperature, the more mixing becomes important (also of surface and bulk) and the lower the surface enrichment will be. The driving force behind surface enrichment is a difference in energy and surface enrichment thus increases at lower temperature. The surface enrichment must come from metal atom diffusion from bulk to surface for the one component and in the reverse direction for the other component. At low temperature this diffusion may become very slow and the actual surface composition may not be in thermodynamic equilibrium. On the other hand, this freezing-in effect will be less important for small metal catalyst particles, in which the diffusion path is not very long.Metal binding In the foregoing section we saw that the difference in binding energies in metals A and B plays a role in determining which element will prefer to be at the surface of the AB alloy. Therefore it is important to be able to predict the binding energies and thus the heats of sublimation of metals. This we can do on the basis of the band model of bonding in metals. The band model is based on molecular orbital theory and the principles are already apparent in the explanation of the bonding of two potassium atoms to a dipotassium molecule and the non bonding of two calcium atoms to a dicalcium molecule. As Figure 3 shows, the two atomic 4s orbitals of the two K atoms can be combined to a bonding and an antibonding molecular orbital with energies - b and b relative to the energy of the separate K 4s orbitals, respectively, where b is a measure of the interaction energy between the 4s orbitals on neighbouring atoms. Whereas in the K atoms each of the two electrons is in an atomic 4s orbital, in the K2 molecule the two electrons can both be put into the level corresponding to the bonding molecular orbital and thus an energy gain of 2b can be realized. According to the Pauli principle, the two electrons in the same orbital must have opposite spin and thus the molecule has no unpaired electrons. The same energy diagram (Figure 3) also holds for the combination of two Ca atoms to a Ca2 molecule. The difference is, however, that each Ca atom donates two 4s electrons and that we can put two of these in the bonding orbital, which gives an energy gain of 2b, but that we have to put the other two electrons in the antibonding molecular orbital, leading to an energy loss of 2b. In total, there thus is no energy gain and the Ca2 molecule is not stable.Figure 3. Energy diagram for the bonding and antibonding molecular orbitals of K2.If one takes three K atoms, one surrounded by two others, one can create three molecular orbitals (MOs) from the three atomic K 4s orbitals: one bonding MO at energy - sqr2.b, one nonbonding MO at energy 0, and one antibonding MO at energy sqr2.b. The three electrons fill the bonding MO completely and the nonbonding MO half and in total there is an energy gain 2sqr2.b relative to the situation with three independent K atoms. Also for more peripheral K atoms bonded to one central K atom only bonding and nonbonding MOs are filled with electrons and thus the Kn molecules are stable. If also the peripheral K atoms are surrounded by K atoms the situation becomes more complex, but it can be shown that the total number of energy levels is equal to the number of participating atomic orbitals (that is equal to the number of K atoms) and that the lowest bonding level cannot go deeper in energy than - nb, where n is the number of direct neighbouring atoms. Likewise, the highest antibonding energy level cannot go higher in energy than nb. That means that if N atoms are all connected to n neighbours there will be N levels between - nb and nb. For a macroscopic piece of K atoms this means that N is of the size of the Avogadro number (6.1023) and that an extremely large number of energy levels will be present between - nb and nb. For fcc metals n = 12 and with b = 2 eV the average energy difference between adjacent energy levels will be in the order of 5.10-22 eV. The distinct energy levels of small molecules become quasi continuous energy bands in solids (Figure 4).Figure 4.Energy diagram of the 4s orbitals of K atoms combining into a 4s energy band in bulk K metal. The energy difference between adjacent energy levels is far smaller than room temperature energy (0.025 eV) and, thus, electrons from the highest bonding orbitals can be excited to empty orbitals without any addition of energy or temperature. The electrons can freely go from one orbital to the next and, since different orbitals have different electron densities on the constituting atoms, this means that electrons can move freely from one atom to the next. In other words, metals conduct electricity without energy cost. The condition for metal behaviour thus is that many atoms are interconnected and that each of the constituting atoms contains unpaired electrons. If the atoms did not contain 4s electrons, then one can create the valence band as described above, but there would be no electrons in the band and no electron could flow from one atom to the next. Also, if the constituting atoms have the maximum number of electrons in their atomic orbital (4s2 for Ca), then the resulting valence band will be completely filled. There would be no empty orbitals close by in energy available for the electrons to be excited to and, again, no electron flow could occur and the material would not behave like a metal but like an isolator.In contrast to the K atom, which has one unpaired electron in a 4s orbital, transition metal atoms have several unpaired electrons in their valence orbitals. For instance, the Ti atom has 4 electrons in the 3d and 4s orbitals and the Ni atom has 10 electrons in the 3d and 4s orbitals. While the 4s orbital can only contain 2 electrons, the maximum number of electrons that can go into the 3d orbital is 10, because there actually are five d orbitals, each of which can contain two electrons (Pauli principle). The first 5 electrons can therefore occupy each of these 5 d orbitals with their spins parallel and are thus unpaired, while the last 5 electrons pair with the first 5 electrons. For Ti this means that the maximum number of unpaired electrons is four, either three in the 3d orbitals and one in the 4s orbital (d3s1) or all four in four 3d obitals (d4). For Ni the maximum number of unpaired electrons is two, independent of the distribution over the d or s orbitals. For the configuration d8s2 there are two unpaired d electrons and for the configuration d9s1 there is one unpaired d electron and one unpaired s electron. Both the d and the s orbitals on the metal atoms can combine to valence bands. Since s orbitals are larger in size than d orbitals, they more strongly interact with each other and b4s b3d. Therefore, the s band is broader (total width 2nb) than the d band. And although the center of the s band lies somewhat higher than that of the d band, the bands overlap in energy (cf. Figure 5). This is the reason that even Ca is a metal, although a single Ca atom contains two electrons in the 4s orbital and contains no 3d electrons. Above we predicted that in solid Ca the 4s band would be completely filled and would have no empty orbitals available for electron transport and that the 3d band would be empty and also would not allow electrons to flow. In the solid state, however, the d and s bands overlap and a fraction of the paired s electrons will go to the d band, so that the s band becomes less than completely filled and the d band obtains free electrons. Now both bands can contribute to electron transport, which explains that Ca behaves as a metal. As we have seen, when there are no electrons in an atomic orbital, there are no electrons in the valence band of the constituting atomic orbitals either and the binding energy between the atoms is zero. This is the case for elements on the left hand side of the Periodic Table. Elements on the right hand side have their atomic orbitals totally filled, and so will be the valence d and s bands of the corresponding solids. The binding energy between these atoms will be zero as well. Going from the left hand side of the Periodic Table to the middle, there will be more and more electrons in d and s orbitals of the atoms, which will fill bonding molecular orbitals in the valence band of the solid. The more electrons there are, the more bonding orbitals will be filled and the stronger the binding between the metal atoms will be. In the mid