Abstract
A thorough thermodynamic analysis of oxidation of tin nanoparticles was performed. Solid tin oxides SnO_{2}, Sn_{3}O_{4} and SnO were considered according to the bulk phase diagram and a number of experimental results on tin nanostructures oxidation were taken into account in the assessment. Two equilibrium models with different spatial configuration, namely two singlecomponent particles and core–shell model were explored. The surface energies for solid SnO and Sn_{3}O_{4} were obtained on the basis of DFT calculations while the interfacial energies at SnO_{2}(s)/Sn(l) and Sn_{3}O_{4}(s)/Sn(l) interfaces were assessed using a broken bond approximation. The opposite influence of nanosizing on stability of SnO_{2} and SnO/Sn_{3}O_{4} oxides is demonstrated. It is due to the surface contribution which is higher for SnO_{2}(s) than Sn(l) while lower for SnO(s) and Sn_{3}O_{4}(s) compared to Sn(l). This situation can explain some experimental findings during oxidation of Sn nanoparticles, namely an increased stability of SnO(s) and Sn_{3}O_{4}(s) with respect to both liquid tin and solid tin dioxide.
Introduction
Oxidation of metal nanostructures is one of size/shape dependent phenomena which are intensively studied now. A huge number of experimental works as well as theoretical studies on this subject have been published recently. Equilibrium thermodynamics provides very useful tools for studying various size/shape effects in the topdown approach.[1,2,3] Whereas the effect of particle size and surface energy on the melting point has been studied both by experimental methods and theoretical modeling for many systems, the phase transitions in partly open systems involving exchange of one or more components with surrounding atmosphere have been less explored. Within the family of metal–oxide systems, Navrotsky et al.[4,5,6] have calculated phase equilibria in the systems CoCoO_{x}O_{2}(g), FeFeO_{x}O_{2}(g) and MnMnO_{x}O_{2}(g) for bulk and nanoparticles forms and have shown substantial influence of nanosizing on stability of various oxide phases. The aim of the present paper is to carry out a similar analysis for the system SnSnO_{x}O_{2}(g).
Thermodynamic Description of the bulk SnO System
A thorough thermodynamic analysis of the SnO system including an assessment of thermodynamic data for tin oxides has been performed by Cahen et al.[7] It should be noted, however, that substantial differences between the optimized and experimental values of entropy at 298.15 K exist for SnO_{2} and Sn_{3}O_{4} oxides. The assessed value S_{m,298}(SnO_{2}) = 73.23 J K^{−1} mol^{−1} is 1.4times higher than the experimental value 51.82 J K^{−1} mol^{−1} obtained from lowtemperature heat capacity measurements.[8] A possible reason for this discrepancy may lie in a variable oxygen stoichiometry of SnO_{2} (oxygen vacancy formation yielding a substoichiometric SnO_{2−δ}) that may stabilize this phase towards higher temperatures and/or lower partial pressures of oxygen. In fact, the Sn_{3}O_{4} phase can be considered as oxygen deficient SnO_{2−δ} with rutile structure and onethird of oxygen vacancies cumulated in (011) planes in an ordered pattern.[9] Based on the assessed data[7] the SnO phase diagram was calculated using the FactSage software[10] as shown in Fig. 1. Three invariant points can be identified in the temperature range 300800 K under the conditions summarized in Table 1.
Tin sesquioxide Sn_{2}O_{3} has been identified as a product of bulk SnO disproportionation reaction[11, 12] as well as a product of nanoSn oxidation.[13] Being an intermediate oxide between SnO and SnO_{2}, its stability has been examined on the basis of firstprinciple calculations.[14, 15] Nevertheless, Sn_{2}O_{3} was not included into the present calculations due to absence of complete thermodynamic data.
Oxidation of NanoSn
Oxidation of nanoSn as well as the stability of nanoSnO_{2} have been subjects of many studies. Although some intermediate oxides, namely SnO, Sn_{3}O_{4} and Sn_{2}O_{3}, have been observed during oxidation, the most stable oxide, SnO_{2}, has been obtained in most cases and the resulting particles frequently adopted a core(Sn)–shell(SnO_{2}) structure.
Using XRD and HRTEM Huh et al.[16] have observed oxide formation on a surface of almost spherical tin nanoparticles with an average diameter 28 nm. Oxidic shells were composed of SnO_{2} which was amorphous at low temperatures while it crystallized in tetragonal rutile structure after heating above 500 K. A metastable (highpressure) orthorhombic form of SnO_{2} has been also observed. In their subsequent study[13] they have observed the tin particles oxidation (640 nm in diameter) in air which resulted in an oxide shell composed of SnO and Sn_{2}O_{3} oxides. Using a simultaneous TG/DSC technique combined with XRD characterization of product oxides Song and Wen[17] have investigated the oxidation of tin (nano)particles with average diameter 110 nm in air. Their results reveal a twostage oxidation process. During a continuous heating (220 K min^{−1}) SnO shell is created first at temperatures 473523 K. The second step in the temperature range 673973 K results in a formation of SnO_{2} coexisting with SnO. At temperatures above 1073 K, only SnO_{2} is present in the oxidic shell of particles. These results have been also confirmed under isothermal heating at various temperatures in the range 4731173 K for 4 h. A similar sequence of oxidation products (Sn → SnO + SnO_{2} → SnO_{2}) has been also observed in the case of tin nanowires.[18] Sn_{3}O_{4} as a product of liquid tin nanodroplets oxidation was identified by Mima et al.[19] who employed TEM for an in situ analysis of Sn(l)/tin oxide boundary during oxidation. Sutter et al.[20] have studied the sizedependent room temperature oxidation of tin nanoparticles in air. SnO has been identified as an oxidation product and using TEM images of core/shell particles with various sizes (3100 nm in diameter) the dependence of SnO shell thickness on particle diameter has been found out. It should be noted that the oxidation of tin nanoparticles is influenced by a lowering of tin melting temperature which has been observed experimentally.[21, 22] This decrease in temperature can be also calculated using the Gibbs–Thomson equation or the Pawlow equation.[23] Using the data from Table 2 one can calculate a decrease as low as 60 K for nanoparticles with a diameter of 5 nm.
Thermodynamic Modeling of NanoSn Oxidation
Equilibrium relations in nanosystems are fundamentally dependent on topology of the system under consideration, e.g. on spatial configuration of coexisting phases. In the following analysis we consider two different configurations, namely (1) two singlecomponent particles (analogy to the Pawlow equation for the melting point decrease) and (2) core–shell geometry, which are both applied on the Sn(l)SnO_{2}(s) and Sn(l)Sn_{3}O_{4}(s) equilibrium. According to the proposed classification scheme[29] the respective configurations represent models of the first and the second generation. Although core–shell model is physically more realistic, simpler two particles model is often used for the thermodynamic modeling of equilibria in oxide systems as well as for evaluation of structural stability of polymorphic oxide nanoparticles.[4,5,6, 30,31,32]
SingleComponent Particles
The spatial configuration of such a system is shown in Fig. 2. The equilibrium condition for the reaction Sn(l) + O_{2}(g) = SnO_{2}(s) is in detail derived in “Appendix” (Eq 23, 26 and 27) yielding Eq 1 with the respective bulk and surface contribution given in Eq 2 and 3:
If the stoichiometric amounts of Sn and SnO_{2} are considered, then
Equation 4 can be used for substituting r_{(SnO2)} into the equilibrium relation resulting from Eq 1 to 3
which represents an implicit function of temperature, oxygen partial pressure and tin nanoparticle size, F(T, p_{(O2),np}, r_{(Sn)}) = 0. It is thus possible to calculate the dependence of p_{(O2)} on r_{(Sn)} at given temperature or the dependence of T on r_{(Sn)} at given oxygen pressure. Similar calculations can be performed for SnSnO and SnSn_{3}O_{4} equilibria.
Using this topology it is convenient to introduce the standard chemical potential of a singlecomponent nanoparticle as
At the temperature of tin fusion, 505.1 K, the surface contributions for tin are: 21.787/r(nm) (kJ/mol) for solid and 19.841/r(nm) (kJ/mol) for liquid Sn. Temperature dependences of the molar volumes and the surface energies were neglected in the following calculations.
The molar volumes, surface energies and surface contribution for tin oxides are summarized in Table 3. Surface energy is the crucial quantity for the surface contribution to the Gibbs energy of nanoparticles. In the case of SnO_{2}, the value γ_{(SnO2,sg)} = 1200 mJ m^{−2}[33] was chosen. This value based on adsorption calorimetry measurements is markedly lower than 1720 mJ m^{−2} as previously obtained at the same laboratory using a combination of hightemperature oxide melt solution calorimetry and water adsorption calorimetry.[34] On the other hand ab initio calculated values for SnO_{2}/vacuum interface for various (hkl) surface planes give mean values 1261 mJ m^{−2},[35] 1438 mJ m^{−2},[36] 1407 mJ m^{−2}.[37] These mean values were calculated according to the expression[38]
where n means the number of (hkl) surface planes for which γ_{(hkl)} is calculated. Only one calculated value γ_{(SnO,sg)} = 392 mJ m^{−2}[37] is available in literature for SnO surface. As in the case of SnO_{2} this mean value was obtained using ab initio γ_{(hkl)} values according to Eq 7. Due to the lack or the complete absence of surface energy data for SnO and Sn_{3}O_{4} we performed DFT calculations for slabs consisting of six and four unit cells stacked along [001] and [010] directions, respectively, with a 15 Å thick vacuum region inserted between the slabs. The calculations were carried out using all electron full potential method as implemented in WIEN2k program (APW + lo basis set, GGAPBE96 functional, R_{MT} * K_{max} in the range 7.58, typical kmesh sampling density 0.8 nm^{3}).[39] As expected, the resulting surface energy values (see Table 3) are much lower compared to rutile structure of SnO_{2} due to the layered character imposing a relatively weak cohesion between the layers in both SnO and Sn_{3}O_{4}.
Using Eq 5 in the form
the ratio of O_{2} equilibrium pressure in Sn(l)SnO_{2}(s)O_{2}(g) nanosystem and bulksystem as a function of temperature for tin nanoparticles of radius 2.5, 5, 10 and 20 nm was calculated. As the surface term for solid SnO_{2} is greater than that for liquid tin, the decrease of nanoparticle size brings about an increase of this ratio (Fig. 3).
Similar calculations were performed for reaction 1.5 Sn(l) + O_{2}(g) = 0.5 Sn_{3}O_{4}(s) using Eq 9
Since the surface contributions for solid Sn_{3}O_{4} are smaller than that for liquid tin, we observe an opposite trend in this case, namely a decrease of the oxygen pressure ratio with a decrease of nanoparticle size (Fig. 4).
Using the surface terms defined by 2V_{m}γ_{sg} values given in Table 3 we can calculate the SnO phase diagram for a selected nanoparticle size. The calculated phase diagram presented in Fig. 5 for tin oxides nanoparticles whose sizes are deduced from solid tin nanoparticles of radius 5 nm exhibits notable differences compared to the bulk version. Indeed, the SnO and Sn_{3}O_{4} stability is substantially enhanced at the expense of SnO_{2} due to the size effect (significantly larger surface energy of SnO_{2}). Furthermore, there is only one invariant point in the temperature range 300800 K, namely Sn(s)SnO(s)Sn_{3}O_{4}(s) at T = 460 K and p(O_{2})/p° = 4.6 × 10^{−58}.
Core–Shell Topology
According to the most experimental findings, the oxidation of Sn nanoparticles leads to a core–shell geometry. This situation is depicted in Fig. 6. The equilibrium condition for the reaction Sn(l) + O_{2}(g) = SnO_{2}(s) is derived in “Appendix” (Eq 23, 27, 29, 30 and 31). Using a material balance (r_{0} = r_{(Sn,l,np)}, r = r_{(Sn,l,core)}):
we obtain for the SnO_{2} shell thickness
Similarly to Eq 1, Δ_{r}G^{np}(T,p_{(O2)np}) = 0 at equilibrium and
Equation 12 represents a relation F(T, p_{(O2),np}, r_{0}, δ) = 0 and it is possible to calculate the dependence of δ on temperature or oxygen pressure at a given value o r_{0}. The x parameter stands for the amount of unoxidized molten tin when 1 mol of SnO_{2} is formed as the shell of nanoparticles: x = n_{(Sn,l,core)}/(n_{(Sn,l,np)} − n_{(Sn,l,core)}), which can be expressed as
It should be noted that for x = 0 (r → 0), Eq 12 is identical with the previously introduced Eq 5 with r + δ = r_{(SnO2,s)} and r_{0} = r_{(sn)}.
One more parameter, namely the interfacial energy γ_{(Sn/SnO2)} at the liquid tin/solid SnO_{2} interface, is included in Eq 12. It can be assessed from contact angle of Sn(l) sessile drop on SnO_{2}(s) substrate measurement, but such a measurement has not been performed yet. Alternatively, it can be calculated from the work of adhesion W_{ad} for Sn(l)/SnO_{2}(s) interface and the surface energies of Sn(l) and SnO_{2}(s) as
Some empirical estimates for the work of adhesion at liquid metal/solid oxide interfaces or just for interfacial energy at liquid metal/solid oxide interfaces have been proposed in literature.[42,43,44,45,46] Here we use a model based on broken bond approximation on the SnO_{2} interface. While the SnO bond energy in bulk SnO_{2} is 229.5 kJ mol^{−1} (obtained from the cohesive energy), the value E_{b}(SnO) = 40.5 kJ mol^{−1} assessed from the surface energy 1200 mJ m^{−2} (Table 3) and the number of broken SnO bonds on (001) surface represents only to 18 per cent of the bulk value. This is likely due to relaxation of both the valence electron density and the atomic positions close to the surface. Considering that oxygen atoms on the surface are lacking this portion of energy to saturate their bonding state and there are enough Sn atoms available for bonding on the Sn(l) counterpart (1.22 Sn per O), we can evaluate the adhesion work as W_{ad} = = 600 mJ m^{−2}. Let us note that this result is in a very good agreement with the values obtained from models[42] (W_{ad} = = 688 mJ m^{−2}) based on partial molar enthalpy of oxygen dissolution in liquid tin, \(\Delta \bar{H}_{{ [ {\text{O]Sn(l)}}}}^{\infty }\) = − 182 kJ mol^{−1}.[47] Substituting our value of W_{ad} into Eq 14 we obtain the corresponding interfacial energy γ_{(Sn/SnO2)} = 1190 mJ m^{−2}.
Equation 12 was used for the calculation of the surface oxide layer thickness δ (expressed as δ/(r + δ)) as a function of relative equilibrium O_{2} pressure (at T = 800 K) and temperature (at p(O_{2},np)/p(O_{2},bulk) = 5) for a given initial tin particle radius r_{0} (Fig. 7 and 8). Due to the large interfacial energy γ_{(Sn/SnO2)} the core–shell structure is not a stable configuration with respect to the complete SnO_{2} particle and thus the oxidation of tin particles proceeds spontaneously under these T and p_{(O2)} conditions. This is imposed by a descending character of the δ/(r + δ) versus log (p(O_{2},np/p(O_{2},bulk)) curve, where for a given oxygen pressure the core–shell structure corresponding to a point on the calculated curve has always higher Gibbs energy compared to a single phase SnO_{2} nanoparticle with δ/(r + δ) = 1. The experimentally observed core–shell structures formed on Sn(l) oxidation are thus the result of kinetic effects (diffusion of oxygen).
Similar calculations were performed for the shell composed of Sn_{3}O_{4} using Eq 15
which was derived in the similar way as the relations for SnO_{2} shell (see “Appendix”). In this case it holds for the surface layer thickness
The contribution of work of adhesion to the interfacial energy is much smaller for the Sn(l)/Sn_{3}O_{4}(s) interface due to very low value of γ_{(Sn3O4)} (Table 3). As mentioned above, Sn_{3}O_{4} exhibits a layered structure which can be derived from rutileSnO_{2} by periodically inserting 1/3 of oxygen vacancies arranged in parallel layers. Hence, only weak SnSn bonds are broken when an interface is formed at this vacancyrich plane, and these interactions are assumed to be reestablished on the interface with liquid Sn. The respective interfacial energy can be thus put equal to that of liquid tin, γ_{(Sn/Sn3O4)} = γ_{(Sn,lg)} = 588 mJ mol^{−2} (Table 2). Using Eq 15 the surface oxide layer thickness δ (expressed as δ/(r + δ)) as a function of relative equilibrium O_{2} pressure (at T = 800 K) and temperature (at p(O_{2},np)/p(O_{2},bulk) = 5) for a given initial tin particle radius r_{0} were calculated (Fig. 9 and 10). The same conclusion regarding the instability of core–shell structure with respect to a single phase Sn_{3}O_{4} nanoparticle can be drawn as in the case of Sn/SnO_{2} particles but with the difference that Sn/Sn_{3}O_{4} core–shell structures are more stable (closer to the equilibrium state) than the Sn/SnO_{2} ones.
Discussion
Thermodynamic calculations whose results are presented here show a substantial influence of the considered system topology. In the case of singlecomponent nanoparticles the system formed by two independent components (tin and oxygen) has three degree of freedom (T, p_{(O2)}, r_{(Sn)}) and it is possible to formulate three distinct dependences: T = f(p_{(O2)}) at [r_{(Sn)}], T = f(r_{(Sn)}) at [p_{(O2)}] and p_{(O2)} = f(r_{(Sn)}) at [T]. Thus, for the constant value of r_{(Sn)}, the oxygen pressure p_{(O2)} is unambiguously determined by temperature (Fig. 3 and 4). On the other hand, core–shell topology brings four degrees of freedom (T, p_{(O2)}, r_{0}, δ). For a given size of tin nanoparticles r_{0}, the thickness of surface oxide layer δ is determined by the oxygen pressure (Fig. 7 and 9) at constant temperature or by the temperature (Fig. 8 and 10) at constant oxygen pressure. Our results, which show the instability of core–shell structure, are perhaps surprising. Using the core–shell model Vegh and Kaptay[29] have shown that in the case of lead melting, a core solid and a liquid shell is found in a finite temperature range below the macroscopic melting point. It is a consequence of the simple relation γ_{(Pb,sg)} > γ_{(Pb,sl)} + γ_{(Pb,lg)}. In analogy, if the relation γ_{(Me)} > γ_{(Me/MeOx)} + γ_{(MeOx)} was satisfied, a coreshell structure might be stable. But it is not the case for many of the liquid metal/solid oxide interfaces including our system under study. By contrast, in an opposite process of SnO_{2} reduction into liquid Sn, the SnO_{2} core—Sn(l) shell structure would be stable for a certain range of conditions. Another aspect that should be mentioned is the sensitivity of the calculated results on the input parameters. While thermodynamic data for pure substances are rather reliable, the values of surface/interface energies are inconsistent. Considering two singlecomponent particles Sn(l), r_{(sn)} = 2.5 nm, and SnO_{2}(s) as an example, the calculated oxygen pressures for γ_{(SnO2)} = 1720 mJ m^{−2}[31] are 228 times higher than those for γ_{(SnO2)} = 1200 mJ m^{−2}[30] in the temperature range 1300300 K. The same applies for the interfacial energy γ_{(Sn/SnO2)} within the core–shell topology. Also other approximations considered in the calculations, e.g. spherical shape of nanoparticles and temperatureindependent molar volumes and surface/interface energy, bring some uncertainty in the results. Therefore, the results of these calculations must be understood as semiquantitative estimate rather than the exact values.
The last issue we would like to mention here is the opposite influence of nanosizing on the stability of SnO_{2} and SnO/Sn_{3}O_{4} oxides within the two singlecomponent nanoparticles model. It is due to the surface contribution which is higher for SnO_{2}(s) than Sn(l) while lower for SnO(s) and Sn_{3}O_{4}(s) compared to Sn(l). This situation can explain some experimental findings during oxidation of Sn nanoparticles, namely an increased stability of SnO(s) and Sn_{3}O_{4}(s) with respect to both liquid tin and solid tin dioxide.
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Acknowledgments
This work was supported by Czech Science Foundation, Grant No. 1713161S.
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This invited article is part of a special issue of the Journal of Phase Equilibria and Diffusion in honor of Prof. Jan Vrestal’s 80th birthday. This special issue was organized by Prof. Andrew Watson, Coventry University, and Dr. Ales Kroupa, Institute of Physics of Materials.
Appendices
Appendix 1
Let us consider a closed heterogeneous system at constant T and V. It follows for the equilibrium from basic thermodynamics relations
F is the Helmholtz energy and superscripts φ and σ stand for coexisting bulk phases and interfaces/surfaces, respectively. Let us now describe the equilibrium in the system in Fig. 2 where the reaction Sn(l) + O_{2}(g) = SnO_{2}(s) takes place ((α) = Sn(l) and (β) = SnO_{2}(s)) in terms of the condition (17). The equilibrium corresponds to coexistence of two singlecomponent particles Sn(l) and SnO_{2}(s) and gaseous oxygen at temperature T and oxygen pressure p_{(O2,g)}:
Summation of these bulk contributions yields
Supposing incompressibility of liquid tin and solid oxide (molar volume does not depend on pressure) and applying material balance conditions in the stoichiometric form \(n_{i} = n_{i}^{o} + \nu_{i} \xi ,\;(\nu_{{({\text{Sn}},{\text{l}})}} =  \,1,\;\nu_{{ ( {\text{O2,g)}}}} =  \,1,\;\nu_{{ ( {\text{SnO2,s)}}}} = 1)\) one can arrange Eq 22 as
The surface terms in Eq 18 can be expressed as
and the summation gives
dF_{syst} = 0 at equilibrium thus
Somewhat more complicated is the derivation of equilibrium condition in the system in Fig. 6 where the same reaction Sn(l) + O_{2}(g) = SnO_{2}(s) takes place ((α) = Sn(l) and (β) = SnO_{2}(s)). The equilibrium corresponds to a coexistence of unoxidized Sn(l) core and SnO_{2}(s) shell in gaseous oxygen at temperature T and oxygen pressure p_{(O2,g)}. Whereas Eq 1923 for bulk contribution hold as in the previous case, the surface contribution is qualitatively different. Due to different interfaces of Sn(l,np) nanoparticle and unoxidized Sn(l,core) the oxidation reaction should be written as (1 + x) Sn(l,np) + O_{2}(g) = SnO_{2}(s) + x Sn(l,core) however, the bulk properties of Sn(l,np) and x Sn(l,core) are the same. Thus the surface contribution should read
To simplify Eq 31 the difference between molar volumes of Sn(l) and SnO_{2}(s) was neglected in the derivative dV_{(core/shell)}/dn_{(core/shell)}. Since (ν_{(Sn,l,np)} = − (1 + x), ν_{(SnO2,s)} = 1, ν_{(Sn,l,core)} = x) the surface contribution has got a form
Equation 27 is valid and applicable for both topologies, two singlecomponent particles and core–shell model, with the difference in surface term being described by Eq 26 and 32, respectively.
Appendix 2
List of symbols
 A :

Surface area
 E _{b} :

Bond energy
 F :

Helmholtz energy
 G :

Gibbs energy
 Δ_{r}G:

The change of Gibbs energy due to reaction (reaction Gibbs energy)
 Δ_{r}G^{o}:

The change of standard Gibbs energy due to reaction (standard reaction Gibbs energy)
 \(\Delta \bar{H}_{{ [ {\text{O]Sn(l)}}}}^{\infty }\) :

The partial molar enthalpy of oxygen dissolution in liquid tin
 n _{ i} :

The number of moles of the species i
 \(n_{i}^{o}\) :

The initial number of moles of the species i
 p :

Pressure
 R :

The gas constant
 r :

Radius of spherical nanoparticle
 T :

Temperature
 T ^{F} :

Temperature of fusion
 V _{m} :

Molar volume
 W _{ad} :

Work of adhesion
 δ :

Thickness of surface layer
 γ :

Surface or interfacial energy
 μ _{ i} :

The chemical potential of the species i
 \(\mu_{i}^{o}\) :

The standard chemical potential of the species i (at pressure p^{o} = 100 kPa)
 ν _{ i} :

The stoichiometric coefficient of the species i
 ξ :

Extent of reaction
 bulk:

Related to bulk properties (bulk contribution)
 np:

Related to nanoparticle
 surf:

Related to surface properties (surface contribution)
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Leitner, J., Sedmidubský, D. Thermodynamic Modeling of Oxidation of Tin Nanoparticles. J. Phase Equilib. Diffus. 40, 10–20 (2019). https://doi.org/10.1007/s1166901806864
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DOI: https://doi.org/10.1007/s1166901806864
Keywords
 surface energy
 thermodynamic modeling
 tin nanoparticles
 tin oxides