lecture3

lecture3 - 1 3.051J/20.340J Lecture 3 Biomaterials Surfaces...

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1 3.051 J / 20 .340 J Lecture 3: Biomaterials Surfaces: Chemistry Surfaces are high-energy regions of materials and thereby facilitate chemical reactions that influence performance of biomaterials. This lecture will focus on 2 classes of surface chemistry relevant to biomaterials: Chemisorption on metals and oxides Aqueous corrosion of metals 1. Chemisorption Strong modifications to electronic structure/ electron density of adsorbate molecule (> 0.5 eV/surface site) Important Examples: a) Metal Oxide Formation on Metals “metals just wanna be oxides” xM + ½yO 2 M x O y G 0 of oxide formation is negative for all but a few metals (e.g., Au) Reaction G 0 (joules) T range (K) 2Cr + 3/2 O 2 = Cr 2 O 3 -1,120,300 + 260T 298-2100 Fe + ½ O 2 = FeO -259,600 + 62.55T 298-1642 2Fe + 3/2 O 2 = Fe 2 O 3 -810,520 + 254.0T 298-1460 Ti + O 2 = TiO 2 -910,000 + 173T 298-2080 from D.R. Gaskell, Intro. To Metallurgical Thermodynamics, McGraw-Hill, 1981
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2 3.051 J / 20 .340 J How does metal oxidation happen? One scenario is… step 1 : physisorption of O 2 ; ~20-25 kJ/mol 1 eV/molec = 96.5 kJ/mol kT 293 0.025 eV step 2 : molecular oxygen dissociates and reduces by chemisorption; ~600 kJ/mol step 3: bond rearrangement; crystallization of oxide layer Resultant reduction in surface energy γ Compare: at 1400 ° C: γ δ -Fe = 1900 dyn/cm FeO = 580 dyn/cm Consider metal oxidation as 2 half reactions: O 2 + ze - Electrons and ions must traverse the oxide layer for rxn to proceed. Across the oxide film, an oxidation potential, E 0 ~ 1V generates an electric field: x~1 nm M M z+ M M z+ + ze - ½z O 2- G 0 = E z F E-Field 1 V/nm = 10 MV/cm 0 Ionic species are “pulled” F= 96,480 C/mol e - through oxide film! 1 J = 1 V-C WHAT HAPPENS AS THE OXIDE CONTINUES TO GROW?
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3 3.051 J / 20 .340 J The E-field decreases. Subsequent oxide growth occurs by thermal diffusion of M z+ to oxide surface or O 2- to metal/oxide interface under the concentration gradient c: 2 l = k t p l Requirements for Passivation : i) small k p (rate const) oxide thickness time k = const D c p ii) adherent oxide Oxide layer must not scale or spall minimize V molar & stress build-up xM + ½yO 2 M x O y ex., Ti (TiO 2 ), Cr (Cr 2 O 3 ), Al (Al 2 O 3 ) (Al metal not used in biomaterials applications due to toxicity) MO = xy Pilling-Bedworth ratio: PB = V oxide ( formed ) ρ M M V metal ( consumed ) x M M M O Want PB ~ 1 (PB > 1) for adherance of oxide to underlying metal—in practice, this rule is marginally predictive, however.
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lecture3 - 1 3.051J/20.340J Lecture 3 Biomaterials Surfaces...

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