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Introductory Inorganic Chemistry

Introductory Inorganic Chemistry. What is Inorganic Chemistry?. As: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3. For more information about these periodic tables visit the site where I obtained the pictures: http://chemlab.pc.maricopa.edu/periodic/default.html.

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Introductory Inorganic Chemistry

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  1. Introductory Inorganic Chemistry What is Inorganic Chemistry?

  2. As: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

  3. For more information about these periodic tables visit the site where I obtained the pictures: http://chemlab.pc.maricopa.edu/periodic/default.html

  4. Classes of Inorganic Substances

  5. Review of Concepts Thermochemistry: Standard state: 298.15 K, 1 atm, unit concentration Enthalpy Change, DH° DH° = SH°products - SH°reactants Entropy Change, DS° Free Energy Change, DG DG = DH - TDS At STP: DG° = DH° - (298.15 K)DS°

  6. Standard Enthalpy of Formation, DH°f DH° for the formation of a substance from its constituent elements Standard Enthalpy of Fusion, DH°fus Na(s) Na(l) Standard Enthalpy of Vapourization, DH°vap Br2(l) Br2(g) Standard Enthalpy of Sublimation, DH°sub P4(s) P4(g) Standard Enthalpy of Dissociation, DH°d ½ Cl2(g) Cl(g) Standard Enthalpy of Solvation, DH°sol Na+(g) Na+(aq)

  7. Na(s) Na(g)  Na+(g) ½ Cl2(g) Cl(g) Cl-(g) DH°sub DH°ie NaCl(s) Lattice Energy, U DH°d DH°ea DH°f Why should we care about these enthalpies? They will provide us information about the strength of bonding in both molecules and extended solids.

  8. Free Energy Change, DG = DH - TDS At STP: DG° = DH° - (298.15 K) DS° The two factors that determine if a reaction is favourable: If it gives off energy (exothermic) DH= SHproducts - SHreactants DH < 0 If the system becomes “more disordered” DS= SSproducts - SSreactants DS > 0 If DG < 0, then reaction is thermodynamically favourable

  9. aA + bB + cC + … hH + iI + jJ + … DG lets us predict where an equilibrium will lie through the relationship: DG = -RT ln K So if DG < 0, then K > 1 and equilibrium lies to the right. There are three possible ways that this can happen with respect to DH and DS.

  10. If both enthalpy and entropy favour the reaction: i.e. DH < 0 and DS > 0 then DG < 0. S(s) + O2(g) SO2(g)DH° = -292.9 kJ/mol TDS° = 7.5 kJ/mol DG° = -300.4 kJ/mol If enthalpy drives the reaction: i.e. DH < 0 and DS < 0, but |DH| > |TDS|, then DG < 0. N2(g) + 3 H2(g) 2 NH3(g)DH° = -46.2 kJ/mol TDS° = -29.5 kJ/mol DG° = -16.7 kJ/mol If entropy drives the reaction: i.e. DH > 0 and DS > 0, but |DH| < |TDS|, then DG < 0. NaCl(s) Na+(aq) + Cl-(aq)DH° = 1.9 kJ/mol TDS° = 4.6 kJ/mol DG° = -2.7 kJ/mol

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