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D H o rxn = [ SD H o f products ] – [ SD H o f reactants]

D H o rxn = [ SD H o f products ] – [ SD H o f reactants]. SD H o f products = [2 x (-242) + (-394) + 0] = -878 kJ. SD H o f reactants = [4 x (-157) + (-74.8)] = -702.8 kJ. [ SD H o f products ] – [ SD H o f reactants] = -878kJ – (-702.8kJ) = -175.2 kJ. 2. 1. 3. 5. 4.

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D H o rxn = [ SD H o f products ] – [ SD H o f reactants]

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  1. DHorxn = [SDHof products] – [SDHof reactants] SDHof products = [2 x (-242) + (-394) + 0] = -878 kJ SDHof reactants = [4 x (-157) + (-74.8)] = -702.8 kJ [SDHof products] – [SDHof reactants] = -878kJ – (-702.8kJ) = -175.2 kJ

  2. 2 1 3 5 4

  3. 4 moles g  2 moles gas DS s  g DS s  solution DS

  4. DH DS spontaneous (DG<0) - + always + - never - - only at low temperatures + + only at high temperatures

  5. DGo = DHo - TDSo DHo = -175.2 kJ (from Q1) DSorxn = [SSo products] – [SSoreactants] SSo products = (2 x 189 + 214 + 4 x 33.2) = 724.8 J/K SSoreactants = (186 + 4 x 42.6) = 356.4 J/K DSorxn = 724.8 J/K – 356.4J/K = 368.4 J/K units !!! DGo = -175.2 kJ - (273+25)K x 368.4 J/K x 1kJ/1000J = -285 kJ

  6. spontenous, DG <0 spontenous, DG <0 spontenous, DG <0

  7. Only configurations that do not follow Pauli’s exclusion principle are physically impossible Configurations that do not obey Hund’s rule are not the lowest energy configurations but are possible, in principle

  8. [Ar]4s23d9 => [Ar]4s13d10 is more stable!

  9. K = [Ar] 4s1 n = 4 l = 0 (valence electron is in s-orbital) ml = 0 (must be, if l=0) ms can be either +1/2 or -1/2

  10. Co: [Ar]4s23d7 Co2+: [Ar]3d7 electrons leave from shell with the highest n first

  11. Ar check valence electron configuration – or simply count all electrons=atomic #

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