 S univ =  S sys +  S surr Must be ____ for spontaneity as dictated by… + …the Second Law of Thermodynamics  S surr =  H / T (at constant P) So that:

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 S univ =  S sys +  S surr Must be ____ for spontaneity as dictated by… + …the Second Law of Thermodynamics  S surr =  H / T (at constant P) So that:  S univ =  S sys -  H / T multiply both sides by –T and rearrange to get… Indicates the units of entropy: J/K

 S univ =  H - T  S  G Hess’s Law Organizational Energy 1 st Law of Thermo Change in Free Energy Put in values that can easily be calculated…  H° - T  S°=  G° This is called: Gibb’s-Helmholtz Equation Standard change in Enthalpy This is called:Standard change in Entropy This is called:Standard change in Free Energy Standard states: (s)- pure (l)- pure (g)- pure, 1 atm (aq)- 1M Elemental- stable form Can be measured from relative standard formation reactions Can be measured from absolute* entropy values

H 2 O (l) + Na 2 O 2 (s) + S (s) → NaOH (aq) + SO 2(g) 224 Use Hess’s Law to get  H°:  H° =  n  H f ° products -  n  H f ° reactants 2(-286 kJ) 2(-515 kJ) 0 kJ 4(-470 kJ) -297 kJ 2(70 J/K) 2(95 J/K) 32 J/K 4(50 J/K) 248 J/K At 25.0°C the following reaction occurs under standard state conditions  H° = kJ kJ  H° = -575 kJ Use absolute S values to get  S°:  S° =  nS° prod. -  nS° react.  S° = 448J/K J/K  S° = 86 J/K Use Gibbs-Helmholtz to get  G°:  G° =  H° - T  S°  G° = -575 kJ – [ (298 K) (86 J/K) (1 kJ/1000 J)]  G° = -600 kJ Redox Reaction: transfer of electrons!!!! O +4 oxidation -2 reduction -2 Spontaneous under standard conditions!

H 2 O (l) + Na 2 O 2 (s) + S (s) → NaOH (aq) + SO 2(g) 224 2(-237 kJ) 2(-451 kJ) 0 kJ 4(-419 kJ) -300 kJ At 25.0°C the following reaction occurs under standard state conditions Use stand.  G f ° values to get  G°:  G° =  n  G f ° prod. -  n  G f ° react.  G° = kJ kJ  G° = -600 kJ Use Gibbs-Helmholtz to get  G°:  G° =  H° - T  S°  G° = -575 kJ – [ (298 K) (86 kJ/K) (1 kJ/1000 J)]  G° = -600 kJ the same !!!!!

The Third Law of Thermodynamics… The entropy of a perfect crystal at Absolute Zero is zero. S = k log W Boltzmann’s Entropy Equation Number of microstates, called the multiplicity The Boltzmann Constant, x J/K