Entropy S Entropy is disorder or randomness 2

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Entropy = S Entropy is disorder or randomness

Entropy = S Entropy is disorder or randomness

2 nd Law of Thermodynamics Suniverse > 0 for spontaneous processes spontaneous = no

2 nd Law of Thermodynamics Suniverse > 0 for spontaneous processes spontaneous = no external intervention Ssystem positional disorder Ssurroundings energetic disorder

Energetic Disorder P. E. a) b) ordered reactants K. E. random a) endothermic reaction

Energetic Disorder P. E. a) b) ordered reactants K. E. random a) endothermic reaction P. E. b) exothermic reaction products Hsystem < 0 Hsurroundings > 0 Ssurroundings > 0

 Ssurr = - Hsystem (J/K) T Ssurr depends on T heat surroundings high

Ssurr = - Hsystem (J/K) T Ssurr depends on T heat surroundings high T low T small effect relatively larger effect

Positional Disorder (1/2)(1/2)(1/2) = 1/64 With 1 mole of gas, Not probable

Positional Disorder (1/2)(1/2)(1/2) = 1/64 With 1 mole of gas, Not probable

Positional Disorder S = k ln W W = degrees of freedom Boltzman ordered

Positional Disorder S = k ln W W = degrees of freedom Boltzman ordered states low probability low S disordered states high probability high S Ssystem Positional disorder Increases with number of possible positions (energy states) Ssolids< Sliquids<< Sgases

Entropy (J/K) [heat entering system at given T] convert q to S System 1

Entropy (J/K) [heat entering system at given T] convert q to S System 1 w = -182 J q = +182 J Pext = 1. 5 atm T = 298 K System 2 Pext = 0 atm w=0 q=0 E = 0

System 3 P 1 = 6. 0 atm P 2 = 1. 5 atm

System 3 P 1 = 6. 0 atm P 2 = 1. 5 atm V 1 = 0. 4 L V 2 = 1. 6 L T 1 = 298 K = T 2 Pext = Pint + d. P reversible process - infinitely slow V wr= - Pext d. V =- n. RT d. V = - n. RT ln (V 2/V 1) V V 2 1

 Ssystem System 1 Pext = 1. 5 atm w = -182 J q

Ssystem System 1 Pext = 1. 5 atm w = -182 J q = +182 J S =3. 77 J/K System 2 Pext = 0 atm w=0 q=0 S =3. 77 J/K Ssurr = - Hsystem T System 3 Pext = Pint + d. P wr = -n. RT ln (V 2/V 1) = - 1120 J qr = + 1120 J S = 3. 77 J/K Ssystem = qr = 1120 J T 298 K

3 rd Law of Thermodynamics Entropy of a perfect crystalline substance at 0 K=

3 rd Law of Thermodynamics Entropy of a perfect crystalline substance at 0 K= 0

Entropy curve solid liquid gas vaporization S qr T fusion 0 0 Temperature (K)

Entropy curve solid liquid gas vaporization S qr T fusion 0 0 Temperature (K)

Entropy At 0 K, S = 0 Entropy is absolute S 0 for elements

Entropy At 0 K, S = 0 Entropy is absolute S 0 for elements in standard states S is a State Function Sorxn = n Soproducts - n Soreactants S is extensive

Increases in Entropy 1. 2. 3. 4. 5. 6. 7. Melting (fusion) Sliquid >

Increases in Entropy 1. 2. 3. 4. 5. 6. 7. Melting (fusion) Sliquid > Ssolid Vaporization Sgas >> Sliquid Increasing ngas in a reaction Heating ST 2 > ST 1 if T 2 > T 1. Dissolution ? Ssolution > (Ssolvent + Ssolute) Molecular complexity number of bonds Atomic complexity e-, protons and neutrons