lecture18_March_24

# lecture18_March_24 - i Determining Ssys What is the entropy...

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i. Determining Δ S sys : What is the entropy of a compound? The entropy of a pure perfect crystal at 0 K is ZERO. standard absolute entropy, Sº (at a given T, 1 bar or 1 M) products (also denoted Δ S o rxn ) 43 Δ S o = n p S o ( p ) n r S o ( r ) reactants Third Law of Thermodynamics: (perfect no defects; 0 K no atomic motion) Sº values tabulated on data sheet and in text (Appendix B). Standard entropy change for a reaction, Δ Sº: Note: S is a state function. Need to compare all compounds to a reference system. Units of S: J mol –1 K –1

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44 ii. Determining Δ S sur : Given amount of heat will produce greater change at lower T | Δ S sur | larger when T is small Δ S sur 1/T System can be a reaction, so Δ H sys = Δ H rxn . Sign depends on direction of heat flow Endothermic: Δ H sys > 0 heat from sur. T sur Δ S sur < 0 Magnitude of Δ S sur depends on T: Δ S sur > 0 Exothermic: Δ H sys < 0 heat to sur. T sur (at constant T and P) Δ S sur −Δ H sys Δ S sur = −Δ H sys T
45 iii. Determining Δ S univ : Using the following data, prove that water does not spontaneously boil at 25 ºC, i.e., show Δ S univ < 0 . Sº (J/Kmol) Δ f (kJ/mol) H 2 O (l) 69.9 285.8 H 2 O (g) 188.8 241.8 Δ S sys = (1mol × 188.8 J/Kmol) (1 mol × 69.9 J/Kmol) Δ S sur = −Δ H rxn / T = 118.9 J/K = [1mol × ( 241.8 kJ/mol) (1mol × ( 285.8kJ/mol)] / 298.15 K = -0.1476 kJ/K = -147.6 J/K = [1 mol × Δ H f °(H 2 O(g)) 1 mol ×Δ H f °(H 2 O(l))]/T H 2 O (l) H 2 O (g) Careful with units of Sº and Δ f

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46 Δ S univ = Δ S sys + Δ S sur Negative forward reaction is not spontaneous.
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