The Second law of thermodynamics "A system and its surroundings spontaneously tend towards increasing disorder” "Natural direction of the universe tends toward increasing disorder." △S =△S niverse sem+△S, The entropy of the universe increases in a spontaneous process and remains unchanged in an irreversible(equilibrium)process. Spontaneous process: (Irreversible) ASuniN 4Ssys ASsurr>0 A process at equilibrium (Reversible) ASUniv=4Ssys 4Ssurr=0
The Second law of thermodynamics “A system and its surroundings spontaneously tend towards increasing disorder” “Natural direction of the universe tends toward increasing disorder.” 0 universe system surrounding ∆ = + ∆ S S S ∆ ≥ The entropy of the universe increases in a spontaneous process and remains unchanged in an irreversible (equilibrium) process. A process at equilibrium : (Reversible) Spontaneous process: (Irreversible) ∆Suniv = ∆Ssys + ∆Ssurr > 0 ∆Suniv = ∆Ssys + ∆Ssurr = 0
Third law of Thermodynamics The entropy of a pure/perfect crystalline substance at absolute zero0K☒is zero. Pure crystalline:atoms OK,SPure Crstalline or molecules in a highly ordered array. S=k.InW S=kIn1=0 At 0 K:thermal energy=0 thermal random motion stops the minimal disorder state Is it possible to measure S(T)for a substance(A)at any temperature Entropy a state function A(O)→A(T) △S=S(T)-S(0)=S(T)
Third law of Thermodynamics The entropy of a pure/ perfect crystalline substance at absolute zero (0K) is zero. 0 , 0 K SPure Crystalline = At 0 K: thermal energy = 0 S k W = ⋅ln S k = ⋅ = ln1 0 Pure crystalline: atoms or molecules in a highly ordered array. At 0 K: thermal energy = 0 thermal random motion stops the minimal disorder state A A T (0) ( ) → ∆ = = S S S S ( )- ( T (0) T) Is it possible to measure S˚(T) for a substance (A) at any temperature ? Entropy : a state function
Standard molar entropy,S 1 atm 25'C Standard molar entropy S):the molar entropy values of substances at their standard states 1.Standard molar entropies of Standard Entropy Values (S)for Some Substances elements are not zero at25°C Substance S(J/K·mol) L2(S) 116.7 2.For a substance in different I2(g) 260.6 C(diamond) 2.4 states. C(graphite) 5.69 S(1 mol)<(1 mol)<<(1 mol) CH4(methane) 186.2 C2H(ethane) 229.5 He(g) 126.1 Ne(g) 146.2
Standard molar entropy, 1. Standard molar entropies of elements are not zero S ° Standard molar entropy ( ): the molar entropy values of substances at their standard states o S 1 atm & 25˚C 2. For a substance in different states: solid liquid gas S (1 mol) < S (1 mol) << S (1 mol) ° ° °
Standard molar entropy S 3.For different substances:entropies increase with the number of atoms in the formula of'substance S Page:1123 Appendix C 图图图器后国男活质图质8质后指图质品活男 4.For different substances: Atomic No 20 56 entropies increase with the Molar mass 9.01 40.08 137.33 increasing molar mass. (g/mol) substance Be(s) Ca(s)Ba(s) S°(J/molK) 9.44 41.4 63.2
Standard molar entropy o S 3. For different substances: entropies increase with the number of atoms in the formula of substance 4. For different substances: entropies increase with the increasing molar mass. CH CH CH (CH ) CH CH (CH ) CH 3 3 3 2 2 3 3 2 4 3 S < S < S ° ° ° Atomic No. 4 20 56 Molar mass (g/mol) 9.01 40.08 137.33 substance Be (s) Ca (s) Ba (s) S˚ (J/mol·K) 9.44 41.4 63.2 Page: 1123 Appendix C
Entropy Changes in reactions(AS) Entropy a state function △Sm= 日1atm&25C aA+bB→cC+dD △Sm=[cS°(C)+dS°(D]-[a·S(A)+b.S(B)] ASgm=∑nS(products)-∑mS((reactants) What is the standard entropy change for the following reaction t250C? 2C0g)+0,(g=2C0,(g) S(J/moK)197.9 205.0 213.6 ASm =2xS'co)-[2xS'co+S △Sm=2×213.6-[2×197.9+205.0]=-173.6J1mol.K
Entropy Changes in reactions(∆Srxn) Entropy : a state function aA + bB cC + dD → S [ ( ) ( )] [ ( ) ( )] rxn c d a b S C S D S A S B ° ° ° ° ° ∆ = ⋅ + ⋅ − ⋅ + ⋅ o S (products) (reactants) rxn o o ∆ = − ∑ ∑ nS mS S ? rxn ° ∆ = 1 atm & 25˚C What is the standard entropy change for the following reaction at 25 0C? 2 2 2CO(g) + O (g) 2CO (g) 2 2 S 2 [2 ] rnx ( ) ( ) ( ) CO CO O S S S ° ° ° ° ∆ = × − × + rnx S 2 213.6 [2 197.9 205.0] 173.6 / J mol K ° ∆ = × − × + = − ⋅ S ( / ) 197.9 205.0 213 J mol K .6 ° ⋅