上海交通大学：《大学化学 Chemistry》教学资源（课件讲稿）Chapter 19 Chemical Thermodynamics（2/3）

Thermodynamics 热力学 Laws of Thermodynamics show that spontaneity is predictable by considering two factors: Enthalpy Entropy 焙H S熵 Enthalpy Change (AH): Entropy(⑨： the total energy difference the randomness factor accounts between reactants and products for energy lost in the process Gibbs Free Energy (G):net amount available Energy for useful work after considering the two factors

Thermodynamics Laws of Thermodynamics show that spontaneity is predictable by considering two factors: Enthalpy & Entropy H S Enthalpy Change (∆H) : Entropy (S) : 热力学 焓 熵 Enthalpy Change (∆H) : the total energy difference between reactants and products Entropy (S) : the randomness factor accounts for energy lost in the process Gibbs Free Energy (G) : net amount available Energy for useful work after considering the two factors

The molecular interpretation of Entropy S=k.InW Statistical thermodynamics volume (pressure) S←-temperature the number of molecules In general,the number of microstates available to a system will increase with an increase in volume,an increase in temperature,or an increase in number of molecules;because any of these changes will increase the possible arrangements of positions and energies of molecules in the system

The molecular interpretation of Entropy S k = ⋅ln W Statistical thermodynamics volume pressure ( ) S temperature the number of molecules   ←    In general, the number of microstates available to a system will increase with an increase in volume, an increase in temperature, or an increase in number of molecules; because any of these changes will increase the possible arrangements of positions and energies of molecules in the system

Entropy s← possible positions of molecules Energies of molecules 1.For any substance: 88 Liquid Solid Liquid Vapor Ssolid STigeid gas Solid Liquid Gas Standard Entropy Values (S)for Some Substances at25°C Substance S(J/K·mol) H20I) 69.9 H20(8) 188.7 Boiling Br2(l) 152.3 Br2(8) 245.3 I2(S) 116.7 2(g) 260.6 Melting Temperature(K)

Entropy 1. For any substance: solid liquid S S possible positions of molecules S Energies of molecules  ←  liquid gas < S S <

Entropy S→ possible positions of molecules Energies of molecules 2.For a substance in the same state: Solid Liquid Gas For liquid water: S.oox>SH.o0x5o _Boiling 3.In dissolving process: Solvent 0000889o .Melting Qoooooooo 0o099990 Temperature(K) Solute Solution Solute:in much more random Water:hydration process is in and disordered state. more ordered state. Net result:depends In general☑y:△S(dissolving)>0

Entropy 2. For a substance in the same state: 2 2 S >S (H O(l))( ) (H 80 C 50 ° ° O(l ( )) ) C 3. In dissolving process: possible positions of molecules S Energies of molecules  →  For liquid water: Solute: in much more random and disordered state. Net result: depends！ In generally: ∆S (dissolving) > 0 Water: hydration process is in more ordered state

Entropy 4.Gases are formed from solid or liquid: CacO(s)->Cao(s)+CO2(g) 5.Mole number of gaseous substances increases in a chemical reaction 2C(s+0,(g)→2CO(8g) C。H,①+1530，(g)→6C0,(g)+3H,00 AgCl(s)Ag"(aq)+Cl(aq) Ca,(P04)2(s)=3Ca2+(aq)+2P0(aq) N2(g)+O2(g)→2NO(g)

Entropy 4. Gases are formed from solid or liquid: 3 2 ( ) ( ) ( ) CaCO s CaO s CO g → + ∆ 5. Mole number of gaseous substances increases in a chemical reaction 2C(s)+O ( ) 2 ( ) 2 g CO g → 6 6 2 2 2 15 C H (l)+ O ( ) 6 ( ) 3 ( ) 2 g CO g H O l → + + - AgCl(s) (aq) +
Ag Cl (aq) 2+ 3 4 2 3- Ca (PO ) (s) ( Ca P 4
3 aq) + (aq) 2 O N (g)+O (g) 2NO(g) 2 2 →

Exercise 1.Give an increasing order of entropy value for I mole of following substances and explain briefly. a)CH;(CH)CH3;b)CH,CH3;c)CH(CHCH d)CHCHCH y Scm.cuScn.cn.cH.<ScH.(CH..cHScu.Cu.cm Exercise 2.Give an increasing order of the entropy value for I mole of following substances and explain briefly. a)Li;b)Cu;c)K;d Fe S(Li)<S(K)<S(Fe)<S(Cu) Atomic number:3,19,26,29 Molar mass(gmol:6.94;39.09;55.85;63.55

S < S <S < S CH CH CH CH CH CH (CH ) CH CH (CH ) CH Exercise 1. Give an increasing order of entropy value for 1 mole of following substances and explain briefly. a) CH3(CH2 ) 4CH3; b) CH3CH3; c) CH3(CH2 ) 2CH3; d) CH3CH2CH3 CH CH CH CH CH CH (CH ) CH CH (CH ) CH 3 3 3 2 3 3 2 2 3 3 2 4 3 S < S <S < S Exercise 2. Give an increasing order of the entropy value for 1 mole of following substances and explain briefly. : a) Li; b) Cu; c) K; d) Fe S( ) < S( ) <S( ) < S Li K Fe (Cu ) Atomic number: 3, 19, 26, 29 Molar mass (g/mol): 6.94; 39.09; 55.85; 63.55

The molecular interpretation of Entropy S=k.Inw The more degrees of freedom a molecule has, the greater its entropy. Which one has the greater entropy value in the following each pair of substances: (1)I mole of HCI(g)or Imole ofAr (g) (2)1 mole of H2 (g)at STP or 1mole of H2 (g)at 100C and 0.5 atm (3)1 mole of NO2 (g)at STP or 1mole of N2O (g)at STP (4)1 mole of H2 (g)at STP or 1mole of NO2 (g)at STP

The molecular interpretation of Entropy S k = ⋅ln W The more degrees of freedom a molecule has, the greater its entropy. Which one has the greater entropy value in the following each pair of substances: (1) 1 mole of (1) 1 mole of HCl (g) or 1mole of (g) or 1mole of Ar (g) (2) 1 mole of H 2 (g) at STP or 1mole of H 2 (g) at 100 °C and 0.5 atm (3) 1 mole of NO 2 (g) at STP or 1mole of N 2 O 4 (g) at STP (4) 1 mole of H 2 (g) at STP or 1mole of NO 2 (g) at STP

Reversible Irreversible Processes 可逆 非可逆 Reversible process:a system is changing in such a way that it and its surrounding can restored to their original states by exactly reversing the change. The system can be completely restored to its original state with no net change to either the system or its surrounding In an isothermal process: 等温 The heat that would be transferred if the process were reversible. l rev The temperature at which the process occurs

Reversible & Irreversible Processes Reversible process: a system is changing in such a way that it and its surrounding can restored to their original states by exactly reversing the change. The system can be completely restored to its original state with no net change to either the system or its surrounding 可逆 非可逆 the system or its surrounding In an isothermal process: rev sys q S T ∆ = The heat that would be transferred if the process were reversible. The temperature at which the process occurs. 等温

Reversible Irreversible Processes I atm,0'C H,O(s)=H,O(1) Is it a reversible process? Melting a mole of ice at 0 'C and I atm to form I mole of liquid water at 0C and 1 atm. ,无限地 By adding the heat infinitely slowly and raising the T ofsurrounding infinitesimally above 0C 无穷小 △T→0 At0C △Sn In an isothermal process: SVS T 等温 H,O(s)=H,OI)）△meling=6.01kJ1mol △H elting 1mol×6.01×103J/mol=22.0J1K T 个 273K S △S 1mol×(-6.01×103J/mol0=-22.0J/K T 273K rr △Sm=AS+ASm=22.0J1K+(-22.0J1K)=0 SVS Reversible!At equilibrium! 。.ee........... 岢逆 平衡

Reversible & Irreversible Processes In an isothermal process: rev sys q S T ∆ = 等温 Melting a mole of ice at 0 ˚C and 1 atm to form 1 mole of liquid water at 0 ˚C and 1 atm. At 0 ˚C H O H O kJ mol ( ) ( ) 6.01 / s
l ∆Η = 1 atm, 0 C ° Is it a reversible process? By adding the heat infinitely slowly and raising the T of surrounding infinitesimally above 0 ˚C . 无穷小 ∆ →T 0 无限地 2 2 H O H O ( ) ( ) s l
等温 rev melting sys sys sys q H S T T ∆ ∆ = ≈ 2 2 ( ) ( ) 6.01 / melting H O H O kJ mol s
l ∆Η = 3 1 6.01 10 / 22.0 / 273 mol J mol J K K × × = = surr r v sur re q S T ∆ − = 3 1 ( 6.01 10 / ) 22.0 / 273 mol J mol J K K × − × = = − 22.0 / ( 22.0 / ) 0 uni sys surr ∆ = ∆ + ∆ = + − = S S S J K J K ∆ = ∆ + ∆ = S S S uni sys surr 0 Reversible! At equilibrium! 可逆 平衡

Reversible Irreversible Processes Ep.:Let's calculate the entropy changes of a system and its surrounding when one mole of ice melts in the palm ofyour hand. Reversible??? H.O(s)H,O) △H melting (0C)=+6.01kjol Irreversible! Hand.3TC “g” Spontaneous or not??? At constant T: △S △Imelting 1mol×6.01×10 mol =22.0J/K T 273K △S 0心 1mol×(-6.01×103J/mol) =-19.4J/K urr T (37+273)K urr △Sm=△S+ASm=+22.0+(-19.4)=+2.6J/K>0" Spontaneous! Irreversible.！ 0,at equelibrium,reversible! AS =AS s+AS 17 >0,spontaneous,irreversible!

At constant T: constant T: (0 ) 6.01 / ∆ ° = + H C kJ mol melting Reversible & Irreversible Processes Ep.: Let’s calculate the entropy changes of a system and its surrounding when one mole of ice melts in the palm of your hand. 2 2 H O( ) ( ) s ←→ H O l Ice C ,0° Hand C ,37° Irreversible! Reversible??? rev melting sys q H S ∆ ∆ = ≈ 3 1 6.01 10 / 22.0 / mol J mol J K × × = = Spontaneous or not??? 自发 22.0 ( 19.4) 2.6 / 0 uni sys sur ∆ = ∆ + S S S J K ∆ = + + − = > + sys sys sys S T T ∆ = ≈ 22.0 / 273 J K K = = surr r v sur re q S T ∆ − = 3 1 ( 6.01 10 37 273 / ) 19.4 / ( ) mol J mol J K K × − × = = − + Spontaneous! Irreversible! 0, , ! 0, tan , ir ! uni sys surr at equelibrium reversible S S S spon eous reversible = ∆ = ∆ + ∆ >