1、Chapter 18 Thermodynamics In this Chapter,we will address several aspects as follows:I)How to predict whether a reaction under a given conditions can occur or not?II)How to make the reaction happen,if it does not proceed under primary conditions?p Spontaneity and Entropy(S).p The Second and Third La
2、ws of thermodynamics.p Free Energy(G)and Equilibrium(K).Spontaneous process:spontaneousnon-spontaneous18.2Go ahead by itself,without help from surroundings.spontaneousNon-spontaneousWhat are the factors controlling the spontaneity?CH4(g)+2O2(g)CO2(g)+2H2O(l)DH0=-890.4 kJH+(aq)+OH-(aq)H2O(l)DH0=-56.2
3、 kJH2O(s)H2O(l)DH0=6.01 kJNH4NO3(s)NH4+(aq)+NO3-(aq)DH0=25 kJH2O Energy Freedom Entropy(S):a measure of system disorder.orderSdisorderSDS=Sf-SiDS 0For a given substance:Ssolid Sliquid 018.2Heating hydrogen gas from 600C to 800CCrystallization from a supersaturated solutionDS 0For a process:Entropy i
4、s also a State function.E1H1S1E2H2S2n1 P1V1T1n2 P2V2T2State 1State 2DE=E2 E1DH=H2 H1DS=S2 S1The value of S is only determined by the state,regardless of how the state is reached.How to quantify the Entropy?MicrostatesW=1W=4W=6SIn 1868,Boltzmann proposedS=k ln WDS=Sf-SiFor a process:DS=k lnWfWiBoltzm
5、ann constant k=1.3810-23 J/KFor CH3OH and C2H5OH:Smethanol 0(1)Spontaneous process:DSuniv=DSsys+DSsurr=0(2)Equilibrium process:In order to predict the direction of a chemical reaction,however,both DSsys and DSsurr need to be calculated,which is a question remaining to be solved.How to calculate DSsu
6、rr,the entropy change in the surroundings?Endothermic DHsys 0DSsurr DHsysD DSsurr 0For a constant-pressure process,it is known that Qp=DH.Exothermic DHsys 0D DSsurr 0for a spontaneous process:DSuniv=DSsys-0surrsysTHDTsurr DSuniv=Tsurr DSsys-DHsys 0-Tsurr DSuniv=DHsys-Tsurr DSsys 0DG 0 non-spontaneou
7、s DG=0 at equilibrium.Definition:G=H-T S G is called Gibbs free energydG=dH T dS S dTat constant P and T:DHsys-Tsys DSsys 0For a process at constant P and T DGsys=DHsys Tsys DSsysNote that S has a true and absolute value,which can be determined by experiment,or by statistics.This is quite different
8、from E and H,the values of which are not available by experiments.For any perfect crystalline substance at 0 K,the entropy S=0.The Third Law of Thermodynamicsp the particles(atom,molecule or ion)arranged in space in a mostly optimized way;p No defect sites both in bulk and on the surface;p Free of i
9、mpurities.Under such conditions,only one way to pack the particles.That is:W=1S=k lnW S=0and 1 atmCalculate the standard entropy change for the following reaction at 250C?2CO(g)+O2(g)2CO2(g)DS0rxn=2 S0(CO2,g)2 S0(CO,g)+S0(O2,g)=427.2 395.8+205.0=-173.6 J/K=2 x(-393.5)2 x(-110.5)+0 =-566.0 kJ=2 (CO2,
10、g)2 (CO2,g)+(O2,g)DH0fDH0fDH0fDH0rxnIs the reaction spontaneous?DGsys=DHsys Tsys DSsysDGorxn=(-566.0 kJ)(298 K)x(-173.6 J/K)Yes,the above reaction is spontaneous at 25 0C.=-514.3 kJ 0,for the production of 1 atm CO2.(2)If the reaction proceeds,DG should be less than zero.Since DG=DH TDS DH TDS SHDDt
11、hen T trans SHDDT trans 1108 K (835 oC)177.8 1000160.5CaCO3(s)CaO(s)+CO2(g)Estimate the boiling point of water?H2O(l)H2O(g)DG =DG (H2O,g)DG (H2O,l)0rxn0 f0f(1)=-228.6(-237.2)=8.6 kJWater can not evaporate into vapor at 25 oC with 1 atm pressure.(2)(2)DG=DH TDST boil 00(298)(298)44.0 1000370 (97)18.8
12、oHHKSSKKCDDDDWater will boil at 97 oC at 1 atm.For a chemical reaction under constant P and constant TaA+bB cC+dD(1)At a standard state(Pi=1 atm,Ci=1 M)and T=298 K.DHrxn(T)DHrxn(298 K),and DSrxn(T)DSrxn(298 K)ooooAssume:The reaction temperature can be estimated first by:(2)At a standard state(Pi=1 a
13、tm,Ci=1 M)and T=a K.(3)At any conditions(T 298 K,Pi 1 atm or Ci 1M)Here R is gas constant(8.314 J K-1 mol-1)DGrxn(T,P)=DGrxn(T)+RT ln Q obBaAdDcCCCCCCCCCCQQ 0000bBaAdDcCPPPPPPPPPQQ 0000P(pa),and Po=101325 paC(mol/L),and Co=1 mol/LaA+bB cC+dDDoes NH4HCO3 decompose in air?DGrxn=DGf(NH3)+DGf (H2O,g)+DG
14、f (CO2,g)-DGf (NH4HCO3,s)ooooO=31.1 kJNo reaction at 25 oC with Pi=1 atmIf in air each occupies 10%,Qp=(PNH3/P0)(PCO2/P0)(PH2O/P0)=0.001 =31.1+8.314 x 0.298 ln(0.001)=-3.1 kJ In general,If DGorxn|40|kJ,It is possible to reverse the reaction by changing the reaction conditions.DGrxn(T,P)=DGrxn(T)+RT
15、ln QoNH4HCO3(s)NH3(g)+H2O(g)+CO2(g)It will occur at 25 oC with Pi=0.1 atmDG=DG0+RT lnQAt an equilibriumDG=0 Q=K0=DG0+RT lnKDG0=RT lnKaA+bB cC+dDequilibrium constantDG0=RT lnKp K can be calculated from DG0 and vice versa;But it does not means that the reaction occurs at a standard state.p K is only a
16、 function of temperature.DGrxn(T)=DHrxn(T)T DSrxn(T)ooo-RT lnK=DHrxn(T)T DSrxn(T)ooRTSRTTHKrxnrxn)()(ln00D D D D D D 2112012)298(lnTTTTRKHKKrxnDHrxn 00T ,K T ,K DHrxn(T)DHrxn(298 K),and DSrxn(T)DSrxn(298 K)ooooAssume:Summary:aA+bB cC+dDDS0rxndS0(D)cS0(C)=+-bS0(B)aS0(A)+DH0rxndDH0(D)fcDH0(C)f=+-bDH0(B)faDH0(A)f+(1)The standard enthalpy(2)The standard entropyDG0rxndDG0(D)fcDG0(C)f=+-bDG0(B)faDG0(A)f+(3)The standard Gibbs free energy(4)DGrxn=DHrxn T DSrxnDG=DG0+RT lnQT rxn)298()298(00KSKHSHDDDDDG0=RT lnKRTSRTTHKrxnrxn)()(ln00DDPractice Exercises:18.11;18.1818.2918.5018.7818.80
