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Chapter 13 (week 8) Thermodynamics and Spontaneous Processes

Chapter 13 (week 8) Thermodynamics and Spontaneous Processes Ch 9, 10, (11-1-11.3, 11.5), 12.-.7, 13(Ex .5 & .8), 14(Ex .4 & .8), 15.-15.3. • Gibbs Free Energy: G= H -TS  G=  H – T  S @ P=const and T=const  G < 0 for Spontaneous Processes H< and S>0 always spontaneous

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Chapter 13 (week 8) Thermodynamics and Spontaneous Processes

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  1. Chapter 13 (week 8) Thermodynamics and Spontaneous Processes Ch 9, 10, (11-1-11.3, 11.5), 12.-.7, 13(Ex .5 & .8), 14(Ex .4 & .8), 15.-15.3 • Gibbs Free Energy: G= H -TS G= H – TS @ P=const and T=const G < 0 for Spontaneous Processes H< and S>0 always spontaneous G = 0 for Equilibrium Processes(S=H/T) G > 0 the Reverse is a Spontaneous Processes So for AB spontaneously G(A)>G(B) @ Equilibrium AB G(A)=G(B) • Equilibrium constant K(T)=exp{-G°/RT} G° is the standard Gibbs free energy for any rxn dD+bB  fF + eE In terms of the activity (a) K(T)=(aF)f(aE)e/(aD)d(aB)b for gases aE = PE(atm) in soln aE = [E](molar), for pure liquids/solids aE = 1 G°= {fGf°[F] + eGf°[E]} – {dGf°[D] + bGf°[B]} also K=[E]e[D]d/[A]a[B]b Law of Mass Action can also be written in terms of PE(atm) etc not @ equilibrium Q=[E]e[F]f/[D]d[B]b Reaction Quotient G=RTln(Q/K) wow! • Le Chateliers Principle: systems at Equilibrium will counteract any change • van’t Hoff Eq: ln(K2/K1) =–(H°/ R)[(1/T2) – (1/T1)] Assumes H° and S° are not functions of T

  2. Heat transfer from the surroundings (Tsur) to the system (Tsys) Universe Surroundings System q If Tsur ≠Tsys energy flows form hot to cold spontaneously? By the 1st Law Energy could flow from cold to hot! But by the 2nd Law Suni > 0,Entropy always increases in a spontaneous processes! So what is Entropy?

  3. POKER ENTROPY State =straight W=10,200 State=3 of a kind W=36=54,912 State = 2 pairs W= 123,552 State= 1 pair W=1,098,240 State= high card W=1,303,540 Total possible 5 Combo 2,598,960 State =Royal Flush AKQJ10 1 suit W=4 micro states State=Straight flush 12345 <3 1 suit W=36=10*4-4 State = 4 of a kind W=624 52*3*2*1*48 4*3*2*1 State=full house W=3744 State= flush W=5108 52*51*50*49*48/5*4*3*2*1 =2,598,960 Hands are indistinguishable 5!= 5 Factorial

  4. Boltzmann S =R lnW S S =R lnW W=W W

  5. S°/4.18 JK-1 mol-1 groups J

  6. An Irreversible Process

  7. Boltzmann Entropy S = N k lnW Difficult to use for quantitative work (at least at this level) Entropy Change for Reversible Isothermal process in an ideal Gas expansion: S = q/T= [nRT ln(V2/V1)]/T q is the heat transferred reversibly! Compare the Boltzmann and Thermodynamic changes Consider one Molecule going from V1 to V2 Using the Boltzmann Entropy S= k ln  Find S! The number of ways that molecule could occupy the Volume is proportional to the size V so V If it’s 1 Dimensional motion,  L The number of possible momentum states is obtained from the internal energy U In 3-D U = (1/2m){(px)2 + (py)2 + (pz)2}, for In 1-D motion it is ±px = U1/2 so  UL

  8. For 3-D ~ [(UL]3= UL3 for one (mono-atomic) particle And now for N such independent particles ~ UL3N But L3= V and since the process is isothermal U=(3/2)NkT = constant since T=const

  9. Going from Volume V1 to V2 and T=const = # UV1) N and = # UV2) N # is some arbitrary collection of constants. Boltzmann Absolute Entropy: S=kln S1 = kln {# U(V1) N} And S2 = k ln {# UV2) N}

  10. Absolute entropy S=kln  S=S2 – S1 = k ln [#(UV2) N]/ [#(UV1) N] S2–S1= kln(V2)N/ (V1)N =kln(V2/V1)N S=Nkln(V2/V1)=nN0kln(V2/V1) R=N0k and n =N/N0 S= n Rln (V2/V1) Same result as the Isothermal Expansion using S=qrev/T consistent S=kln 

  11. Reversible Isothermal Expansion Of an Ideal Gas q = nRTln(V2/V1) S=qrev / T = nRln(V2/V1) Irreversible Expansion dV V1 V V+dV V2

  12. Consider the Reversible Isothermal Expansion of an Ideal Gas: So T=const for the system. U = q + w = 0 q = -w = (PextV) Heat must be transferred from the Surroundings to the System And the System does work on the surroundings If the process is reversible, Pext ~ P=nRT/V Which means done slowly and changes are infinitesimal V dV: The gas expands from V1 to V2 then q ~ PdV= (nRT/V)dV = (nRT)(dV/V)=(nRT)dlnV since dlnV=dV/V Integrating from V1 to V2q = nRT[lnV2 - lnV1]=nRTln(V2/V1) (q) Heat transferred in the Reversible Isothermal Expansion of an Ideal Gas q = nRTln(V2/V1)

  13. Electron Translation and Atomic Vibrations(phonons) Store Energy in Metals By the 3rd Law S 0 as T 0 Standard Entropy with phase transition and P=const T S°(T) =0∫ ncpdT/T + nSfus + nSvap where Svap= Hvap/TB and Sfus= Hfus/TB

  14. S=S2– S1 for Heat Transfer Processes at P=const and V=const Much easier to use S =∫dqrev/T than S=kln For Isothermal Processes: S =(1/T) ∫dqrev=(1/T) qrev qrev=∫dqrev For heat transfer at V=const with no reactions or phase transitions dqrev = ncvdT and S =∫ ncvdT/T = ∫ ncvd(lnT) If cv≠f(T) then S =ncv lnT2/T1 For heat transfer at P=const with no reactions or phase transitions dqrev = ncpdT and S =∫ ncpdT/T = ∫ ncpd(lnT) If cv≠f(T) then S =ncplnT2/T1

  15. So in a Spontaneous Process the heat absorbed by the system from the surroundings is always less than TS The Clausius Inequality S=qrev / T S > qirr / T S ≥ q / T where q is the heat transferred to the system reversibly or irreversibly So for an Isolated system, e.g., the Universe, since q=0 S ≥ 0 for all processes

  16. Consider an Isothermally process @ T with P=const Since quni=0 and Suni = Ssys + Ssur ≥ 0 That is even if quni≠0 Suni≥ quni /T ≥ 0 Suni> 0 for Irreversible Processes(Spontaneous) Suni= 0 for Reversible Processes(equilibrium) Suni< 0 not possible(reverse process is spontaneous) Ssur = qsur /T= – qsys /T = – Hsys /T Ssys– Hsys /T ≥ 0 or [Hsys – TSsys]/T ≤ 0 Define the Gibbs Free energy as G=H - TS @ P and T constant G=[Hsys –TSsys] ≤ 0 or G ≤ 0)

  17. For the Universe Suni> 0 for Irreversible Processes(Spontaneous) Suni= 0 for Reversible Processes(Equilibrium) Suni< 0 not possible(reverse process is spontaneous) For the System G< 0 for Irreversible Processes(Spontaneous) G= 0 for Reversible Processes(Equilibrium) G> 0 not possible(Reverse Process is Spontaneous)

  18. Entropy of Mixing(H=0) System A/B separated by an impermeable membrane: No flow of matter until after the membrane is removed. The surroundings maintain the fixed T and P. Once the membrane is removed A and B will mix spontaneously by diffusion S=S(mixed) – S(unmixed) > 0 G=[H –TS]<0, andH=0 for simple mixing of rare gases for example! So the final state is more disordered than the initial state, which Explains colligative properties like Osmosis G=–TS<0 since S>0 Now cases for H≠0 phase transitions Melting , Boiling etc Thermodynamic Universe

  19. Spontaneous Phase Transitions G(g) G(l) liquid gas G < 0 G < 0 G gas liquid Temperature

  20. Phase Transitions Equilibrium G(g) G(l) G liquid gas Tb Temperature

  21. G=0 for Coexistence Curves. Gas and Liquid phase have the same G above the critical pt. and all 3 phase @ Triple pt . Now Consider H2O(l)H2O(s) H= - Hfus H< 0 S= H/Tm= – Hfus/Tm<0 S= – Sfu< 0 G = -Hfus – T(-Sfu) G = -Hfus + TSfu

  22. Chemical Reactions, the Standard Free energy G° Change can be calculate using Hess’ Law us the standard free energy of formation Gf° which is zero for elements in their standard states Gf° [N2(g)]=0, etc.. G°=H° - TS° For the combustion reaction: CH4(g) + 2O2 (g)  CO2(g) + 2H2O(l) G° =? G°={2Gf°[H2O(l)]+Gf°[CO2(g)]} - {Gf°[CH4(g)]+ 2Gf°[O2(g)]} Gf°[O2(g)]=0 And H°={2Hf°[H2O(l)]+Hf°[CO2(g)]} - {Hf°[CH4(g)]+ 2Hf°[O2(g)]} Hf°[O2(g)]=0 However, the entropy is not zero for the elements in their standard states! S°[O2(g)]=S°[O2(g)]=205.3 JK-1≠ 0 S°={2S°[H2O(l)]+S°[CO2(g)]} - {S°[CH4(g)]+ 2S°[O2(g)]}

  23. Assuming again that H° and S° are Independent T If both H° and S° are Negative or positive There is a T*=H°/S° where G° =0 if H° and S° < 0 The rxn is spontaneous T<T* etc. H° and S° > 0 rxn is Spontaneous T>T*. H°> 0 and S°< 0 Rxn is never spontaneous! H°< 0 and S° > 0 the Rxn is always spontaneous!

  24. The Law of Mass Action: aA(g)bB(g) @ Equilibrium K= (PB)b/(PA)a P(atm) aA(aq)bB(aq) @ Equilibrium K=kf/kr= [B]b/ [A]a [molar] Which can be determined from the rate laws of the overall forward and reverse reactions. Can also be determined from thermodynamics K(T)=exp{-G°/RT}

  25. Consider the Reaction dD(g)bB(g) At a pressure PD and PB for products dD(g)bB(g) but H° and G° etc are for P=1 atm need H° at others pressures. This can be determined since enthalpy is a state function! G bB(PB) dD(PD) G2 G1 dD(PD=1 atm) bB(PB=1 atm) G° G= G1 +G° + G 2

  26. Reaction Among Ideal Gases at P≠1 atm, i,e, Not at Standard Pressure conditions G= H – TS G= – TS, since H = 0 S=q/T and q= nRTln(V2/V1) G= –nRTln(V2/V1) Now with P=nRT/V G= –nRTln(P1/P2) G= nRTln(P2/P1) Let P1= Pref G= nRTln(P/Pref ) If Pref=1 atm G= nRTln(P ) P1 P2

  27. dD (PD) bB (PB) dD (PD=1atm) bB (PB=1atm) dD(g) bB(g) G3= bTRln(PB/1) G3= RTln(PB)b G1= dRTln(1/PD) G1= Rln(1/PD)d G2= G°= bGf°[B] - dGf°[D] G= G1 + G° + G3 =G°+RT{ln(1/PD)d + ln(PB)b}= G° +RT{ln(PB)b/(PD)a} G = G° +RT{ln(PB)b/(PD)d} Let Q= (PB)b/PD)d be the reaction Quotient So the Reaction not at Equilibrium G = G° +RT{lnQ}≠0 Fig. 14-4, p. 581

  28. Reaction not at EquilibriumdD(g)  bB(g) G = G° +RT{lnQ} Reaction at EquilibriumdD(g)  bB(g) G =0= G° +RT{lnQ} and G° = -RTlnK From Kinetics at Equilibrium K= (PBeq)b/PDeq)d Therefore at Equilibrium Q=K and G° = -RTlnK or K(T) = exp{G°/RT}= (PBeq)b/PAeq)a

  29. In GeneralaA(g)  bB(g) G = G° +RT{lnQ} and Q= (PB)b/PA)a is the Reaction Quotient G° = -RTlnK then G = -RTlnK+RT{lnQ} = RTln(Q/K) G = RTln(Q/K) If Q<K, Q/K<1 and G <0 aA(g)  bB(g) Spontaneous If Q=K, Q/K=1 and G=0 aA(g)  bB(g) Equilibrium If Q>K, Q/K>1 and G >0 bB(g)  aA(g) Spontaneous

  30. G G>0 G<0 G=0 Gmin Q=K Q<K Q>K

  31. K Small: Reactants favored at Equilibrium Equilibrium to the Left G G<0 G>0 G=0 Gmin Q=K Q>K Q<K Q

  32. K Large: Product Favored Equilibrium to the Right G>0 G<0 G=0 G Gmin Q>K Q<K Q=K Q

  33. For a generic overall Gas Phase Reaction dD + bB  eE + fF Q= (PE)e(PF)f/ (PD)d(PB)b And at Equilibrium K= (PEeq)e(Pfeq)f/ (Pdeq)d(PBeq)b Pressure Measured in atm! G° = eGf°[E] + fGf°[F] - bGf°[B] - dGf°[D]

  34. Generalize to Solids, Liquids and Heterogeneous mixed phase reactions. Lets define the activity “a”a=P/Pref SoG= nRTln(P/Pref) goes G= nRTln(a) For none ideal gases ai=i (Pi/Pref)=i Pi Pref= I atm where I = activity the activity coefficient is different For different gases, Ar ≠ Ar Reactions in Solution G= nRTln(a) ai=i ([A]i/[A]ref)= i [A] [A]ref =1 Molar In any case a=1 for pure Solids and Liquids K= (aE)e(aF)f/(aD)d(aB)b

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