CHE 330 Thermodynamics March 10 Friday Instructors: Prof. Nicholas A. Kotov, email@example.com Bean Getsoian, firstname.lastname@example.org Raghu Kainkaryam, email@example.com Homework Homework 6.10 and 6.21, 6.24 due March 17 Example for Chapter 6 C6H6 + 3H2 ? C6H12 Standard heat of reaction? Let?s calculate it from the heats of combustion. C6H6 + 3H2 ? C6H12 3,267,620 J/mol 3,919,906 J/mol 285,840 J/mol ?Hrxn = -?H(C6H12)+ ?H(C6H6)+ 3?H(H2) -205,234 J/mol New Concepts From the Last Class ?Horxn (250C, 1 bar) = [2Hof,HNO3 + Hof,NO -3Hof,NO2-Hof,H2O ](250C, 1 bar) ?Gorxn (250C, 1 bar) = ?vi?Gof,i (250C, 1 bar) ?Sorxn (250C, 1 bar) = ?vi?Sof,i (250C, 1 bar) ?Horxn (250C, 1 bar) = ?vi?Hof,i (250C, 1 bar) ?Horxn (T, 1 bar) = ?vi?Hof,i (250C, 1 bar) + + ?vi? CoP,idT Criteria for Equilibrium in Multicomponent Systems dU/dt = . Q + Ws ? P(dV/dt) . dS/dt = Q/T + Sgen . . S = maximum for equilibrium at constant M, U, and V A = minimum for equilibrium at constant M, T, and V G = minimum for equilibrium at constant M, T, and P Closed Adiabatic System, V = const S = maximum Ni = NiI + NiII dS = (1/T)dU + (P/T)dV ?(1/T)?Gi dNi i=1 L dUII = -dUI dVII = -dVI dNII = -dNI dS = dSI + dSII = 0 dU = TdS - PdV + ?GidNi j=1 L Criteria for Equilibrium (?S/?UI)Vi, N i I = 0 TI = TII (?S/?VI)Ui, N i I = 0 PI = PII (?S/?NiI)Ui,VI, N I i?j = 0 GiI = GiII dS = (1/T)dU + (P/T)dV ?(1/T)?Gi dNi i=1 L T, P = Const What if we have constant T, P in a closed system ? dG = dGI + dGII = 0 GiI = GiII dG|T,P =?Gi dNi i=1 L dNII = -dNI Case of Chemical Reaction G =?Ni Gi i=1 L (Ni,0 + vX)Gi 0 =?vi Gi i=1 L 0 =?vi,j Gi i=1 L G =? i=1 L Closed System with T=const, P = const (?G/?X)T,P = 0 L 0 = ?xi dGi i=1 Many Independent Reactions G =?Ni Gi i=1 L Ni = Ni,0 +?vijXj j = 1 M dG = 0 L 0 = ?xi dGi i=1 Gibbs-Duhem Equation 0 = (?G/?Xj)T, P, X i?j 0 =?vi,j Gi i=1 L For many reactions 0 =?vi Gi i=1 L For one reactions Combined Phase and Chemical Equilibrium The equilibrium will establish in a closed system (T = const, P= const) when the equilibriums between all the reactions and all the phases are established. Gibbs Phase Rule The thermodynamic equilibrium in a single-phase system is set by the values of two intensive variables and (L-1) concentrations. If there are P phases, then the number of variables appears to be P(L+1) Temperature, pressure and chemical potential Should be the same in all phases. New restrictions: P-1 (temperature) P-1 (pressure L(P-1) chemicals in phases M chemical reactions F = P(L+1) ? [2(P-1) + L(P-1) + M] = L ? M ? P + 2 Chapter 7, Gibbs Energy in Mixtures Ideal Gas Mixture PV IGM =(N1 + N2 + ?)RT = (?Ni)RT = NRT j=1 L UIGM (T,N) = (?Ni)UjIG(T) j=1 L UiIGM (T,x) = ?UIGM(T,N)/?Ni|T,P,Nj?I = = ?/?Ni|T,P,Nj?I ?NiUjIG(T) = UjIG(T) j=1 L ViIGM (T,P,x) = ?VIGM(T,P,N)/?Ni|T,P,Nj?I = = ?/?Ni|T,P,Nj?I ?NiRT/P =RT/P = = VjIG(T,P) j L PiIGM (N,V,T,x) = NiRT/V =RT/P = = PjIG(Ni,T,P) SiIGM (T,P,x) ? SiIG(T,P) = -Rln(Pi/P) = = -Rlnxi SiIGM (T,V,x) ? SiIG(T,Vi) = -Rln(V/Vi) = = -Rlnxi Consequently, ?SmixIGM = ?Ni[(SiIGM(T,P,x) ? SiIG(T,P)] = -R?Nilnxi i=1 L i=1 L
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