Stoichiometry, Chemical Equilibrium

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1 Lecture 3 Stoichiometry, Chemical Equilibrium

2 Stoichiometry = Constraints for Kinetics Q?: for a prescribed amount of reactants, how can the composition evolve towards equilibrium? 1 E a 1... n elem E a nelem ns N i i i =! a N Reac =! a N Products =1 ns =1 This system has non-trivial non zero solution of the form: S! Ns = {! } =1 Nr N! N Reactants = #! Reactants S =1 stoichiometric vector of the -th reaction stoichiometric coefficient of the -th species in the -th reaction N, N Reactants number of moles, of reactants, Reactants Ns i a N Products! N Reac = 0 i=1,ne =1 Homogeneous linear system with Ne eqns Ns>Ne unnowns # $ Ns-Ne solutions # 0 rate of progress of the -th reaction, at the initial composition Ns$ Ne i { S } =1 : Basis of Null Space of the matrix A= a i=1,ne { } =1,Ns

3 Stoichiometry = Constraints for Kinetics Q?: for a prescribed amount of reactants, how can the composition evolve towards equilibrium? N! N Reac = nreac # S $ N! N Reac =1 nreac = # =1 N! N,Reac = #, =1,Nr; =1,Ns 1! dn dt d! dt = d dt nreac # = N! N,Reac =1 dn dt = r p,t, N = r f r b Constitutive law for rate of progress Mole change always proportional to!, =1,Nr; =1,Ns =! S d =1,Nr dt Rate Equation for Composition

4 How to find the stoichiometric coefficients 2 elements, 8 species An example: H 2 /O 2 Mixture H 2 O 2 H O OH H 2 O HO 2 H 2 O 2 N H = 2n 1 + n 3 + n 5 + 2n 6 + n 7 + 2n 8 N O = 2n 2 + n 4 + n 5 + n 6 + 2n 7 + 2n 8 dn H = 0 = 2dn 1 + dn 3 + dn 5 + 2dn 6 + dn 7 + 2dn 8 dn O = 0 = 2dn 2 + dn 4 + dn 5 + dn 6 + 2dn 7 + 2dn 8 ' dn 1 + dn 2 dn 3! $ dn 4 # , = Ax = 0 dn 5 dn 6 dn 7 dn 8 -

5 How to find the stoichiometric coefficients ' dn 1 + dn 2 dn ! $ dn 4 # , = Ax = 0 dn 5 dn 6 dn 7 dn 8 - i e = { e } = # 1 se i = =1,8-2=6 $ 0 se i! ' dn 1 # $ dn 2 ' = 2 0, # 0 $ dn 2+ = e d/ , dn 2+ 1 Ns Ne=6 solutions An example: H 2 /O 2 Mixture ' dn 3. dn.! 2 0$ ' dn 1 # dn 2, = -! $. dn 5. # dn 6.. dn 7... dn 8, dn 1 =! 1 # 2 d1 dn 2 = 0 $ ' 1 dn 3 = +d 1 2 H 2 = H Negative sign = reactant Positive sign = product 8-2=6 linearly independent reactions

6 Condition of Chemical Equilibrium 2 dg = Reversible zero entropy process dg =! µ dn = 0! µ d#! Stoichiometric constraint dn = d# $! =! d#! µ = 0 d#, ' 0,! µ = 0- + $ Free Enthalpy Gibbs is stationary Chemical equilibrium Nreactions. =1

7 Law of Mass Action / Equilibrium Constant Equilibrium Condition µ T, p! = 0 =1,Nreactions µ T, p :=!H T # T!S T, p i!s T, p :=!S 0 T # $ Log p µ T, p :=!H T # T!S 0 T # $ Log p ' ' =!H T # T!S 0 T + $T Log p = µ 0 T + $T Log p µ 0 T + $T Log p ' '! = 0 + µ 0 T! = # $T Log p ' '!!H T # T S! 0 T! = #$T! Log p = #$T Log p!,. Exp # 1!H T # T!S 0 T $T! = Log p!,. = Exp Log p! /. $$$$$$ # $$$$$$ 2. /. 2. = p! 3 K p T 3 Law of Mass Action K p T = p!,. Definition K p T 4 Exp -# 1 /. $T!H T # T!S 0 T!

8 Alternative Form of Equilibrium Constants Law of Mass Action K p T = p! + - with K p T # Exp, $ 1.- T Alternative form of law of mass actions: K! c = c ; K! X = X c = p / T ; X = p / p '!H T $ T S! 0 T! / = Exp + $ 2G o T /, 0. T 1 2G o T # H! ' T $ T S! 0 T! Relations among different form of equilibrium constants: K p T = K c T! $! ' = K X p! $! ' K p T = FT ; K c = FT ; K X = K p T p $! ' $! 2! #! $! ' = FT, p $ 2! K c, K p are dimensional K x is nondimensional

9 Equilibrium is a Thermodynamic Condition! Equilibrium Condition Define Equilibrium Constant Law of Mass Action The equilibrium constant of a reaction is a pure thermodynamic function!! It depends on T, p and on thermochemical properties of the species NB: pressure affects equilibrium only if µ T, p! = 0 ' K p T # Exp $ G o T + T, K X T, p # K p T / p! $! ' - = X! Equilibrium state depends on thermodynamics only!!!! ' 0 # $ '

10 Small Displacement from Equilibrium LogK p =! G o #T G o = 0 $ K p = 1 reactants products in equilibrium p = p =1,Nreac G o > 0 $ K p < 1 reaction favors reactants p > p =1,Nreac =1,Nprod G o < 0 $ K p > 1 reaction favors products p < p =1,Nreac =1,Nprod =1,Nprod

11 How to find the equilibrium state 1 Method of equilibrium constants: Easy to grasp but of difficult convergence 2 Method of minimization of Gibbs free enthalpy: Complex formulation, but fast and robust

12 Method of the Equilibrium Constants Q.?: Find Equilibrium Condition of an Adiabatic H=const, Isobaric p=const System {N } =1,8, N, T: unnown values need 10 equations to be uniquely identified H 2 O 2 H O OH H 2 O HO 2 H 2 O 2 2 Atomic Mass Conservation Equilibrium Condition for 6 N H = 2N 1 + N 3 + N 5 + 2N 6 + N 7 + 2N 8 lin. indep. reactions N O = 2N 2 + N 4 + N 5 + N 6 + 2N 7 + 2N 8 m '! p K m p T $ = # m=1,6 ; =1,8 Total number of Moles N =! =1,8 X := N N N p = N N p! N K m p T p $! = # N = #! K m p T # p $ ' m = ' m! # N N p $ ' m $ ' m! # N N Conservation of Absolute Entalpy!H T, N :=! H! T N $ ' m!h T 0, N, 0 reactants =! H T, N products Equilibrium State T, N, N p = given

13 Method of Minimization of Gibbs Free Enthalpy Stoichiometric constraint N i =! a i N i=1, Ne ; =1, Ns # Condition of Minimal Gibbs Free Enthalpy G p,t =! µ N $ ' Adoint Form Lagrange Multipliers G := G + p,t p,t! idn i i=1,ne Constrained Minimum G p,t = 0 G p,t = 0 and dni =0 for i=1, Ne After some manipulations the system is written in the unnowns i, log N, log N, p, T 1 A Ne x Ne system is solved to find the Ne multipliers 2 The remaining unnowns are found as a funtion of the multipliers PROs: the system to solve is only Ne x Ne instead of Ns x Ns as for the method of the equlibrium constants The convergence properties of this method are excellent p,t = 0

14 Enthalpy of Reaction Heat of Combustion For an isobaric, isothermal system, but NOT adiabatic, wherein one reaction transforms the composition from X 0 to X, the first principle reads dp=0! Q! = d H! Vdp! T = d H! with H! T, X = H! ref # X + # X!c, p T dt! Q! = d H! dt =0 T, X = H! ref # dx + # X!c, p T dt = H! ref # dx Enthalpy of Reaction H! Reaction = H! T, X H! T, X 0 = H! ref # X H! ref 0 # X Heat of Combustion!Q Comb =! H Reaction J ' mol +!H Reaction < 0!Q Comb > 0 Exothermic Reactions!H Reaction > 0!Q Comb < 0 Endothermic Reactions Ĥ Reaction,Fuel =!H Reaction W Fuel KJ ' Kg-Fuel + Ĥ Reaction,Mix =!H Reaction W Mix $ T ref KJ ' Kg-Mix +

15 Equilibrium Adiabatic, Isobaric Temperature For an isobaric and adiabatic system, wherein many reactions transform the composition from X 0 to X, the first principle reads!!q,dp=0! Q! = d H! Vdp! T # d H! = 0 with H! T, X = H! ref $ X + $ X!c, p T dt 0 = H! T, X H! T 0, X 0 = H! ref $ X + $ X ' Change in sensible enthalpy T T ref T ref!c, p T dt + H! ref $ X $ X ' T T 0 0 $ X!c, p T dt $ X!c, p T dt + = H! ref $ X 0 H! ref $ X + ' T ref T ' # $ ref # $ T, p Adiabatic Isobaric Temperature Heat of Combustion T 0 T ref!c, p T dt +

16 Equilibrium Adiabatic, Isocoric Temperature For an isocoric and adiabatic system, wherein many reactions transform the composition from X 0 to X, the first principle reads!!q,d!v =0! Q! = d U! + pd V! d!u = 0 0 =!UT, X #!UT 0, X 0 #!H T 0, X 0 # $T 0 =!H T, X # $T = H! T, X # H! T 0, X 0 # $ # $ T # T 0 Enthalpy of Combustion T,v Adiabatic Isocoric Temperature

UNIT 15: THERMODYNAMICS

UNIT 15: THERMODYNAMICS ENTHALPY, DH ENTROPY, DS GIBBS FREE ENERGY, DG ENTHALPY, DH Energy Changes in Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures.