Statistical Thermodynamics. Lecture 8: Theory of Chemical Equilibria(I)

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1 Statistical Thermodynamics Lecture 8: Theory of Chemical Equilibria(I)

2 Chemical Equilibria A major goal in chemistry is to predict the equilibria of chemical reactions, including the relative amounts of the reactants and products from their atomic masses and structure properties. partition functions chemical potential equilibrium constant K reactant A B product Two-state equilibria include: chemical isomerization; the folding of a protein from an unfolded to folded state The equilibrium constant K is the ratio of numbers of particles in each of the two states at equilibrium. K = N B /N A The quantity that predicts the chemical equilibria is the chemical potential

3 Equilibria Condition Consider the Gibbs free energy of the system: dg = SdT +Vdp + µ A dn A + µ B dn B At constant T and P: dg = µ A dn A + µ B dn B = 0 Every molecule is in either state A or B: N A + N B = N total = constant dn A + dn B = 0 dn B = dn A dg = µ A dn A µ B dn A = (µ A µ B )dn A = 0 Condition for equilibria: µ A = µ B

4 t Partition Function for Chemical Reactions q = exp( ε i / kt ) = exp( ε 0 / kt )+ exp( ε 1 / kt ) exp( ε t / kt ) i=0 Define a reduced partition function with the ground-state term factored out: q reduce = exp(ε 0 / kt )q =1+ exp[ (ε 1 ε 0 ) / kt ]+ exp[ (ε 2 ε 0 ) / kt ] exp[ (ε t ε 0 ) / kt ] Partition function for chemical potential µ = kt ln( q N )

5 µ A = kt ln( q A N A ) µ B = kt ln( q B N B ) q From condition for equilibria: µ A = µ A B = q B N A N B K = N B = q B = q reduce,b exp( ε 0,B / kt ) N A q A q reduce,a exp( ε 0,A / kt ) = q reduce,b exp[ (ε 0,B ε 0,A ) / kt ] q reduce,a The above equation gives a way to compute K from the atomic properties of A and B through their partition functions. Takes no account of the interactions of one molecule with another, it applies only to isolated particles such as those in the gas phase.

6 More Complex Equilibria K aa+bb cc a, b and c indicate the stoichiometries of species A, B and C. At constant T and P: dg = µ A dn A + µ B dn B + µ C dn C = 0 The equilibrium is now subject to two stoichiometric constraints: N A a + N C c = constant dn A = a c dn C N B b + N C c = constant dn B = b c dn C dg = a c µ AdN C b c µ BdN C + µ C dn C = 0 a c µ A b c µ B + µ C = 0

7 Condition for equilibria: cµ C = aµ A + bµ B c[ kt ln( q C N C )] = a[ kt ln( q A N A )]+ b[ kt ln( q B N B )] ( q C N C ) c = ( q A N A ) a ( q B N B ) b To express the relative numbers of particles of each species present at equilibrium, a natural definition of the equilibrium constant K arises from rearranging the above equation. K = N c C N a AN b B = qc C q a Aq b B = q c reduce,c q a reduce,aq b reduce,b exp[ (cε 0,C aε 0,A bε 0,B ) / kt ] difference in ground-state energies: Δε 0

8 Finding Ground-State Energies To define the ground state energies further, we must resolve a matter of the vibrational ground state. Vibrational energies: ε vib = (n )hν n is the quantum number, n = 0, 1, 2, 3 ν = (1/ 2π )(k s / m) 1/2 is the vibrational frequency of the harmonic oscillator Define Ground-State energy as: dissociated state with energy ε dis ε 0 = ε (well-bottom state) ε(dissociated state) = ε dis Define dissociation energy as: D = ε(dissociated state) ε(zero-point state) = ε dis (1/ 2)hν zero-point state with energy (1/2) hν well-bottom state with energy 0 D = ε 0 (1/ 2)hν

9 q vib = exp( hν / 2kT ) 1 exp( hν / kt ) exp( hν / 2kT ) q vib exp( ε 0 / kt ) = exp[(d + hν / 2) / kt ] 1 exp( hν / kt ) 1 = 1 exp( hν / kt ) exp(d / kt ) = q VZ exp(d / kt ) K = q c reduce,c q a reduce,a qb reduce,b exp[ (Δε 0 ) / kt ] q reduce = q tran q rot q vib = q c tran,c q c rot,c q c VZ,C exp[(cd C ad A bd B ) / kt ] q a tran,aq b tran,b q a rot,aq b rot,b q a VZ,Aq b VZ,B = q c reduce VZ,C q a reduce VZ,A qb reduce VZ,B K = K t K r K v exp(δd / kt ) exp(δd / kt ) ΔD: difference between the dissociation energies of all the products and all the reactants

10 Example: Estimation of K K H 2 +D 2 2HD The dissociation energies are kj/mol for H 2 ; kj/mol for D 2 ; and kj/mol for HD K t = ΔD = 2D HD D H2 D D2 = 0.6kJ / mol So at T = 300K: exp(δd / RT ) = 0.79 [(2πm HD kth 2 ) 3/2 ] 2 [(2πm H2 kth 2 ) 3/2 ] 2 [(2πm D2 kth 2 ) 3/2 ] 2 = ( m 2 HD m H2 m D2 ) 3/2 = ( )3/2 =1.19 [(8π 2 I K r = HD kt ) / (σ HD h 2 )] 2 [(8π 2 I H2 kt ) / (σ H2 h 2 )] 2 [(8π 2 I D2 kt ) / (σ D2 h 2 )] = (σ H 2 σ D2 2 σ HD )( I 2 HD I H2 I D2 ) = 4 ( ) = 3.56

11 K v = [1 exp( hν HD / kt )] 2 =1 [1 exp( hν H2 / kt )] 1 1 [1 exp( hν D2 / kt )] Since at T = 300K, (hv)/kt >> 1 There, K = K t K r K v exp(δd / kt ) = = 3.35 Rotation part contributes most, specially, the change in rotational symmetry is the main contributor to this equilibrium. The reaction is driven by the gain in entropy due to the rotational asymmetry of the products.

12 Pressure-Based Equilibrium Constants Because pressures are easier to measure than particle numbers for gases, it is often more convenient to use equilibrium constants K p based on pressures rather than equilibrium constant K based on the particle numbers K = N = pv / kt N c C N a A N b B = (p C V / kt ) c (p A V / kt ) a (p B V / kt ) b = q c reduce VZ,C q a reduce VZ,A qb reduce VZ,B exp(δd / kt ) Multiple both sides by (V / kt ) a+b c K p = pc C p a A pb B (q = (kt ) c a b reduce VZ,C /V) c exp(δd / kt ) (q reduce VZ,A /V) a b (q reduce VZ,B /V)

13 Chemical Potentials in terms of Partial Pressures µ = kt ln( q N ) = kt ln( q pv / kt ) = kt ln( p p 0 int ) = µ 0 + kt ln p Here p 0 int = qkt V µ 0 = kt ln( qkt V ) It divides the chemical potential into a part that depends on pressure ktlnp, and a part that does not µ 0 µ 0 is called the standard-state chemical potential, which is depend on temperature

14 Le Chatelier s Principle Any perturbation away from a stable equilibrium state must increase the free energy of the system. The system will respond by moving back toward the state of equilibrium. K A B Suppose a fluctuation changes the number of B molecules by an amount dn B. dg = (µ B µ A )dn B ζ = N B / (N A + N B ) Define a reaction coordinate the system has proceeded to B. The total number of particles is fixed: dg = (µ B µ A )Ndζ, the factional degree to which N A + N B = N, so N B = Nζ dn B = Ndζ To move toward equilibrium, dg 0, so (µ B µ A ), dζ must have opposite signs. µ B > µ A dζ If, then the direct toward equilibrium is < 0, N B will decrease. Le Chatelier s principle refers to the tendency of system to return to equilibrium by moving in a direction opposite to that caused by an external pertuibation

15 Temperature Dependence of Equilibrium Measure K(T) at different temperatures, learn the enthalpy and entropy of the reaction, which is useful for constructing or testing microscopic models. K A B At constant T and P, the condition for equilibrium is µ A = µ B The pressure based equilibrium constant is K p = p B p A µ 0 A + kt ln p A = µ 0 B + kt ln p B Recall ln K p = ln( p B ) = (µ 0 B µ 0 A) p A kt µ 0 = kt ln( qkt V ) = Δµ 0 kt, which depends on temperature

16 G = H TS Chemical potential can be divided into partial molar enthalpy and entropy components. ( ln K p T µ = ( G N ) T,P = ( H N ) T,P T( S N ) T,P = h Ts Δµ 0 = Δh 0 TΔs 0 ) = 0 T (Δµ kt ) = TΔs 0 T (Δh0 kt If Δh 0 = h 0 0, are independent of temperature: B h A Δs 0 = s 0 0 B s A ) ( ln K p T ) = Δh0 kt 2

17 ( ln K p T ) = Δh0 kt 2 ( ln K p (1/ T ) ) = Δh0 k The relation provides a useful way to plot data to obtain Δh 0 van t Hoff relation van t Hoff plots show lnk versus 1/T, the slope is Δh 0 / k Water is more dissociated at higher temperatures For dissociation, Δh 0 >0, is the characteristic of bond-breaking processes The enthalpy change is positive, so dissociation must be driven by entropy The plot illustrates a common feature: they are often linear, so Δh 0 is independent of temperature H 2 O H 2 +(1/ 2)O 2

18 ( ln K p (1/ T ) ) = Δh0 k integration ln( K p (T 2 ) K p (T 1 ) ) = Δh0 k ( 1 T 2 1 T 1 ) Can be used to find how K p depends on temperature if know Δh 0 or determine Δh 0 if K p (T) is measured Example: calculate enthalpy of dissociation of water: At T = 1500K, lnk p = ; at T = 2257 K, lnk p = -6.4 Δh 0 = Rln( K p(t 2 ) K p (T 1 ) ) / ( 1 1 ) = JK 1 mol ( 6.4) T 2 T 1 1/1500 1/ 2257 At T = 1500K: Δµ 0 = RT ln K p = (8.1314JK 1 mol 1 )(1500K)( ) =164kJ / mol = 249kJ / mol Δs 0 = Δh0 Δµ 0 T = ( )kJ / mol 1500K = 56.7JK 1 mol 1

19 Gibbs-Helmholtz Equation Generalize beyond chemical equilibria and the gas phase to any dependence of a free energy G(T) on temperature H = G +TS S = ( G / T ) p H = G T( G T ) p d(u / v) = (vuʹ uv ʹ) / v 2 with v= T and u = G to express the temperature derivative of G/T as ( (G / T ) ) p = T ( G T ( (G / T ) ) p = H(T ) T )T G( T T ) T 2 ( (F / T ) T 2 T Gibbs-Helmholtz Equation = 1 T ( G T ) p G T = 1 [G T( G 2 2 T T ) p] ) V = U(T ) T 2

20 Statistical Thermodynamics Lecture 8: Theory of Chemical Equilibria(II)

21 Equilibrium between Liquid and Gas Consider a system of liquid in equilibrium with its vapor at constant T and P Free energy depends only on the chemical potentials and numbers of particles in the two phase dg = SdT +Vdp + µ v dn v + µ l dn l = µ v dn v + µ l dn l If total number of molecules is conversed: N v + N l = N total = constant dn v + dn l = dn total = 0 dn v = dn l dg = (µ v µ l )dn v The condition for equilibrium is dg = 0, so µ v = µ l

22 Lattice Model µ l How to compute Model liquid as a lattice of particles of a single type, liquid particles occupied the crystalline lattice, with every site occupied by one particle The translational entropy of the lattice particles is zero, since pairs of particles trade positions, the rearrangement can t be distinguished, S = 0 Two particles of type A in terms of bond energy: w AA < 0 This energy applies to every pair of particles that occupy neighboring lattice sites Consider the liquid system has total of N particle, each particle has z nearest neighbors (coordinate number of lattice) U = Nzw AA 2 F =U TS = Nzw AA 2 µ l = ( F N ) T,V = zw AA 2

23 Lattice Model Vapor Pressure µ v = kt ln( p Recall: p 0 int Equilibrium condition: kt ln( p p 0 int ) ) = zw AA 2 p = p 0 int exp( zw AA 2kT ) This describes the vapor pressure P of the molecule A escape the liquid The vapor pressure is a measure of the density of vapor-phase molecule As the AA bonds are made stronger, w AA becomes more negative, P decreases because molecule prefer to bond together in the liquid rather than to escape to the vapor phase If w AA is fixed, increasing the temperature increases the vapor pressure

24 Clapeyron Equation In last example, we see at fixed T and P, the equilibrium between liquid and vapor Two points (P 1,T 1 ) and (P 2,T 2 ) at which the liquid and vapor are in equilibrium µ v (T 1, p 1 ) = µ l (T 1, p 1 ) µ v (T 2, p 2 ) = µ l (T 2, p 2 ) The chemical potential at point 2 involves a small perturbation from point 1 µ v (T 2, p 2 ) = µ v (T 1, p 1 )+ dµ v (T, p) µ l (T 2, p 2 ) = µ l (T 1, p 1 )+ dµ l (T, p) Total differential equation dµ v (T, p) = dµ l (T, p) dµ(t, p) = ( µ T ) p,n dt + ( µ p ) T,N dp

25 Based on Maxwell relation ( µ T ) p,n = ( S N ) T, p = s ( µ p ) T,N = ( V N ) T, p = v dµ v = s v dt + v v dp = dµ l = s l dt + v l dp Rearrange dp dt = s v s l v v v l = Δs Δv Δs = s v s l is partial molar change of entropy Δv = v v v l is partial molar change of volume At phase equilibrium: Δµ = Δh TΔs = 0 Δh = TΔs with Δh = h v h l is partial molar change of enthalpy dp dt = Δs Δv = Δh TΔv Clapeyron equation

26 The molar volume of the gas phase is much larger than the molar volume of the liquid, so Δv = v v v l v v = RT / p dp dt = Δh TΔv dp dt = pδh RT 2 d ln p dt = Δh RT 2 Integrate the equation, when Δh is independent of p and T: Clausius-Clapeyron equation p 2 dln p = p 1 Δh T 2 RT dt ln p 2 = Δh T 2 1 p 1 R ( 1 1 ) T 2 T 1

27 Vapor pressure of benzene versus 1/T lnp is linearly proportional to 1/T, and therefore Δh is independence of T and p Δh can be obtained from slope and used to predict a boiling point with (p 2,T 2 ) if another boiling point (p 1, T 1 ) is known

28 Example: Apply of Clausius-Clapeyron Equation If water boils at T 1 = 373 K and P 1 = 1 atm. At a high altitude where P 2 = ½ atm, what is the boiling temperature T 2? Δh = kj/mol ln p 2 p 1 = Δh R ( 1 T 2 1 T 1 ) 1 T 2 = 1 T 1 R Δh ln p 2 p 1 1 T 2 = K (8.314JK mol 1 )ln J / mol 2 T 2 = 354K = 81 C Water boils at lower temperature at higher altitudes

29 Mixture: Entropy Suppose there are N A molecules of species A, and N B molecules of species B. Together, they completely fill a lattice of N lattice sites N = N A + N B The multiplicity of states is the number of spatial arrangements of the molecules: W = S AB = k ln( N! N A!N B! S = k lnw N! N A!N B! ) = k(n A ln N + N B ln N N A ln N A N B ln N B ) Define the mole fraction: x A = N A / N x B = N B / N S AB = k(n A ln x A + N B ln x B ) To simplify further: let x = x A, so 1-x = x B S AB / NK = x ln x (1 x)ln(1 x)

30 Mixture: Energy In the lattice model, the total energy is the sum of the noncovalent bonds (or contact) of all the pairs of nearest neighbors There are three possible types noncovalent bonds in the mixture: AA, BB, AB The total energy of the system: U AB = m AA w AA + m BB w BB + m AB w AB m AA is the number of AA bonds, m BB is the number of BB bonds, m AB is the number of AB bonds, w AA, w BB, w AB are the corresponding noncovalent interactions Each lattice site has z sides, and every contact involves two sides. The total number of sides of type A can be expressed in terms of numbers of contacts: zn A = 2m AA + m AB zn B = 2m BB + m AB

31 Solve the two equation to get m AA and m AB m AA = zn A m AB 2 m BB = zn B m AB 2 U AB = ( zn A m AB 2 )w AA + ( zn B m AB 2 )w BB + m AB w AB = ( zw AA 2 )N A + ( zw BB 2 )N B + (w AB w + w AA BB 2 m AB )m AB In this expression, the only unknown is, the number of AB contact, we need to use some approximation to estimate it

32 The Mean-Field Approximation The mean-field approximation: for any given numbers N A and N B, the particles are mixed as randomly and uniformly as possible. Consider a specific site next to an A molecule, based on the approximation, the probability that a B molecule occupy the neighboring site is equal to the probability that any site is occupied by B: p B = N B N = x B =1 x There are z nearest-neighbor sites for each A molecule, the average number of AB contacts made by that particle A molecule: zp B = z N B N The total number of A molecules is N A, so m AB = N A zp B = zn A N B N

33 U AB = ( zw AA 2 )N A + ( zw BB 2 )N B + (w AB w AA + w BB 2 = ( zw AA 2 )N A + ( zw BB 2 )N B + z(w AB w AA + w BB 2 = ( zw AA 2 )N A + (zw BB 2 )N B + kt χ AB N A N B N )m AB ) N AN B N where we define a dimensionless quantity called the exchange parameter: χ AB = z kt (w AB w + w AA BB 2 )

34 The Free Energy and Chemical Potential of Mixture F =U TS The free energy of a mixed solution of N A number of A molecules and N B number of B molecules: F AB =U AB TS AB = ( zw AA 2 )N A + ( zw BB 2 )N N B + kt χ A N B AB N + ktn A ln( N A N )+ ktn B ln( N B N ) The chemical potential for A is found by taking the derivative of F with respective to N A, holding N B constant: µ A = ( F AB N A ) NB,T = kt ln x A + zw AA 2 + kt χ AB(1 x A ) 2 µ B = ( F AB N B ) NA,T = kt ln x B + zw BB 2 + kt χ AB(1 x B ) 2 µ = kt ln x + other terms The chemical potential of one component in solution depends on mole fraction

35 The more general case: γ µ Activity and Standard State µ = µ + kt lnγ x is the activity coefficient is the standard state chemical potential In terms of molar concentration µ = µ + kt ln γc c c is the actual concentration with unit mol/l, c o is the standard concentration. Usually, c o is set to 1 mol/l. At low concentration of c, γ 1

36 Equilibrium in Solutions aa+bb cc+dd The equilibrium condition : aµ A +bµ B = cµ C + dµ D a(µ 0 A + kt ln c A c 0 )+b(µ 0 B + kt ln c B c 0 ) = c(µ 0 C + kt ln c C c 0 )+ d(µ 0 D + kt ln c D c 0 ) ΔF 0 cµ 0 C + dµ 0 D aµ 0 A bµ 0 B = kt ln (c C / c0 ) a (c D / c 0 ) b ΔF 0 =cµ 0 C + dµ 0 D aµ 0 A bµ 0 B (c A / c 0 ) c (c B / c 0 ) d = kt ln K c is the standard free energy change of the reaction K c is the concentration based equilibrium constant

37 Statistical Mechanical Consideration µ i = F i,n F 0,N = kt ln Z i,n Z 0,N µ i is the chemical potential of species i in a solution of N water molecules F i,n F 0,N Z i,n is the free energy of solution with one molecule i and N water is the free energy of solution with only N water is the partition function with one molecule I and N water Z 0,N is the partition function of solution with only N water µ 0 i = µ i kt ln c i c 0 = kt ln c i c 0 We use the relation: c i =1/V c 0 =1/V 0 Z i,n = kt ln V 0 Z 0,N V Z i,n Z 0,N ΔF 0 =cµ 0 C + dµ 0 D aµ 0 A bµ c 0 B= kt ln[( Z C,NZ c d D,N Z a A,NZ b B,N )( VZ 0,N V 0 ) a+b c d ]

The Second Law of Thermodynamics (Chapter 4)

The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made