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Lecture 8: Theory of Chemical Equilibria (I)

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

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Lecture 8: Theory of Chemical Equilibria (I)

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

  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. reactant product partition functions chemical potential equilibrium constant K • 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 = NB/NA • The quantity that predicts the chemical equilibria is the chemical potential

  3. Equilibria Condition • Consider the Gibbs free energy of the system: • At constant T and P: • Every molecule is in either state A or B: • Condition for equilibria:

  4. Partition Function for Chemical Reactions • Define a reduced partition function with the ground-state term factored out: • Partition function for chemical potential

  5. From condition for equilibria: • 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 • a, b and c indicate the stoichiometries of species A, B and C. • At constant T and P: • The equilibrium is now subject to two stoichiometric constraints:

  7. Condition for equilibria: • 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. difference in ground-state energies:

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

  9. ΔD: difference between the dissociation energies of all the products and all the reactants

  10. Example: Estimation of K K • The dissociation energies are 431.8 kJ/mol for H2; 439.2 kJ/mol for D2; and 435.2 kJ/mol for HD So at T = 300K:

  11. Since at T = 300K, (hv)/kT >> 1 There, • 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 Kp based on pressures rather than equilibrium constant K based on the particle numbers • Multiple both sides by

  13. Chemical Potentials in terms of Partial Pressures • Here • It divides the chemical potential into a part that depends on pressure kTlnp, and a part that does not μ0 • μ0is 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 • Suppose a fluctuation changes the number of B molecules by an amount dNB. • Define a reaction coordinate , the factional degree to which the system has proceeded to B. • The total number of particles is fixed: , so • To move toward equilibrium, dG ≤ 0, so , must have opposite signs. • If , then the direct toward equilibrium is < 0, NB 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 • At constant T and P, the condition for equilibrium is • The pressure based equilibrium constant is • Recall , which depends on temperature

  16. Chemical potential can be divided into partial molar enthalpy and entropy components. • If , are independent of temperature:

  17. van’t Hoff relation • The relation provides a useful way to plot data to obtain • van’t Hoff plots show lnK versus 1/T, the slope is • Water is more dissociated at higher temperatures • For dissociation, Δh0 >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 Δh0 is independent of temperature

  18. integration • Can be used to find how Kp depends on temperature if know Δh0 or determine Δh0 if Kp(T) is measured • Example: calculate enthalpy of dissociation of water: • At T = 1500K, lnKp = -13.147 ; at T = 2257 K, lnKp = -6.4 At T = 1500K:

  19. Gibbs-Helmholtz Equation • Generalize beyond chemical equilibria and the gas phase to any dependence of a free energy G(T) on temperature with v= T and u = G to express the temperature derivative of G/T as Gibbs-Helmholtz Equation

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

  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 • If total number of molecules is conversed: • The condition for equilibrium is dG = 0 , so

  22. Lattice Model • 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: wAA < 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)

  23. Lattice Model – Vapor Pressure • Recall: • Equilibrium condition: • 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, wAA becomes more negative, P decreases because molecule prefer to bond together in the liquid rather than to escape to the vapor phase • If wAA 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 (P1,T1) and (P2,T2) at which the liquid and vapor are in equilibrium • The chemical potential at point 2 involves a small perturbation from point 1 • Total differential equation

  25. Based on Maxwell relation • Rearrange • is partial molar change of entropy • is partial molar change of volume • At phase equilibrium: • with is partial molar change of enthalpy Clapeyron equation

  26. The molar volume of the gas phase is much larger than the molar volume of the liquid, so Clausius-Clapeyron equation • Integrate the equation, when is independent of p and T:

  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 (p2,T2) if another boiling point (p1, T1) is known

  28. Example: Apply of Clausius-Clapeyron Equation • If water boils at T1 = 373 K and P1 = 1 atm. At a high altitude where P2 = ½ atm, what is the boiling temperature T2 ? Δh = 40.66 kJ/mol • Water boils at lower temperature at higher altitudes

  29. Mixture: Entropy • Suppose there are NA molecules of species A, and NB molecules of species B. Together, they completely fill a lattice of N lattice sites • The multiplicity of states is the number of spatial arrangements of the molecules: • Define the mole fraction: • To simplify further: let x = xA, so 1-x = xB

  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: is the number of AA bonds, is the number of BB bonds, is the number of AB bonds, , , 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:

  31. Solve the two equation to get and 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 NA and NB, 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: • There are z nearest-neighbor sites for each A molecule, the average number of AB contacts made by that particle A molecule: • The total number of A molecules is NA, so

  33. where we define a dimensionless quantity called the exchange parameter:

  34. The Free Energy and Chemical Potential of Mixture • The free energy of a mixed solution of NA number of A molecules and NB number of B molecules: • The chemical potential for A is found by taking the derivative of F with respective to NA, holding NB constant: other terms • The chemical potential of one component in solution depends on mole fraction

  35. Activity and Standard State • The more general case: is the activity coefficient is the standard state chemical potential • In terms of molar concentration c is the actual concentration with unit mol/L , co is the standard concentration. Usually, co is set to 1 mol/L . At low concentration of c,

  36. Equilibrium in Solutions • The equilibrium condition : • is the standard free energy change of the reaction • is the concentration based equilibrium constant

  37. Statistical Mechanical Consideration is the chemical potential of species i in a solution of N water molecules 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 is the partition function of solution with only N water We use the relation:

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