PHYSICAL CHEMISTRY ERT 108 Semester II 2011/2012 Huzairy Hassan School of Bioprocess Engineering UniMAP

Material Equilibrium

Introduction Material equilibrium: -In each phase of the closed system, the number of moles of each substance present remains constant in time. Reaction equilibrium: Is equilibrium with respect to conversion of 1 set of chemical species to another set. Phase equilibrium: Is equilibrium with respect to transport of matter between phases of the system without conversion of 1 species to another.

Phase equilibrium: - involves the same chemical species present in different phases - Ex: C 6 H 12 O 6 (s) ↔ C 6 H 12 O 6 (aq) Reaction Equilibrium: -Involves different chemical species, which may or may not be present in the same phase. - Ex: CaCO 3 (s) CaO (s) + CO 2 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g)

Thermodynamic Properties of Non-equilibrium Systems Overview: Deals with systems (non-thermodynamical equilibrium) in which a chemical rxn or transport of matter from 1 phase to another is occurring.

1 st consider systems not in phase equilibrium: Figure 4.1 When the partition is removed, the system is not in phase equilibrium Initially, P and T are held fixed, then the U and S: U = U soln + U NaCl S = S soln + S NaCl

Now, the partition is removed: - the removal is done reversibly and adiabatically, - q and w for the partition removal are zero, - therefore, ΔU and ΔS are zero. After the partition is removed  no phase equilibrium - as the crystals (solids) of NaCl start to dissolve in the unsaturated solution  values for U and S are changing.

2 nd consider systems not in reaction equilibrium: Suppose H 2, O 2, and H 2 O gases are mixed. Provided no catalyst is present and the T is moderate  the gases will not react when mixed. -- 1) we can use First Law to measure the ΔU of the mixing (ΔU = q + w), -- 2) by using semi-permeable membranes, can do the mixing reversibly and hence measure the ΔS for the mixing process. However, the mixture is not necessarily at reaction equilibrium. Then, add catalyst, the gases react – changing the mixture composition. We stop the reaction by catalyst removal – we get new composition of mixture. ---  U & S can be determined by measurement done in a reversible separation process.

Therefore; Values of U and S can be assigned to a system that is in mechanical and thermal equilibrium and has a uniform composition in each phase, even though the system is not in material equilibrium. -- such systems also have well-defined values of P, V, & T

Entropy & Equilibrium Overview: Consider an isolated system that is not at material equilibrium; - the spontaneous chemical rxn or transport of matter between phases in this system are irreversible processes that increase the entropy (S). - the processes continue until the S is maximized  once the S is maximized, further processes can only decrease S, thus violate 2 nd Law.  criteria for equilibrium in an isolated system is the maximization of the system’s entropy S.

Closed system (in material equilibrium); - not isolated - can exchange heat and work with its surroundings. By considering the system itself plus the surroundings with which it interacts to constitute an isolated system; - then, the condition for material equilibrium in the system is maximization of the total entropy of the system plus its surroundings: S syst + S surr a max at equilibrium

Reaction equilibrium: 1)Rxn that involve gases – e.g.: chemicals put in container of fixed volume, and the system is allowed to reach equilibrium at constant T and V in a constant-T bath. 2) Rxn in liquid solutions – e.g.: the system is usually held at atmospheric P and allowed to reach equilibrium at constant T and P.

Surroundings at T System at T Impermeable wall Rigid, adiabatic, impermeable wall Figure 4.2 A closed system that is in mechanical and thermal equilibrium but not in material equilibrium. The system and surroundings are isolated from the rest of the world  The system: not in material equilibrium, but is in mechanical and thermal equilibrium  The surroundings: are in material, mechanical and thermal equilibrium.

Let heat dq syst flow into the system as a result of the changes that occur in the system during an infinitesimal time period. For ex; Endothermic chemical rxn: dq syst is +ve.  dq surr = - dq syst since system and surrounding are isolated from the rest of the world dS univ = dS syst + dS surr > 0 since the chemical rxn or matter transport within nonequilibrium system is irreversible dS surr = dq surr / T the surroundings are in thermodynamic equilibrium – heat transfer is reversible However, the system is not thermodynamic equilibrium  irreversible change in system  dS syst ≠ dq syst / T dS > dq irrev / T thus, at material equilibrium  dS = dq rev / T (4.7) and dS ≥ dq/T

For the 1 st law for a closed system is; dq = dU – dw Multiply with T, dq ≤ T dS  dU – dw ≤ T dS or  dU ≤ T dS + dw (4.9)

GIBBS & HELMHOLTZ Energies Material equilibrium in a system held at constant T and V (dV = 0, dT= 0) Now, add and subtract S dT: since

Since dw = - P dV State function

The condition for material equilibrium in a closed system capable of doing only P-V work and held constant T and V is minimization of the system’s state function U – TS  A = Helmholtz free energy or (Helmholtz free energy or Helmholtz function) (unit= J or cal)

At constant T and P (capable doing only P-V work); -Add and substract S dT and V dP: State function

Since H ≡ U + PV (Eq. 2.45) G = Gibbs free energy (unit= J or cal)

In a closed system (P-V work), the constant T and V material-equilibrium condition is the minimization of the Helmholtz function A, and the constant T and P material- equilibrium condition is the minimization of the Gibbs function G In summary

Example 4.1 Calculate ΔG and ΔA for the vaporization of 1.00 mol of H 2 O at 1.00 atm and 100 °C.

Solution: G ≡ H –TS,T is constant and ΔG = G 2 – G 1 = H 2 –TS 2 – (H 1 – TS 1 ) = H 2 – H 1 – T(S 2 – S 1 ) = ΔH - T ΔS, constant T The process is reversible and isothermal;  dS = d q/T ΔS = q/T Since P is constant and only P-V work is done; ΔH = q P = q (Eq. 2.46) Therefore, ΔG = q – T(q/T) = 0  since reversible (equilibrium) process at constant T and P

From A ≡ U – TS,  ΔA = ΔU -T ΔS at constant T. From ΔU = q + w, and ΔS = q/T  ΔA = q + w – q = w (1) The work is reversible P-V work at constant P, so; w = ---- (2) Given, molar volume of H 2 O (l) at 100 ºC = 18.8 cm 3 /mol, Therefore, from ideal-gas law V m = RT/P = 30.6 x 10 3 cm 3 /mol  ΔV = 30.6 x 10 3 cm 3 From (2), w = ( x 10 3 cm 3 atm) (8.314 J) / (82.06 cm 3 atm) = kJ = ΔA - (from (1))

 The decrease in G syst as the system proceeds to equilibrium at constant T and P corresponds to a proportional increase in S univ. The occurrence of rxn is favored by +ve ΔS syst and +ve ΔS surr  -ve ΔH syst (exothermic rxn) since, the heat transfered to the surroundings increases entropy (ΔS surr = - ΔH syst /T)

“Work function” - for a closed system in thermal and mechanical equilibrium  dA ≤ -S dT + dw, for constant T, dA ≤ dw, for finite isothermal process, ΔA ≤ w work done by the system on its surroundings;  w by = -w, and ΔA ≤ -w by for an isothermal process. therefore, w by ≤ - ΔA, constant T, closed system (4.22) - (1) the equality sign holds for the reversible process (2) w by can be greater or less than – ΔU (int. energy decrease of syst) (3) w by = – ΔU + q (recall Carnot cycle, ΔU = 0 and w by > 0)

G From G = A + PV dG = dA + P dV + V dP From d(U – TS) ≤ -S dT + dw, for dA gives; dG ≤ -S dT + dw + P dV + P dV, for closed syst. in thermal and mechanical equilibrium.  dG ≤ dw + P dV, constant T and P, closed system ---- (4.23) If P-V work is done

If P-V work is done in mechanically reversible ; dw = - P dV + dw non-P-V Eq. (4.23) becomes dG ≤ dw non-P-V or ΔG ≤ w non-P-V = -w by,non-P-V Therefore, w by,non-P-V ≤ -ΔG  The quantity – ΔG equals the max. possible non-expansion work output w by,non-P-V done by a system in a constant-T-and-P process – “free energy” “Ex. of non-expansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses.”

When a closed system capable of only P-V work is held at constant T and V, the condition for material equilibrium (phase & reaction) is that the Helmholtz function A (A ≡ U – TS) is minimized. When such a system is held at constant T and P, the material-equilibrium condition is the minimization of the Gibbs function G ≡ H – TS. In summary

Thermodynamic Relations for a System in Equilibrium Overview: Investigation of the properties of A and G ---- Reversible process through equilibrium state.

Basic Equation: 1)dU = T dS – P dV, closed syst, reversible, P-V work 2)H ≡ U + PV 3)A ≡ U - TS 4)G ≡ H – TS 5)C V =, closed syst, in equilibrium, P-V work 6)C P =, closed syst, in equilibrium, P-V work

In term of S; C V = T and, C P = T, closed syst. In equilibrium Therefore, we can find the rates of change of U, H, and S with respect to Temperature. dU = T dS – P dV -does not apply to irreversible composition changes (ex.: by adding or removing 1 or more substance in a mixture, or chemical rxn or transport of phase from 1 to another) in a closed system. -Can apply to reversible composition changes (ex.: mixture of N 2, H 2 & NH 3 with catalyst and reversibly vary T or P).

The Gibbs Equations;

Relation between C p, α (thermal expansivity or cubic expansion coefficient), and κ (isothermal compressibility - kappa) =

Euler Reciprocacity Relation

The Maxwell Relations

Dependence of State Functions on T, P, and V: 1)Volume Dependence of U 2) Temperature Dependence of U 3) Temperature Dependence of H

4) Pressure Dependence of H 5) Temperature Dependence of S

6) Pressure Dependence of S 7) Temperature and Pressure Dependences of G

Joule-Thomson Coefficient

Heat-capacity Difference

Example 4.2 (C P -C V ) For water at 30 ºC and 1 atm: α = 3.04 x K -1 κ = 4.52 x atm -1 = 4.46 x m 2 /N, C P,m = 75.3 J/(mol K), V m = 18.1 cm 3 /mol Find C V,m of water at 30 ºC and 1 atm.

Assignment 1 For water at 90.0 ºC and 1 atm: α = x K -1 κ = 4.81x bar -1 C P = 1.02 cal/(g K) and ρ = g/cm 3. Find C V for water at 90.0 ºC and 1 atm.

Example 4.3

Assignment 2 Derive Equation as follow: You may refer to Chapter 1, 4 and 8 (text book).

Calculation of Changes in State Functions 1)Calculation of ΔS - The system’s entropy is a function of T and P; S = S (T, P) dS = dT + dP = dT – αV dP ΔS = S 2 – S 1 = dT (4.60)

a) T is held constant at T 2, P is changed from P 1 to P 2 dP = 0; Eq becomes ΔS a = dT, constant P = P (4.61) b) P is held constant, C P in Eq. (4.61) depends only on T dT = 0; Eq becomes ΔS b = -, constant T = T (4.62) ΔS = ΔS a + ΔS b Figure 4.3: Path for calculating ΔS and ΔH

Example 4.4 ΔS when both T and P change Calculate ΔS when 2.00 mol of water goes from 27 ºC and 1 atm to 37 ºC and 40 atm. Given; C P,m = 75.3 J/(mol K), V m = 18.1 cm 3 /mol α = 3.04 x K -1

2) Calculation of ΔH and ΔU ΔH = + Figure 4.3 Path for calculating ΔS or ΔH From Equations 4.30 and 4.48 See Figure 4.4 Alternative calculation of Δu and Δs using u-u tr,l and s-s tr,l for H 2 O, curves can be found from Eq and 4.63.

3) Calculation of ΔG and ΔA For an isothermal process, G ≡ H – TS; Alternatively, Δ G for isothermal process that does not involve an irreversible composition change; from Eq. 4.51: Special case, when ΔH = q and ΔS = q/T;

For ΔA, Consider the processes with ΔT = 0. Since ΔA is undefined if T changes, so, we use; ΔA = ΔU - T ΔS or for isothermal process

Material Equilibrium-Phase and Reaction Reaction Equilibrium: -Involves different chemical species, which may or may not be present in the same phase. - Ex: CaCO 3 (s) CaO (s) + CO 2 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g)

Phase equilibrium: - involves the same chemical species present in different phases - Ex: C 6 H 12 O 6 (s) ↔ C 6 H 12 O 6 (aq)

Let a reaction be; Where A 1, A 2, … = reactants A m, A m+1, … = products a, b, …, e, f, … = coefficients For example; then, Transposing the reactants to the right side of equation;

Now let; Where the stoichiometric coefficients v i are negative for reactants and positive for products.

For example;  And the stoichiometric coefficients are: ⨳

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