SCHOOL OF BIOPROSES ENGINEERING PRT 140 PHYSICAL CHEMISTRY PROGRAMME INDUSTRIAL CHEMICAL PROCESS SEM 1 2013/2014 MATERIAL EQUILIBRIUM BY PN ROZAINI ABDULLAH SCHOOL OF BIOPROSES ENGINEERING FTK RY 20 2013

MATERIAL EQUILIBRIUM Material equilibrium means that in each phase of the closed system, the number of moles of each substance present remains constant in time. Material equilibrium is subdivided into: Reaction Equilibrium which is equilibrium with respect to conversion of one set of chemical species to another set Phase Equilibrium which is equilibrium with respect to transport of matter between phases of the system without conversion of one species to another FTK RY 20 2013

Helmholtz energy A≡ H - TS You will be introduced with 2 new state function: Helmholtz energy A≡ H - TS Gibbs energy G≡ H - TS Sec 4.3 It turns out that the conditions for reaction equilibrium and phase equilibrium are most conveniently formulated in terms of state functions called the chemical potentials, which are closely related to G. FTK RY 20 2013

Sec 4.4 & 4.5 FTK RY 20 2013 First Law Second Law to derive expressions for thermodynamic quantities in terms of readily measured properties Sec 4.4 & 4.5 Sec. 4.6 : Material Equilibrium Sec. 4.7 : Phase Equilibrium Sec. 4.8 : Reaction Equilibrium FTK RY 20 2013

- involves the same chemical species present in different phases Phase equilibrium: - involves the same chemical species present in different phases - Ex: C6H12O6 (s) ↔ C6H12O6 (aq) Reaction Equilibrium: Involves different chemical species, which may or may not be present in the same phase. - Ex: CaCO3 (s) CaO (s) + CO2 (g) N2 (g) + 3H2 (g) 2NH3 (g) FTK RY 20 2013

ENTROPY AND EQUILIBRIUM #Consider not in Material Equilibrium System Energy Surrounding Matter 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 2nd Law.  criteria for equilibrium in an isolated system is the maximization of the system’s entropy S. FTK RY 20 2013

# Closed system (in material equilibrium): Energy Surrounding Matter not ordinary isolated can exchange heat and work with its surroundings. which it interacts to constitute an isolated system Surrounding System the condition for material equilibrium in the system is then maximization of the total entropy of the system plus its surroundings: Ssyst + Ssurr a maximum at equilib.    (4.1)* FTK RY 20 2013

Reaction equilibrium is ordinarily studied under one of two conditions: # involved in gas The system is allowed to reach equilibrium at constant T and V in a constant temperature bath. Fixed Volume # involved in liquid The system is usually held at atmospheric pressure and allowed to reach equilibrium at constant T and P. FTK RY 20 2013

dSuniv = dSsyst + dSsurr > 0 (2) #Consider a system at T The system is not in material equilibrium but is in mechanical and thermal equilibrium The surroundings are in material, mechanical and thermal equilibrium System and surroundings can exchange energy (as heat and work) but not matter Since system and surroundings are isolated , we have dqsurr= -dqsyst (1) Since, the chemical reaction or matter transport within the non equilibrium system is irreversible, dSuniv must be positive: dSuniv = dSsyst + dSsurr > 0 (2) FTK RY 20 2013

Therefore, the heat transfer is reversible, and dSsurr= dqsurr/T (3) The surroundings are in thermodynamic equilibrium throughout the process. Therefore, the heat transfer is reversible, and dSsurr= dqsurr/T (3) The systems is not in thermodynamic equilibrium, and the process involves an irreversible change in the system, therefore dSsyst ≠dqsyst/T (4) FTK RY 20 2013

dSsyst > -dSsurr = -dqsurr/T = dqsyst/T (5) Therefore Equation (1) to (3) give dSsyst > -dSsurr = -dqsurr/T = dqsyst/T (5) Therefore dSsyst > dqsyst/T dS > dqirrev/T (6) closed syst. in them. and mech. equilib. dqsurr= -dqsyst (1) dSuniv = dSsyst + dSsurr >0 (2) dSsurr= dqsurr/T (3) FTK RY 20 2013

Thus, at material equilibrium we have, ds = dqrev/T (7) When the system has reached material equilibrium, any infinitesimal process is a change from a system at equilibrium to one infinitesimally close to equilibrium and hence is a reversible process. Thus, at material equilibrium we have, ds = dqrev/T (7) Combining (6) and (7): ds ≥dq/T (8) material change, closed syst. in them & mech. Equilib FTK RY 20 2013

The first law for a closed system is dq = dU – dw (9) Eq 8 gives dq≤ TdS Hence for a closed system in mechanical and thermal equilibrium we have dU – dw ≤ TdS Or dU ≤ TdS + dw (10) ds ≥ dq/T (8) FTK RY 20 2013

THE GIBSS & HELMHOLTZ ENERGIES A spontaneous process at constant-T-and-V is accompanied by a decrease in the Helmholtz energy, A. A spontaneous process at constant-T-and-P is accompanied by a decrease in the Gibbs energy, G. dA = 0 at equilibrium, const. T, V dG = 0 at equilibrium, const. T, P FTK RY 20 2013

dw = -P dV for P-V work only Helmholtz free energy A  U - TS Consider material equilibrium at constant T and V dU  TdS + dw dU  TdS + SdT – SdT + dw dU  d(TS) – SdT + dw d(U – TS)  – SdT + dw dw = -P dV for P-V work only d(U – TS) – SdT - PdV at constant T and V, dT=0, dV=0 d(U – TS)  0 Equality sign holds at material equilibrium FTK RY 20 2013

dw = -P dV for P-V work only Helmholtz free energy A  U - TS Consider material equilibrium at constant T and V dU  TdS + dw dU  TdS + SdT – SdT + dw dU  d(TS) – SdT + dw d(U – TS)  – SdT + dw dw = -P dV for P-V work only d(U – TS) – SdT - PdV at constant T and V, dT=0, dV=0, closed system in therm & mech. Equlibirum; P-V work only d(U – TS)  0 Equality sign holds at material equilibrium FTK RY 20 2013

d(U-TS)=0 at equilibrium Helmholtz free energy For a closed system (T & V constant), the state function U-TS, continually decrease during the spontaneous, irreversible process of chemical reaction and matter transport until material equilibrium is reached d(U-TS)=0 at equilibrium FTK RY 20 2013

Gibbs free energy dU  d(TS) – SdT – d(PV) + VdP G  H – TS  U + PV – TS Consider material equilibrium for constant T & P, into with dw = -P dV dU  T dS + dw dU  T dS + S dT – S dT - P dV - V dP + V dP dU  d(TS) – SdT – d(PV) + VdP d(U + PV – TS)  – SdT + VdP d(H – TS) – SdT + VdP at constant T and P, dT=0, dP=0; material chamge;closed system in mechanical & therm. Equli.; P-V work only d(H – TS)  0 FTK RY 20 2013

Gibbs free energy d(H – TS)  0 G = H – TS = U + PV - TS the state function H-TS, continually decrease during material changes (constant T and P) , until material equilibrium is reached. This is the minimization of Gibbs free energy. d(H – TS)  0 GIBBS FREE ENERGY, G=H-TS G = H – TS = U + PV - TS FTK RY 20 2013

GIBBS FREE ENERGY G  H – TS  U + PV – TS dGT,P  0 G Constant T, P Equilibrium reached Constant T, P Time G G decreases during the approach to equilibrium, reaching minimum at equilibrium dGT,P  0 FTK RY 20 2013

As G of the system decrease at constant T & P, GIBBS FREE ENERGY As G of the system decrease at constant T & P, Suniv increases. WHY? Consider a system in mechanical and thermal equilibrium which undergoes an irreversible chemical reaction or phase change at constant T and P. closed syst., const. T, V, P-V work only The decrease in Gsyst as the system proceeds to equilibrium at constant T and P corresponds to a proportional increase in S univ FTK RY 20 2013

Closed system, in thermal &mechanic. equilibrium const. T const. T, closed syst. It turns out that A carries a greater significance than being simply a signpost of spontaneous change: The change in the Helmholtz energy is equal to the maximum work the system can do: FTK RY 20 2013

G  H – TS  U + PV – TS G  U– TS + PV  A + PV const. T and P, closed syst. If the P-V work is done in a mechanically reversible manner, then or const. T and P, closed syst. FTK RY 20 2013

For a reversible change The maximum non-expansion work from a process at constant P and T is given by the value of -G (const. T, P) FTK RY 20 2013

Thermodynamic Reactions for a System in Equilibrium 6 Basic Equations: closed syst., rev. proc., P-V work only dU = TdS - PdV H  U + PV A  U – TS G  H - TS closed syst., in equilib., P-V work only closed syst., in equilib., P-V work only FTK RY 20 2013

Key properties Basic Equations Heat capacities closed syst., in equilib. The rates of change of U, H, and S with respect to T can be determined from the heat capacities CP and CV. Heat capacities Key properties (CP CV ) FTK RY 20 2013

How to derive dH, dA and dG? The Gibbs Equations dU = TdS - PdV dH = TdS + VdP closed syst., rev. proc., P-V work only dA = -SdT - PdV dG = -SdT + VdP How to derive dH, dA and dG? FTK RY 20 2013

The Gibbs Equations dH = ? H  U + PV dH = TdS + VdP dH = d(U + PV) dU = TdS - PdV = dU + d(PV) = dU + PdV + VdP = (TdS - PdV) + PdV + VdP dH = TdS + VdP FTK RY 20 2013

dA = ? A  U - TS dA = -SdT - PdV dG = ? G  H - TS dA = d(U - TS) dU = TdS - PdV dA = d(U - TS) = dU - d(TS) = dU - TdS - SdT = (TdS - PdV) - TdS - SdT dA = -SdT - PdV dG = ? G  H - TS dH = TdS+VdP dG = d(H - TS) = dH - d(TS) = dH - TdS - SdT = (TdS + VdP) - TdS - SdT dG = -SdT + VdP FTK RY 20 2013

The Power of thermodynamics: The Gibbs equation dU= T dS – P dV implies that U is being considered a function of the variables S and V. From U= U (S,V) we have (dG = -SdT + VdP) The Power of thermodynamics: Difficultly measured properties to be expressed in terms of easily measured properties. FTK RY 20 2013

The Euler Reciprocity Relations If Z＝f(x，y)，and Z has continuous second partial derivatives, then That is FTK RY 20 2013

The Maxwell Relations (Application of Euler relation to Gibss equations) dU = TdS - PdV The Gibbs equation (4.33) for dU is dU＝TdS－PdV dS＝0 dV＝0 Applying Euler Reciprocity, FTK RY 20 2013

These are the Maxwell Relations The first two are little used. The last two are extremely valuable. The equations relate the isothermal pressure and volume variations of entropy to measurable properties. FTK RY 20 2013

Dependence of State Functions on T, P, and V We now find the dependence of U, H, S and G on the variables of the system. The most common independent variables are T and P. We can relate the temperature and pressure variations of H, S, and G to the measurable Cp,α, and κ FTK RY 20 2013

Volume dependence of U The Gibbs equation gives dU＝TdS－PdV For an isothermal process dUT＝TdST－PdVT Divided above equation by dVT, the infinitesimal volume change at constant T, to give T subscripts indicate that the infinitesimal changes dU, dS, and dV are for a constant-T process From Maxwell Relations FTK RY 20 2013

Pressure dependence of H Temperature dependence of U Temperature dependence of H Pressure dependence of H From Basic Equations from Gibbs equations, dH＝TdS＋VdP From Maxwell Relations FTK RY 20 2013

Temperature dependence of S The equations of this section apply to a closed system of fixed composition and also to a closed system where the composition changes reversibly Temperature dependence of S From Basic Equations Pressure dependence of S From Maxwell Relations Temperature and Pressure dependence of G The Gibbs equation (4.36) for dG is dG ＝ -SdT + VdP dT＝0 dP＝0 FTK RY 20 2013

From pressure dependence of H Joule-Thomson Coefficient (easily measured quantities) from (2.65) From pressure dependence of H FTK RY 20 2013

From volume dependence of U Heat-Capacity Difference (easily measured quantities) From volume dependence of U FTK RY 20 2013

Heat-Capacity Difference As T  0, CP  CV CP  CV (since  > 0) CP = CV (if  = 0) FTK RY 20 2013

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Internal Pressure Ideal gases Solids, Liquids, & Non-ideal Gases Solids 300 J/cm3 (25 oC, 1 atm) Liquids 300 J/cm3 (25 oC, 1 atm) Strong intermolecular forces in solids and liquids. FTK RY 20 2013

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Calculation of Changes in State Function Calculation of ΔS Suppose a closed system of constant composition goes from state (P1,T1) to state (P2,T2), the system’s entropy is a function of T and P FTK RY 20 2013

For step (a), dP=0 and gives Integration gives: Since S is a state function, ΔS is independent of the path used to connect states 1 and 2. A convenient path (Figure 4.3) is first to hold P constant at P1 and change T from T1 to T2. Then T is held constant at T2, and P is changed from P1 to P2. For step (a), dP=0 and gives For step (b), dT=0 and gives FTK RY 20 2013

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ΔU = ΔH – Δ (PV) 2. Calculation of ΔH ΔU can be easily found from ΔH using : ΔU = ΔH – Δ (PV) Alternatively we can write down the equation for ΔU similar to: FTK RY 20 2013

3. Calculation of ΔG For isothermal process: Alternatively, ΔG for an isothermal process that does not involve an irreversible composition change can be found as: A special case: [Since ] FTK RY 20 2013

Phase Equilibrium A phase equilibrium involves the same chemical species present in different phase. [ eg:C6H12O6(s) C6H12O6(g) ] - - Phase equilib, in closed syst, P-V work only FTK RY 20 2013

- For the spontaneous flow of moles of j from phase to phase Closed syst that has not yet reached phase equilibrium - FTK RY 20 2013

One EXCEPTION to the phase equilibrium, Then, j cannot flow out of (since it is absent from ). The system will therefore unchanged with time and hence in equilibrium. So the equilibrium condition becomes: Phase equilib, j absent from FTK RY 20 2013

Reaction Equilibrium A reaction equilibrium involves different chemical species present in the same phase. Let the reaction be: products reactants a, b,…..e, f….. Are the coefficients FTK RY 20 2013

Adopt the convention of of transporting the reactant to the right side of equation: are negative for reactant and positive for products During a chemical reaction, the change Δn in the no. of moles of each substance is proportional to its stoichometric coefficient v. This proportionality constant is called the extent of reaction (xi) For general chemical reaction undergoing a definite amount of reaction, the change in moles of species i, , equals multiplied by the proportionality constant : FTK RY 20 2013

The condition for chemical-reaction equilibrium in a closed system is Reaction equilib, in closed system., P-V work only FTK RY 20 2013

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