1. SCHOOL OF BIOPROSES ENGINEERING
PRT 140 PHYSICAL CHEMISTRY PROGRAMME INDUSTRIAL CHEMICAL PROCESS SEM /2014
MATERIAL EQUILIBRIUM BY
PN ROZAINI ABDULLAH
SCHOOL OF BIOPROSES ENGINEERING
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2. 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
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3. 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.
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4. 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
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5. - 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) NH3 (g)
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6. 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.
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7. # 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)*
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8. 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.
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9. 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)
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10. 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)
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11. 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)
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12. 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
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13. 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)
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14. 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
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15. 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
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16. 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
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17. 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
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18. 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
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19. 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
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20. 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
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21. 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
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22. 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:
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23. 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.
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24. 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)
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25. 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
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26. 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 )
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27. 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?
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28. 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
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29. 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
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30. 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.
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31. The Euler Reciprocity Relations
If Z＝f(x，y)，and Z has continuous second partial derivatives, then
That is
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32. 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,
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33. 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.
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34. 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
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35. 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
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36. 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
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37. 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
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38. From pressure dependence of H
Joule-Thomson Coefficient
(easily measured quantities)
from (2.65)
From pressure dependence of H
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39. From volume dependence of U
Heat-Capacity Difference
(easily measured quantities)
From volume dependence of U
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40. Heat-Capacity Difference
As T  0, CP  CV
CP  CV (since  > 0)
CP = CV (if  = 0)
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41. FTK RY

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

44. 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
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45. 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
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46. FTK RY

47. 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:
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48. 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 ]
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49. 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
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50. - For the spontaneous flow of moles of j from phase to phase
Closed syst that has not yet reached phase equilibrium -
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51. 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
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52. 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
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53. 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 :
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54. The condition for chemical-reaction equilibrium in a closed system is
Reaction equilib, in closed system., P-V work only
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55. FTK RY

56. Thank you