Chemical Chemical EquilibriumEquilibrium
Part 1Part 1
Week Topic Topic Outcomes3-4 Chemical equilibrium
The Gibbs energy & Helmholtzenergy
The Gibbs energy of a gas in a mixture
The thermodynamic equilibrium constant, Kp for a mixture of ideal gases
The dependence of eq on to T & P
It is expected that students are able to:
Interpret the spontaneity of a gas mixture.
Calculate the Gibbs energy and the equilibrium partial pressure of mixing for ideal gases.
Topic Outcomes
(3)
How fast we get the product How much we get the product
Introduction
In each phase of the closed system, the number of molesof 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.
Definition
Reaction Equilibrium:Involves different chemical species, which may or may not be present in the same phase.
E.g.: CaCO3 (s) CaO (s) + CO2 (g)
N2 (g) + 3H2 (g) 2NH3 (g)
Phase equilibrium:Involves the same chemical species present in different phases
E.g.: C6H12O6 (s) C6H12O6 (aq)
Example
Entropy & Equilibrium
Consider an isolated system that is not at material equilibrium;
The spontaneous chemical rxn / transport / matter btw 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, i.e., violating the 2nd Law.
Criteria for equilibrium in an isolated system is the maximization of the systems entropy, S.
Material equilibrium in Closed System
Ordinarily 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:
a maximum at equilibriumgsurroundinsystem SS +
Reaction Equilibrium
1. Rxn that involve gases Chemicals put in container of fixed V,
System is allowed to reach equilib. at const. T & V in a const.-T bath.
2. Rxn in liquid solutions E.g.: The system is usually held at atm P
Allowed to reach equilib. at constant T & P.
The Gibbs The Gibbs Energy and Energy and HelmholtzHelmholtz
EnergyEnergy
(10)
New State Functions
Spontaneity is discussed in context of the approach to equilibrium of a reactive mixture of gas.
2 new state functions are introduced that express spontaneity in terms of the properties of the systemonly.
Previously: the direction for an arbitrary process in predicted by Ssys + Ssurr > 0.
Reactant Product
Systems at const. V & T Systems at const. P & T
Criterion for determining if the reaction mixture will evolve toward reactants or products:
Kp derived Predict equilibrium conc.
The Helmholtz Energy, AMaterial equilib. in a system held at const. T &V (dV= 0, dT= 0)
dU TdS + SdT SdT + dw
dU d(TS) SdT + dw
d(U TS) SdT + dw
at equilibriumdU TdS + dw
Add & subtract SdT:
Since d(TS) = SdT + SdS
d(U TS) SdT PdV
d(U TS) 0Const. T & V, dT=0, dV=0, closed syst.In therm. & mech. Equilib.,P-V work only
At const. T & V, dT=0, dV=0
Since dw = P dV
State function
Helmholtz free energy
A U - TS
The Gibbs EnergyConsider for const.T & P, dw = PdV into dU TdS + dw
dU d(TS) SdT d(PV) + VdP
d(H TS) SdT VdPd(U + PV TS) SdT + VdP
dU TdS + SdT SdT - PdV - VdP + VdP
d(H TS) 0At const. T and P, dT=0, dP=0
G H TS U + PV TSGibbs free energy,
dAT,V 0 dGT,P 0
Equilibrium reached
Const. T, P
Time
G
Equilibrium State
G decreasesSuniv increase
In a closed system capable of doing only P-V work,
Const. T & V material-equilibrium condition is the minimization of the A,
Const. T & P material-equilibrium condition is the minimization of the G.
dA = 0
dG = 0
at equilib., const. T, V
at equilib., const. T, P
Relationship Gequilib & SunivConsider a system in mechanical & thermal equilib.which undergoes an irreversible chemical reaction / phase change at const. T & P.
/ TG / TSTH S/ THSSSsystsystsyst
systsystsystsurruni v
==+=+=
/ TGS systuni v = Closed syst., const. T, V, P-V work onlyGsyst as the system proceeds to equilib. at const. T&P corresponds to a proportional Suniv.The occurrence of rxn is favored by +ve Ssyst and +veSsurr
Work Function: A
at const. T( ) dwSdTTSUd +dwdAwA
dwSdTdA +
wwby =Awby Const. T, closed syst.
Closed system in thermal & mechanical equilibrium
Finite isothermal process
Work done by the system on its surroundings
A carries a greater significance than being simply a signpost of spontaneous change:
Awmax =Change in the Helmholtz energy is equal to the max. work the system can do:
Equality sign reversible proc. wby can be greater or less than U (int. energy of syst.)
wby = U + q (E.g. Carnotcycle, U = 0 & wby>0)
Work Function: G
TSPVUTSH G +PVAPVTSU G ++ VdPPdVdAdG ++=
VdPPdVdwSdTdG +++
( ) dwSdTTSUd +dwSdTdA +
PdVdwdG + const. T & P, closed syst.
From
and H=U+PV
d(PV)=PdV+VdP
VPnondwPdVdw +=
VPnondwdG VPnonby,VPnonwwG
=
VPnondwdG Gw VPnonby, const. T & P, closed syst.
If the P-V work is done in a mechanically reversible
PdVdwdG +Note:
Awby Earlier for A;
From dw
wA Compare A
Compare A
Gw maxV,Pnon =
( )PVddwdqdH ++=
( ) ( )PVddwTdSPVddwTdSdG rev +=++=
dwdqdU +=
( ) VdPdwVdPPdVdwPdVdG VPnonVPnon +=+++=
revdwdw=TdSdqdq rev ==
The max. non-expansion work from a process at cont. P & T is given by the value of G
(const. T, P)Reversible change, equality sign holds & wby,non-P-V= G
From dU
dH=dU+d(PV)
dS=dQ/T
dG=dH-TdS
dH
What G means: +G : not spontaneous Zero G : at equilibrium G : spontaneous
What A means: +A : not spontaneous Zero A : at equilibrium A : spontaneous
Second law of thermodynamics:
Entropy must always increase!
Summary
Gibbs Free Energy Helmholtz Free EnergySTHG =
STUA =
Initial & final states of system at the same values of
P & T
Initial & final states of system at the same values of
V & T
Working Session
For a phase change on vaporization of 1.0 mol water at 1 atm and 100C, calculate G and A.
Given: molar volume of H2O (l) at 100C is 18.8 cm3/mol, R=8.314 J/mol.K=82.06 cm3.atm/mol.K
Example 1You wish to construct a fuel cell based on the oxidation of hydrocarbon fuel. The 2 choices for a fuel are methane & octane.The reactions are,
Methane: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)Octane: C8H18 (l) + 25/2 O2 (g) 8CO2 (g) + 9H2O (l)
Calculate the maximum work and non-expansion work available through the combustion of these 2 hydrocarbons, on a per mole and a per gram basis at 298.15 K and 1 bar pressure.
Given at 298.15K: Hcombustions(CH4,g)=891kJ/mol;Hcombustions(C8H8,l)=5471 kJ/mol; Sm(CH4,g)=186.3 J/mol.K; Sm(C8H8,l)=361.1 J/mol.K; Sm(O2,g)=205.2 J/mol.K; Sm(CO2,g)=213.1 J/mol.K; Sm(H2O,l)=70 J/mol.K
Differential Differential Forms of Forms of UU, , HH, ,
AA and and GG
All thermodynamic state-function relations can be derived from six basic equations,
1st basic equation it combines the 1st & 2nd Laws
Next 3 basic equations are the definitions of H, A, and G
Finally, the 2 equations are for the CP and CV equations
PdVTdSdU = Closed syst., P-V work onlyPVUH += TSUA = TSHG =
Closed syst., in equilib., P-V work only
VV T
UC
=
PP T
HC
=
Closed syst., in equilib., P-V work only
The Gibbs Equations
Closed syst., rev. proc., P-V work only
VdPTdSdH +=
PdVTdSdU =
SdTPdVdA =SdTVdPdG =
Equivalent Expressions
VPHandTS
HSP =
=
PV
U and TSU
SV =
=
VPG and ST
GSP =
=
PV
A and SSA
TV =
=
VT TP
VS
=
PT TV
P S
=
PS SV
PT
=
SV VT
S P
=
The last two are extremely valuable.
Maxwell Relations
The first two are little used.
The equations relate the isothermal P & V variations of entropy to measurable properties.
Dependence of G & A on P, V, T
Most reactions of interest to chemist are carried out under const. P rather than const. V conditions.
VPG and ST
GSP =
=
G, T
G ,P
Macroscopic change in P & T, 2nd expression integrate
( ) ( ) ==P
P
P
PVdP'PT,GPT,GdG
oo
oo Initial P, P = 1 bar
This equation takes on different form for liq. & solids & for gaseous.
Liquids & Solids
Independent of P over a limited range in P,
( ) ( ) ( ) ( )oooooo
PPVPT,GVdP'PT,GPT,G PP
++=
Systems changes appreciably with P
( ) ( ) ( ) ( ) oooooo
PPl nTGdP'P'
nRTTGVdP'TGPT,G PP
P
PnRT+=+=+=
Gaseous
Gibbs-Helmholtz EquationMore useful to obtain an expression for T dependence of G/T than for T dependence of G,
[ ] [ ]
2T
HT
TSGTG
TS
TG
TG
T1
dT1/ TdGT GT
1T
G/ T
222P
PP
=+==
=
+
=
[ ][ ]
[ ][ ] ( )
HTTH
1/ TddT
TG/ T
1/ TG/ T 2
2PP ==
=
Can also be written [ ] 2T1
dT1/ Td
=
Preceding equation also applies to the change in G & H associated with a process
= 2
1
2
1
TT
TT THdT
Gd 1
( ) ( ) ( )
+= 1211
12
2 TTTHTTG
TTG
11 Assumption: H is independent of T over the T interval of interest
Integral must be evaluated numerically using tabulated Hf & T-dependent expressions of Cp,mfor reactants & products
Example 2
The value of Gf for Fe(g) is 370.7 kJ mol1 at 298.15 K, and Hf for Fe(g) is 416. 3 kJ mol1 at the same temperature. Assuming that Hf is constant in the interval 250 400 K. Calculate Gf for Fe(g) at 400 K.
Gibbs Energy of Gibbs Energy of Gas MixtureGas Mixture
G of a Gas in a MixtureConditions for equilibrium in a reactive mixture of ideal gases, Derived in terms of the i of the chemical
constituents.Chemical potential of a reactant/product species changes as its conc. in the reaction mixture changes.
2 important processes occur The mixing of reactants The conversion of
reactants to products
Barrier removedR1 R2 R1R2
R1
R2
R1R2
R2
R2
R1
R1
Barrier
Equilibrium Condition
Pd membrane Permeable to H2, not to Ar
H2 pressure (not total P) is same on both sides of the membrane,Equilib. reached with respect to the conc. of H2
mi xt ureH
pureH 22
= at equilibrium
of a Gas in Mixture( ) ( ) ( )oo PPl n nRTTGPT,G +=
( ) ( ) ( )
+== oo P
Pl n RTTPT,PT, 222222
HHH
mi xtureHH
pureH
of a gas in a mix. depends logarithmically on its partial P, PA = xAP
( ) ( ) ( ) AAAAmi xture A xRTl nPPl n RTT xl n RTP
Pl n RTTPT, +
+=+
+= oooo
( ) ( ) Apure Ami xture A xl n RTPT,PT, +=
Amixture < Apure Substance flows from to Diffusion continue until partial P both
sides of the barrier are equal Mixing are spontaneous if no barrier.
Conclusion
xA < 0
( ) ( ) Apure Ami xture A xl n RTPT,PT, +=
G of a Gas of Mixing for Ideal Gas
Chemical potential of a reactant/product species changes through mixing with other species.
Quantitative relationship between Gmixing & the mole fractions of individual constituents of the mixture.
Consider Xe, Ar, Ne, He
Compare G initial state & final state (uniformly distributed)
Calculation the G
Initial state Final state
Xem,Xem, ArArNem,NeHem,He
XeArNeHeiGnGnGnGn
GGGGG+++=
+++=
( ) ( )( )
( )XeXem,XeArm, ArAr
NeNem,NeHeHem,Hef
RTl nxGn RTl nxGn
RTl nxGnRTl nxGnG
++++
+++=
Gibbs energy of mixing(Gf-Gi)
xi < 1, each term in the last expression is veGmixing < 0 mixing is spontaneous process
Gibbs Energy of Binary Mixture
Consider mixture of A (xA = x) & B (xB = 1 x),
T = 298.15 K
Gmixing = 0 (xA=0 & xA=1) Pure substance
Gmixing = min (xA=0.5) Largest decrease in G arises
from mixing (A & B equal amount)
( ) ( )[ ]x1l nx1xl nxnRTGmi xi ng +=
Entropy of Binary MixtureEntropy of mixing,
=
=
i
iiP
mi xi ngmi xi ng l nxxnRT
GS T = 298.15 K
xA=0 xA=0.5, increase Each components of mixture
expands to larger final volume Smixing arises from the
dependence of S on V at const. T
Each component contributes equally to S
Example 3
Consider the system consists of 4 separate
subsystems containing He, Ne, Ar & Xe.
Assume that the separate compartments
contain 1.0 mol of He, 3.0 mol of Ne, 2.0 mol of
Ar & 2.5 mol of Xe at 298.15 K. The pressure in
each compartment is 1 bar.
a. Calculate Gmixingb. Calculate Smixing
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