Post on 19-Jan-2016
Chemical Equilibrium
By Doba Jackson, Ph.D.
Outline of Chpt 5• Gibbs Energy and Helmholtz Energy• Gibbs energy of a reaction mixture
(Chemical Potential)• Gibbs energy of mixing• Calculating Gibbs energy of reactions• Equilibrium constant• Equilibrium reactions with phase mixtures
Introduction to Free Energy
Gibbs Free Energy Change (∆G)
Entropychange
∆G =
Enthalpy ofreaction
Temperature(Kelvin)
∆H ∆S- T
∆G < 0 Process is spontaneous
∆G = 0 Process is at equilibrium
∆G > 0 Process is non-spontaneous
Enthalpy term islarger than Entropy
Entropy term islarger than Enthalpy
Types of Expansion Work• Irreversible Process
exw = -P ΔV
• Reversible Processf
i
V
V
nRTw = - dV
V
Comparison of work done upon expansion of an Ideal gas
0revdw dw
0revdq dq
ReversibleIrreversibledU dq dw
revdw dw
An irreversible process produces less work and heat than a reversible process
U is a state function. It does not depend on the path(reversible or irreversible)
Two Isothermal processes
0revdq dq
Isothermal, Reversible
Isothermal, Irreversible
revdw dw
An irreversible process produces less work and less heat than areversible process
revrev
revrev
dwdq
dwdq
dwdqdU
dwdqdU
0
0
0 dqdqrev
dqdqrev
The inequality of Clausius
0revdq dq
Isothermal, Reversible
Isothermal, Irreversible
revdq dq
T T
dqdqrev 0 dqdqrev
revrev dqTdST
dqdS ;
dqTdS Clausius Inequality
Reversible, NonspontaneousReversible Process: The thermodynamic process in which a system can be changed from its initial state to its final state then back to its initialstate leaving all thermodynamic variables for the universe (system + surroundings) unchanged.
A truly reversible change will:
- Occur in an infinite amount of time - All variables must be in equilibrium with each other at every stage of the change
IrreversibleReversible
Irreversible, SpontaneousIrreversible Process: The thermodynamic process in which a system that is changed from its initial state to its final state then back to its initialstate will change some thermodynamic variables of the universe.
A truly irreversible change will:
- Occur in an finite amount of time - All variables will not be in equilibrium with each other at every stage of the change
IrreversibleReversible
Clausius Inequality leads to equations for spontaneity
0revdw dw
0revdq dq
ReversibleIrreversible
revdq dq
T T
dqdS
T
Constant Volume
TdS dq
VTdS dq TdS dU
Helmholtz Energy
dA = 0; Equilibrium
dA < 0; Irreversible, Spontaneous
TdSdUdA
STUA sSpontaneou-Non;0A
Clausius Inequality leads to equations for spontaneity
Reversible
IrreversibleConstant Pressure
TdS dq
pTdS dq TdS dH
Gibbs Energy
dG = 0; Equilibrium
dG < 0; Irreversible, Spontaneous
STHG
TdSdHdG
sSpontaneou-Non;0G
Properties of the Gibbs Energy
G H TS ( )dG dh d TS
( ) ( )dG d U PV d TS
( ) ( )dG dU d PV d TS
dG dU PdV VdP TdS SdT
dU q w dU TdS PdV
q TdSw PdV
dG TdS PdV PdV VdP TdS SdT
dG VdP SdT
Gibbs energy is a function of only pressure and temperature
Properties of the Gibbs Energy
dG VdP SdT
Gibbs energy is a function of only pressure and temperature
P
GS
T
T
GV
P
T P
G GdG = dP - dT
P T
Meaning of Gibbs energy
• The Gibbs energy is the maximum work (non-expansion) that can be performed at a given state (T,P).
• Because S is always >0, G will decrease as temperature is increased.
• As S increases, the slope of G verses T becomes more negative (see graph).
P
GS
T
P
GS
T
Meaning of Gibbs energy• The Gibbs energy is the
maximum work (non-expansion) that can be performed at a given state (T,P).
• Because V is always >0, G will increase as pressure is increased.
• As V increases, the slope of G verses P becomes more positive (see graph).
T
GV
P
Calculate the change in Gibbs Energy when pressure increases
dG VdP SdT
If temperature is constant:
dG VdP2
1
P
PG VdP
If volume change is small (liquid, solids):
G V P If volume change is not small (Gases):
2
1
PG = nRTLn
P
Problem:Calculate the entropy and enthalpy changes for this chemical reaction at 298K .
CO2 (g) + H2O (l) ------------ C6H12O6 (s) + O2 (g) 6 6 6
ΔfH= (CO2) = -393.5 kJ/mol
ΔfH= (H2O) = -285.8 kJ/mol
ΔfH= (C6H12O6) = -1273.1 kJ/mol
ΔSm= (CO2) = 213.8 J/mol
ΔSm= (H2O) = 70.0 J/mol
ΔSm= (C6H12O6) = 209.2 J/mol
ΔSm= (O2) = 205.2 J/mol
Sec 6.4-6.6: Gibbs energy of a mixing (Chemical Potential)
Chemical Potential
• Chemical Potential (μ)- is a measure of the potential that a substance has to undergo a change in state or composition
• Molar Gibbs energy (Gm)- is the maximum energy that a system has available for non-expansion work.
For all Pure substances: μ = Gm
Partial Molar Quantities
• Partial Molar Volume- the change in volume of a substance per mole of volume added to the larger mixture.
*, ,
jj P T n
VV
n
A A B BdV = V dn + V dn , , B
AA P T n
VV
n
, , A
BB P T n
VV
n
• Partial Molar Gibbs- the change in Gibbs energy of a substance per mole of volume added to the larger mixture.
, , '
jj P T n
G
n
A A B BdG = dn + dn , , B
AA P T n
G
n
, , A
BB P T n
G
n
Partial Molar Gibbs Energy
A A B BG = n + n This assumes the temperature and pressure is constant.
, , B
AA P T n
G
n
, , A
BB P T n
G
n
A A B BdG = VdP - SdT + n + n Fundamental Equation which includes composition
,max A A B Bw = n + nadd The Gibbs energy is the maximum additional non-expansion work. This depends on composition.
Thermodynamics of Mixing• Why does mixing occur spontaneously?
2
1
PG = RTLn
Pm From Second Law
2,2 ,1
1
PG = RTLn
Pm mG
22 1
1
P= RTLn
P Gm = μ
P= RTLn
P
PΘ = 1 bar
= RTLn P P = in bar units
This is the chemical potentials of the pure gasses before mixing
Why does mixing occur spontaneously
= RTLnA A AP
= RTLnB B BP
A A B BG = n + n Mixing A & Bat a constant T, P
A BG = RTLn n + RTLn ninit A BP P
A BG = RTLn n + RTLn nfinal A A B BP P
mix final initialG G G
A B A BG = RTLn n + RTLn n RTLn n + RTLn nmix A A B B A BP P P P
-mix A A B B A BG RTLnP n RTLnP n RTLnP n RTLnP n
-mix A A B B A BG RTLnP n RTLnP n RTLnP n RTLnP n
A Bmix A B
P PG RTLn n RTLn n
P P
Dalton’s Law of Partial Pressures
A AP P AA
P
P
mix A A B BG RTLn n RTLn n
AA
T
n
n
A A Tn n
mix A A T B B TG RTLn n RTLn n
mix T A A B BG n RT Ln Ln
Remember this equation
Enthalpy and Entropy of mixing Gases
mix T A A B BG n RT Ln Ln
For a Perfect Gas:
0mixH
S mix T A A B Bn R Ln Ln
Problem 5.8A container of volume 5.0 dm3 that is divided into two equal compartments. One side contains H2 gas and the other contains N2 gas in equal molar anounts. Both containers are at 25ºC and 1 atm. Calculate the Gibbs energy and Entropy of mixing when the partition is removed.
Sec 6.4-6.6: Gibbs energy of a chemical reaction
Standard rxn Gibbs energies (ΔrGº) and
Standard Gibbs energies of formation (∆fGº)
H2O(l)H2(g) + ½ O2(g)
ReactantsProducts
∆rG° = { (1)∆fGo(H2O)} – {(1)∆fG
o(H2) + (.5)∆fGo(O2)}
Reaction Gibbs energies (ΔrGº) can be determined by the difference of the product Gibbs energies of formation and the reactant Gibbs energies of formation.
0 0 0r
Products Reactants
Δ = v - vf fG G G
Example:
Sec 6.8-6.10: Gibbs energy of a chemical reaction mixture