lecture 4: equilibrium - aalborg universitethomes.nano.aau.dk/lg/physchem2010_files/physical... ·...
TRANSCRIPT
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Lecture 4: Equilibrium28-09-2010
• Lecture plan:– equilibrium
• equilibrium and Gibbs free energy• description of equilibrium• response of equilibrium to conditions (P, T, pH)
– equilibrium electrochemistry• representing redox reactions in terms of half-reactions• electrochemical cells• the Nernst equation• standard potentials and electrode calibration
– problems
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EQUILIBRIUM
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Chemical Equilibrium
• Chemical reaction tend to move towards a dynamic equilibrium in which both reactants and products are present but have no tendency to undergo net change
A B C D⎯⎯→+ +←⎯⎯
The question: How to predict the composition of mixture at various condition
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The Gibbs energy minimum• Spontaneous change at const P and T happens towards lower values of
the Gibbs energy• Let’s consider reaction A B⎯⎯→←⎯⎯
A
B
dn ddn d
ξξ
= −= +
If some amount of A changed into B: dξextent of the reaction
Reaction Gibbs energy (definition):,
rP T
GGξ
⎛ ⎞∂Δ = ⎜ ⎟∂⎝ ⎠
( )A A B B A B B AdG dn dn d d dμ μ μ ξ μ ξ μ μ ξ= + = − + = −
,r B A
P T
GG μ μξ
⎛ ⎞∂Δ = = −⎜ ⎟∂⎝ ⎠
At equilibrium 0rGΔ =
Difference between chemical potentials of the products and the reactants at the composition fo the reaction mixture
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The Gibbs energy minimum
• Spontaneity reaction at const P, T
0rGΔ <
0rGΔ =
0rGΔ >
Forward reaction is spontaneous, reaction exergonic (work-producing)
Reverse reaction is spontaneous, reaction endergonic i.e.required work to go in forward reaction
Reaction at equilibrium
Exergonic reaction,e.g. glucose oxidation Endergonic reaction,
e.g. protein synthesis
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The description of equilibrium• Perfect gas equilibrium
0 0
0
( ln ) ( ln )
ln
r B A B B A A
Br
A
G RT p RT ppG RTp
μ μ μ μΔ = − = + − + =
= Δ +
Q – reaction quotient
At equilibrium: 0
0
0 ln
ln
Br r
A
r
pG G RTp
RT K G
Δ = = Δ +
= −Δ
K- equilibrium constant
Why reaction doesn’t go till the end:
( ln ln )mix A A B BG nRT x x x xΔ = +
dimensionless: 0Ap
p
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The description of equilibrium
• General case of a reaction 2 3A B C D+ ⎯⎯→ +
0 3 2C D A B= + − −
0 lnr rG G RT QΔ = Δ +0 0 0
products reactantsr f fG G Gν νΔ = Δ − Δ∑ ∑
activitiesof productsactivitiesof reactants
Q =jv
jj
Q a=∏
3
2C D
A B
a aQa a
=For example, for the reaction above:
standard Gibbs free energies of formation
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The description of equilibrium
0ln rRT K G= −Δ
jvj
j equilibrium
K a⎛ ⎞
= ⎜ ⎟⎝ ⎠∏At equilibrium:
Example: Find degree of dissociation of water vapour at 2300K and 1 bar if standard Gibbs energy for decomposition is 118 kJ/mol
2 2 21( ) ( ) ( )2
H O g H g O g⎯⎯→ +←⎯⎯
0 33118*10ln 2.08*10
8.3*2300GK KRTΔ
= − = =
2 2
2
12 3 2 1 2
1 2(1 )(2 )H O
H O
p p pKp
αα α
= =− + 0.0205α =
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The description of equilibrium0ln rRT K G= −Δ
0 0 0/ /r r rG RT H RT S RK e e e−Δ −Δ Δ= =Increase with reaction entropy
decrease with reaction enthalpy
Boltzmann distribution interpretation:
/
/
i
i
E kT
i E kT
i
ep Ne
−
−=∑
Due to higher density of energy levels (i.e. higher S), B is dominant at equilibrium
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The description of equilibrium
• Relation between equilibrium constants
• Using biological standard state
C D C D C Db
A B A B A B
a a b bK K Ka a b b γ
γ γγ γ
= = × =
At low concentration: bK K≈
2( ) ( ) ( ) ( )NADH aq H aq NAD aq H g+ ++ ⎯⎯→ +
If a biological reaction involves H+ ions, we have to take into account that standard biological condition is at log 7
HpH a += − =
0
3
7 ln10
21.8 / 16.1 8.3 10 / 310 19.7 /r rG G RT
kJ mol kJ K mol K kJ mol
⊕
−
Δ = Δ + × =
= − + × × × =
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The response of equilibria to the conditions• Equilibria will respond to temperature, pressure and
concentration changes
Pressure dependence:
0T
KP
∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠
Depends on standard (standard pressure)
0rGΔ
• Pressure increase by injecting inert gas: no change as partial pressures of reactants and products stay the same .• Pressure increase by compression: system will adjust partial pressures so the constant stays the same.
2
0( ) 2 ( ) B
A
pA g B g Kp p
⎯⎯→ =←⎯⎯
0ln rRT K G= −Δ
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The response of equilibria to the conditions• Le Chatelier principle:
A system at equilibrium, when subjected to disturbance responds in a way that tends to minimize the effect of disturbance
2
0( ) 2 ( ) B
A
pA g B g Kp p
⎯⎯→ =←⎯⎯(1 )nα−
Extent of dissociation, α:
2 nα
Mole fractions at equilibrium:
2 2 2 2
2
½
0
(1 ) 1 2(1 ) 2 1 1
41
11 4
A B
B B
A A
nn n
p p pKp p
p Kp
α α ακ κα α α ακ ακ α
α
− −= = =
− + + +
= = =−
⎛ ⎞= ⎜ ⎟+⎝ ⎠
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The response of equilibria to the conditions• Temperature response
Equilibrium will shift in endothermic direction if temperature is increased and in exothermic direction if temperature is lowered.
Van’t Hoff equation:
0ln rRT K G= −Δ0 0 0
2
( / ) ( / )ln 1 r r rd G T d G T Hd KdT R dT dT T
Δ Δ Δ= − = −
Gibbs-Helmholtz equation
0 0
2
ln ln(1/ )
r rH Hd K d KdT RT d T R
Δ Δ= = −
i.e. for exothermic reaction:
0 ln0 0rd KH
dTΔ < <
0
2 12 1
1 1ln ln r HK KR T T
⎛ ⎞Δ− = − −⎜ ⎟
⎝ ⎠So, we can predict the equilibrium constant at another temperature:
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The response of equilibria to the conditions
• Boltzmann distribution interpretation
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The response of equilibria to the conditions
• Noncalorimetric measuring reaction enthalpy
0
slope: r HR
Δ
0ln(1/ )
r Hd Kd T R
Δ=
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The response of equilibria to the conditions
• Value of K at different temperatures
0ln(1/ )
r Hd Kd T R
Δ=
2
1
1/ 00
2 12 11/
1 1 1ln ln (1/ )T
rr
T
HK K H d TR R T T
⎛ ⎞Δ− = Δ = −⎜ ⎟
⎝ ⎠∫
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Equilibria and pH
3log
H OpH a += −
at low concentration equal to molarity
3
32
2 3
142
2 ( ) ( ) ( )
10 298H O OHw H O OH
H O
H O l H O aq OH aqa a
K a a at Ka
+ −
+ −
+ −
−
⎯⎯→ +←⎯⎯
= = =
• Dissociation of water (autoprotolysis)
Ionic dissociation constant of water
For pure water:3
710H O OH
a a+ −−= =
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The response of equilibria to pH
Acidity constant, Ka:
32 3( ) ( ) ( ) ( ) H O A
aHA
a aHA aq H O l H O aq A aq K
a+ −+ −⎯⎯→+ + =←⎯⎯ loga apK K= −
Basicity constant, Kb:
2( ) ( ) ( ) ( ) HB OHb
B
a aB aq H O l HB aq OH aq K
a+ −+ −⎯⎯→+ + =←⎯⎯ logb bpK K= −
• Arrhenius acid: increases concentration of H3O+ in solutionArrhenius base: increases concentration of OH- in solution
3 2 4
( ) ( ) ( )
( ) ( ) ( ) ( )
NaOH aq Na aq OH aq
NH aq H O l NH aq OH aq
+ −
+ −
⎯⎯→ +←⎯⎯⎯⎯→+ +←⎯⎯
Can be done via donation of OH- or removing of H+.
Conjugate base
Conjugate acid
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Example: dissociation of formic acid
Answer: 33[ ] 1.22 10
2.91H OpH
+ −= ×
=
• Example: pK of formic acid is 3.77 at 298K. What is pH of 0.01M solution?What would happen if it were strong acid?
2 3( )HCOOH H O l HCOO H O− +⎯⎯→+ +←⎯⎯2 4
2b b acx
a±− ± −
=
[ ]3 41.695 10a
H O HCOOK
HCOOH
+ −−
⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= = × 2 4 61.695 10 1.695 10 0x x− −+ × − × = 31.22 10x −= ×
• What would be dissociation degree at pH=4 and pH=10?
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Equilibrium electrochemistry
• Any redox reaction can be expressed as difference of two half-reactions, which are conceptual reactions showing gain of electrons
Ox Redeν −+ →
2 2( ) ( ) ( ) ( )Zn s Cu aq Zn aq Cu s+ ++ ⎯⎯→ +
2( ) ( ) 2Zn s Zn aq e+⎯⎯→ +2 ( ) 2 ( )Cu aq e Cu s+ + ⎯⎯→
+
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Equilibrium electrochemistry
1 1Ox Redeν −+ →
The electrode where oxidation occurs is called anode, the electrode where reduction occurs is called cathode.
Electrolyte
• Two half-reactions will run in the opposite directions in two half cells
Cathode
2 2Red Ox eν −→ +Anode
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Electrochemical cells
4 4( ) | ( ) ( ) | ( )Zn s ZnSO aq CuSO aq Cu sM 4 4( ) | ( ) || ( ) | ( )Zn s ZnSO aq CuSO aq Cu s
Notation:
Liquid junction Liquid junction potential assumed eliminated
Phase boundary
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Types of half-cells
• Metal in a solution of it’s ions
• Gas in contact with a solution of it’s ions
• Two different oxidation states of the same species
• Metal in contact with its insoluble salt
( ) ( )AgCl s e Ag s Cl− −+ → +
3 2Fe e Fe+ − ++ →22 2H e H+ −+ →
2 2Zn e Zn+ −+ →
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The Nernst equation• A cell where overall cell reaction hasn’t reached chemical
equilibrium can do electrical work as the reaction drives electrons through an external circuit
,maxe r
r
w GFE Gν
= Δ
− = ΔFaraday constant AF eN=
rGEFν
Δ= −
Cell emf:
00ln lnrG RT RTE Q E Q
F F Fν ν νΔ
= − − = −
• As there is no potential difference at equilibrium:0
ln FEKRT
ν=
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Standard potentials and SHE• Although it’s not possible to measure potentials of electrodes separately,
we can define a particular electrode as having zero potential at all temperatures.
Standard Hydrogen Electrode (SHE) at aH+=1 (pH=0) and p=1bar
02( ) | ( ) | ( ) 0Pt s H g H aq E+ =
22 2H e H+ −+
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Standard potentials and SHE
2
01/ 2
0 0 2 2
( / , ) ln
ln ln ln
H Cl
H
H Cl
a aRTE E AgCl Ag ClF a
RT RT RTE E a a E bF F F
γ
+ −
+ −
−
±
= −
= − = − −
• All other electrodes can be calibrated using SHE:(“Harned cell” – calibration of Ag/AgCl electrode)
2
2
( ) | ( ) | ( ) | ( ) | ( ) | ( )1 ( ) ( ) ( ) ( )2
Pt s H g H aq HCl aq AgCl s Ag s
H g AgCl s HCl aq Ag s
+
+ → +
1202 lnRTE b E Cb
F+ = +For experimental calibration:
Standard efm can be found from the offset
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Electrochemical series• Cell emfs are convenient source for data on equilibrium constants, Gibbs
energies etc.
0 0 01 1 2 2 2 1Red ,Ox ||Red ,Ox E E E= −
Red1 has thermodynamic tendency to reduce Ox2 if: 0 02 1E E>
low reduces high
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Species selective electrodes
• Ion-selective electrode is an electrode that generates a potential in response to the presence of a solution of specific ions
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Determination of thermodynamic functions by emf
• By measuring emf Gibbs energy can be determined:0 0
rG FEνΔ = −
00rSdE
dT FνΔ
=
• The temperature coefficient of standard emf gives standard entropy of the reaction:
doesn’t depend on pressure
00 0 0 0
r r rdEH G T S F E TdT
ν⎛ ⎞
Δ = Δ + Δ = − −⎜ ⎟⎝ ⎠
• and therefore provides non-calorimetric way to measure enthalpy
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Application: Batteries• Lead-acid rechargeable battery (inv. 1859)
2 02 4 4 2
2 04 4
4 ( ) ( ) 2 ( ) 2 1.685
( ) ( ) ( ) 2 0.356
PbO H aq SO aq e PbSO s H O E
Pb s SO aq PbSO s e E
+ − −
− −
+ + + → + =
+ → + = −
• During the charging, the reactions are reversed
• Life time is limited due to mechanical stress due to formation and dissolution of solid material
02
02 2 2 3
( ) 2 ( ) ( ) 2 1.1
2 ( ) 2 ( ) 2 ( ) 0.76
Zn s OH aq ZnO s H O e E
MnO s H O e Mn O s OH aq E
− −
− −
+ → + + =
+ + → + = −
• Alkaline cell:
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Structure of metal-electrolyte interface• Formation of electrical double layer due to
specifically and non-specifically adsorbed ions
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Measuring absolute half-cell potential• The energy diagram of the cell:
Fermi levels of two metals should be offset by the Gibbs energy
work function in solution
double layer
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Measuring absolute half-cell potential• Gomer and Tryson experiment (J.Chem.Phys 66(1977), 4413):
variable DC voltage is applied to the gold-electrode capacitor with a vibrating plate, AC voltage is measured
Vdc=0
Vdc=Vm2Au
2Au M Au AuE V φ= −• Absolute half cell potential for gold air electrode can be measured
• Absolute SHE potential 0 4.73SHEE V= −
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Class problems:• Atkins 6.8b: In the gas phase reaction A+B=C+2D it was found
that when 2mol A, 1mol B and 3 mol D were mixed and allowed to come to equilibrium at 25C, the mixture contained 0.79mol of C at 1 bar. Calculate mol fraction of every species at equilibrium, Kx, K and ∆rG0.
• Atkins 6.21(a) Devise cells in which the following are the reactions and calculate the standard emf in each case:(a) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)(b) 2 AgCl(s) + H2(g) → 2 HCl(aq) + 2 Ag(s)(c) 2 H2(g) + O2(g) → 2 H2O(l)
• Atkins 6.22(a) Use the Debye-Huckel limiting law and the Nernst equation to estimate the potential of the cell
Ag|AgBr(s)|KBr(aq,0.050mol/kg)||Cd(NO3)2(aq, 0.010mol/kg) |Cd
at 25°C.