ib chemistry on gibbs free energy, equilibrium constant and cell potential
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http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
Tutorial on Gibbs Free Energy Change, Equilibrium and Cell Potential
cellnFEG
Relationship betweenEnergetics and Equilibrium
cKRTG ln STHG Enthalpy change
Entropy change
Equilibrium constant
Gibbs free energy change
HG
Relationship bet ∆G, Kc and E cell
cellnFEG STHG cKRTG ln
cK
Relationship betweenEnergetics and Cell Potential
G cellE
Gibbs free energy change
Cell potential
F = Faraday constant (96 500 Cmol-1)
n = number electron
Relationship bet ∆G, Kc and Ecell
ΔGθ Kc Eθ/V Extent of rxn
> 0 < 1 < 0 No ReactionNon spontaneous
ΔGθ = 0 Kc = 1
0 EquilibriumMix
reactant/product
< 0 > 1 > 0 Reaction completeSpontaneous
ΔGθ Kc Eq mixture
ΔGθ = + 200
9 x 10-36 Reactants
ΔGθ = + 10 2 x 1-2 Mixture
ΔGθ = 0 Kc = 1 Equilibrium
ΔGθ = - 10 5 x 101 Mixture
ΔGθ = - 200 1 x 1035 Products
Relationship bet ∆G and Kc
shift to left (reactant)
shift to right (products)
cellE
G
cK
∆G θ = -nFE θ cell ∆Gθ = -R
T ln K c
KnFRTE cell ln
Magnitude of Kc
Extend of reaction
How far rxn shift to right or left?
Not how fast
cK
Position of equilibrium
cKTemp
dependentExtend of rxn
Not how fast
Shift to left/favour
reactant
Shift to right/favour
product
cKRelationship between
Equilibrium and Energetics
cKRTG ln STHG
Enthalpy change
Entropy change
Equilibrium constant
Gibbs free energy change
HG cK
GEnergetically
Thermodynamically
Favourable/feasible
ΔGθ ln K Kc Eq mixture
ΔGθ -ve < 0
Positive ( + )
Kc > 1
Product(Right)
ΔGθ +ve > 0
Negative
( - )
Kc < 1
Reactant(left)
ΔGθ = 0 0 Kc = 1
Equilibrium
Measure work available from system
Sign predict spontaneity of rxn
Negative (-ve)
spontaneous
Positive (+ve) NOT spontaneous
veG veG
NOT
favourable
Energetically favourable
Product formation
NO product
cKRTG ln
Magnitude of Kc
Extend of reaction
How far rxn shift to right or left?
Not how fast
cK
Position of equilibrium
cKTemp
dependentExtend of rxn
Not how fast
Shift to left/favour
reactant
Shift to right/favour
product
cKRelationship between
Equilibrium and Energetics
cKRTG ln STHG
Enthalpy change
Entropy change
Equilibrium constant
Gibbs free energy change
HG cK
ΔGθ ln K Kc Eq mixture
ΔGθ -ve < 0
Positive ( + )
Kc > 1
Product(Right)
ΔGθ +ve > 0
Negative
( - )
Kc < 1
Reactant(left)
ΔGθ = 0 0 Kc = 1
Equilibrium
cKRTG ln STHG ∆Hsys ∆Ssys ∆Gsys Description
- + ∆G = ∆H - T∆S ∆G = - ve Spontaneous, All Temp
+ - ∆G = ∆H - T∆S ∆G = + ve Non spontaneous, All Temp
+ + ∆G = ∆H - T∆S ∆G = - ve Spontaneous, High ↑ Temp
- - ∆G = ∆H - T∆S ∆G = - ve Spontaneous, Low ↓ Temp
Relationship bet ∆G and Kc
GEnergetically
Thermodynamically
Favourable/feasible
Sign predict spontaneity of rxn
veG veG
NOT
favourable
Energetically
favourable
Product formation
NO product
KRTG ln
Predict will rxn occur with ΔG and Kc cK
Very SMALL Kc < 1
Shift to right/favour product
Shift to left/favour
reactant
Very BIG Kc > 1 veG veG
KRTG ln
1cK 1cK
Negative (-ve)spontaneous
Positive (+ve) NOT spontaneous
Relationship bet ∆G and Kc
ΔGθ Kc Eq mixture
ΔGθ = + 200
9 x 10-
36Reactant
ΔGθ = + 10 2 x 1-2 Mixture
ΔGθ = 0 Kc = 1 Equilibrium
ΔGθ = - 10 5 x 101 Mixture
ΔGθ = - 200 1 x 1035 Products
shift to left (reactant)
shift to right (product)
G, Gibbs free energy
A
Mixture composition
B
100% A 100% B
∆G decreases ↓
30 % A70 % B
Equilibrium mixture
∆G < 0
∆G = 0 (Equilibrium) ↓
Free energy minimum
∆G < 0∆G < 0
∆G = 0
Free energy system is lowered on the way to equilibrium
Rxn proceed to minimum free energy ∆G = 0System seek lowest possible free energy
Product have lower free energy than reactant
∆G < 0 productreactant
GEnergetically
Thermodynamically
Favourable/feasible
Sign predict spontaneity of rxn
veG veG
NOT
favourable
Energetically
favourable
Product formation
NO product
KRTG ln
cK
Very SMALL Kc < 1
Shift to right/favour product
Shift to left/favour
reactant
Very BIG Kc > 1 veG veG
KRTG ln
1cK 1cK
Negative (-ve)spontaneous
Positive (+ve) NOT spontaneous
Relationship bet ∆G, Q and Kc
G, Gibbs free energy
A
B
100% A 100% B
∆G decreases ↓
30 % A70 % B
Equilibrium mixture
∆G < 0
∆G = 0 (Equilibrium) ↓
Free energy minimum
∆G < 0∆G < 0
∆G = 0
∆G < 0 productreactant
G, Gibbs free energy
reactant product∆G < 0A
B
∆G decreases ↓
100% A 100% B30 % A70 % B
∆G = 0 Q = K
∆G < 0 Q < K
∆G > 0
∆G < 0 Q > K
∆G > 0
A ↔ B A ↔ B
Equilibrium mixture
Predict will rxn occur with ΔG and Kc
Relationship bet ∆G and Kc
G, Gibbs free energy
A
B
100% A
100% B
∆G decreases ↓
30 % A70 % B
Equilibrium mix close to product
∆G < 0
∆G = 0 (Equilibrium) ↓
Free energy minimum
∆G < 0∆G < 0
∆G = 0
∆G < -10
Kc > 1
A ↔ B A ↔ BG, Gibbs free energy
A
B
∆G decreases ↓
∆G < -100
100% A
100% B
∆G = 0 (Equilibrium) ↓
Free energy minimum
Kc > 1Equilibrium mix close to product
10 % A90 % B
∆G < 0
∆G < 0 ∆G = 0
∆G very –ve → Kc > 1 → (more product/close to completion)∆G –ve → Kc > 1 → (more product > reactant)
A ↔ BG, Gibbs free energy
100% A
100% B
A
B
∆G +ve → Kc < 1 → (more reactant > product)
∆G > +10
∆G = 0 (Equilibrium) ↓
Free energy minimum
Kc < 1
∆G increases ↑
70 % A30 % B
Equilibrium mix close to reactant
∆G < 0∆G = 0
A ↔ BG, Gibbs free energy
∆G more +ve → Kc < 1 → (All reactant / no product at all)
A
∆G = 0 (Equilibrium) ↓
Free energy minimum
Kc < 1100% A
100% B
Equilibrium mix close to reactant/ No reaction.
∆G > +100B
90 % A10 % B
∆G increases ↑
∆G = 0∆G < 0
reactant
reactant
reactant
reactant
productproduct
product product
Relationship bet ∆G and Kc
shift to left (reactant)
shift to right (product)
G, Gibbs free energy
A
B
100% A
100% B
∆G decreases ↓
30 % A70 % B
Equilibrium mixture
∆G < 0
∆G = 0 (Equilibrium) ↓
Free energy minimum
∆G < 0∆G < 0
∆G = 0
Free energy system is lowered on the way to equilibrium
Rxn proceed to minimum free energy ∆G = 0System seek lowest possible free energy
Product have lower free energy than reactant
∆G < -10
Kc > 1
A ↔ B A ↔ BG, Gibbs free energy
A
B
∆G decreases ↓
∆G < -100
100% A
100% B
∆G = 0 (Equilibrium) ↓
Free energy minimum
Kc > 1Equilibrium mixture
10 % A90 % B
∆G < 0
∆G < 0 ∆G = 0
∆G very –ve → Kc > 1 → (All product/close to completion)∆G –ve → Kc > 1 → (more product > reactant)
∆G
∆G = 0
∆G > 0
∆G < 0
No reaction/most reactantsKc <1
Complete rxn/Most productsKc > 1
Kc = 1 (Equilibrium)Reactants = Products
reactant reactant
ΔGθ Kc Eq mixture
ΔGθ = + 200
9 x 10-36 Reactant
ΔGθ = + 10 2 x 1-2 Mixture
ΔGθ = 0 Kc = 1 Equilibrium
ΔGθ = - 10 5 x 101 Mixture
ΔGθ = - 200 1 x 1035 Products
298314.8)212000(ln
RTGK c
Zn ↔ Zn2+ + 2e Eθ = +0.76Cu2+ + 2e ↔ Cu Eθ = +0.34Zn + Cu2+ → Zn 2+ + Cu Eθ = +1.10V
Zn half cell (-ve)Oxidation
Cu half cell (+ve)Reduction
Anode Cathode
Zn(s) | Zn2+(aq) || Cu2+
(aq) | Cu (s)
Cell diagram
Anode Cathode
Half Cell Half Cell(Oxidation) (Reduction)
Salt Bridge Flow electrons
Zn/Cu Voltaic Cell
-e -e
Zn/Cu half cells
Eθcell = Eθ
(cathode) – Eθ (anode)
Eθcell = +0.34 – (-0.76) = +1.10V
Zn 2+ + 2e ↔ Zn (anode) Eθ = -0.76VCu2+ + 2e ↔ Cu (cathode) Eθ = +0.34V
Std electrode potential as std reduction potential
Find Eθcell (use reduction potential)Find Eθ
cell (use formula)
Eθcell = Eθ
(cathode) – Eθ(anode)
Zn 2+ + 2e ↔ Zn Eθ = -0.76VCu2+ + 2e ↔ Cu Eθ = +0.34V
Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ 1/2H2 + OH- -0.83Zn2+ + 2e- ↔ Zn - 0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14Pb2+ + 2e- ↔ Pb -0.13H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4
2- + 4H+ + 2e- ↔ H2SO3 +0.17Cu2+ + 2e- ↔ Cu + 0.341/2O2 + H2O +2e- ↔ 2OH- +0.40
+
+1.10 V
Eθ Zn/Cu = 1.10V
Cu2+
----
Zn Cu++++
cellnFEG
E cell with ∆G
F = Faraday constant (96 500 Cmol-1)
n = number electron
cellnFEG
kJJG
G
212212300
10.1965002
∆G –ve, E +ve, K > 1∆G <0, E > 0, K > 1
↓Rxn SpontaneouscKRTG ln
Equilibrium constant
Gas constant, 8.314
∆G with Kc
cKRTG ln 37103.1 cKFavour products
Zn ↔ Zn2+ + 2e Eθ = +0.762Ag++2e ↔ 2Ag Eθ = +0.80Zn + Ag+ → Zn 2+ + Ag Eθ = +1.56V
Zn half cell (-ve)Oxidation
Ag half cell (+ve)Reduction
Anode Cathode
Zn(s) | Zn2+(aq) || Ag+
(aq) | Ag (s)
Cell diagram
Anode Cathode
Half Cell Half Cell(Oxidation) (Reduction)
Salt Bridge Flow electrons
Zn/Ag Voltaic Cell
-e -e
Zn/Ag half cells
Eθcell = Eθ
(cathode) – Eθ (anode)
Eθcell = +0.80 – (-0.76) = +1.56V
Zn 2+ + 2e ↔ Zn (anode) Eθ = -0.76VAg + + e ↔ Ag(cathode) Eθ = +0.80V
Std electrode potential as std reduction potential
Find Eθcell (use reduction potential)Find Eθ
cell (use formula)
Eθcell = Eθ
(cathode) – Eθ(anode)
Zn 2+ + 2e ↔ Zn Eθ = -0.76VAg+ + e ↔ Ag Eθ = +0.80V
Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ 1/2H2 + OH- -0.83Zn2+ + 2e- ↔ Zn - 0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14Pb2+ + 2e- ↔ Pb -0.13H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4
2- + 4H+ + 2e- ↔ H2SO3 +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54Fe3+ + e- ↔ Fe2+ +0.77Ag+ + e- ↔ Ag + 0.801/2Br2 + e- ↔ Br- +1.07
+
+1.56 V
Ag
Eθ Zn/Ag = +1.56V
Ag+
----
++++
Zn
E cell with ∆G
cellnFEG
n = number electron F = Faraday constant (96 500 Cmol-1)
cellnFEG
kJJG
G
301301000
56.1965002
∆G with Kc
cKRTG ln
Gas constant, 8.314 Equilibrium constant
cKRTG ln
298314.8)301000(ln
RTGK c
52105.3 cK
∆G –ve, E +ve, K > 1∆G <0, E > 0, K > 1
↓Rxn Spontaneous
Favour products
Mn ↔ Mn2+ + 2e Eθ = +1.19Ni2+ + 2e ↔ Ni Eθ = -0.26Mn + Ni2+ → Mn2+ + Ni Eθ = +0.93V
Mn half cell (-ve)Oxidation
Ni half cell (+ve)Reduction
Anode Cathode
Mn(s) | Mn2+(aq) || Ni2+
(aq) | Ni (s)
Cell diagram
Anode Cathode
Half Cell Half Cell(Oxidation) (Reduction)
Salt Bridge Flow electrons
Mn/Ni Voltaic Cell
-e -e
Mn/Ni half cells
Eθcell = Eθ
(cathode) – Eθ (anode)
Eθcell = -0.26 – (-1.19) = +0.93V
Mn 2+ + 2e ↔ Mn (anode) Eθ = -1.19V
Ni2+ + 2e ↔ Ni (cathode) Eθ = -0.26V
Std electrode potential as std reduction potential
Find Eθcell (use reduction potential)Find Eθ
cell (use formula)
Eθcell = Eθ
(cathode) – Eθ(anode)
Mn 2+ + 2e ↔ Mn Eθ = -1.19VNi2+ + 2e ↔ Ni Eθ = -0.26V
Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ 1/2H2 -0.83Zn2+ + 2e- ↔ Zn -0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni - 0.26Sn2+ + 2e- ↔ Sn -0.14Pb2+ + 2e- ↔ Pb -0.13H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4
2- + 4H+ + 2e- ↔ H2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54+
+0.93 V
Eθ Mn/Ni = +0.93V
Ni2+
----
NiMn++++Mn2+
E cell with ∆G
cellnFEG
n = number electron F = Faraday constant (96 500 Cmol-1)
cellnFEG
kJJG
G
179179490
93.0965002
cKRTG ln
298314.8)179000(ln
RTGK c
cKRTG ln
∆G with Kc
Gas constant, 8.314 Equilibrium constant
∆G –ve, E +ve, K > 1∆G <0, E > 0, K > 1
↓Rxn Spontaneous
31102.2 cKFavour products
Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ H2 + OH- -0.83Zn2+ + 2e- ↔ Zn -0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14
H+ + e- ↔ H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4
2- + 4H+ + 2e- ↔ H2S +0.17Cu2+ + 2e- ↔ Cu +0.34
Cu ↔ Cu2+ + 2e Eθ = -0.342H+ + 2e ↔ H2 Eθ = +0.00Cu + 2H+→ Cu2+ +H2
Eθ = -0.34V
Rxn bet Cu + H+
Will it happen ?
Eθ = -0.34V (NON spontaneous) О
Cu(s) | Cu2+(aq) || H+
H2 | Pt (s)
(Oxidation) (Reduction)
Anode Cathode
Find Eθcell (use formula)
Eθcell = Eθ
(cathode) – Eθ (anode)
Eθcell = 0.00 – (+0.34) = -0.34V
Eθ = -0.34V (NON spontaneous)
О Rxn not feasible
Determine spontaneity rxn. Will it HAPPEN ?
Find Eθcell (use reduction potential)
Eθ Cu/H+ = - 0.34VE cell with ∆G
cellnFEG
n = number electron F = Faraday constant (96 500 Cmol-1)
cellnFEG
kJJG
G
6565620
34.0965002
cKRTG ln
Gas constant, 8.314 Equilibrium constant
∆G with Kc
cKRTG ln
298314.8)65000(ln
RTGK c
∆G +ve, E -ve, K < 1∆G >0, E < 0, K < 1
↓Rxn Non Spontaneous
12104 cKFavour reactants
-0.34 V
acid
copper
Predicting will rxn occur with ΔG, E cell and Kc
+
Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ H2 + OH- -0.83Zn2+ + 2e- ↔ Zn -0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14
H+ + e- ↔ H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4
2- + 4H+ + 2e- ↔ H2S +0.17Cu2+ + 2e- ↔ Cu +0.34
Au3+ + 3e- ↔ Au +1.58
Rxn bet Au + H+
Will it happen ?
Eθ = -1.58 V (NON spontaneous)
ОAu(s) | Au3+
(aq) || H+ H2 | Pt (s)
(Oxidation) (Reduction)
Anode Cathode
Find Eθcell (use formula)
Eθcell = Eθ
(cathode) – Eθ (anode)
Eθcell = 0.00 – (+1.58) = -1.58V
Eθ = - 1.58 V (NON spontaneous)
О Rxn not feasible
Determine spontaneity rxn. Will it HAPPEN ?
Find Eθcell (use reduction potential)
Eθ Au/H+ = - 1.58VE cell with ∆G
cellnFEG
n = number electron F = Faraday constant (96 500 Cmol-1)
cellnFEG
kJJG
G
914914820
58.1965006
cKRTG ln
Gas constant, 8.314 Equilibrium constant
∆G with Kc
cKRTG ln
298314.8)914000(ln
RTGK c
∆G +ve, E -ve, K < 1∆G >0, E < 0, K < 1
↓Rxn Non Spontaneous
50104 cKKc too small – No reaction at all
-1.58 V
acid
gold
2Au ↔ 2Au3+ + 6e Eθ = -1.586H+ + 6e ↔ 3H2 Eθ = 0.002Au + 6H+ → 2Au3+ + 3H2
Eθ
= -1.58V
Predicting will rxn occur with ΔG, E cell and Kc
+
Eθ = - 0.20 V (NON spontaneous)
(Oxidation) (Reduction)
Anode Cathode
Find Eθcell (use formula)
Eθcell = Eθ
(cathode) – Eθ (anode)
Eθcell = 0.34 – (0.54) = - 0.20V
Eθ = - 0.20 V (NON spontaneous)
Determine spontaneity rxn. Will it HAPPEN ?
Find Eθcell (use reduction potential)
Eθ Cu2+/I- = - 0.20VE cell with ∆G
cellnFEG
n = number electron F = Faraday constant (96 500 Cmol-1)
cellnFEG
kJJG
G
3838600
20.0965002
cKRTG ln
Gas constant, 8.314 Equilibrium constant
∆G with Kc
cKRTG ln
298314.8)38000(ln
RTGK c
∆G +ve, E -ve, K < 1∆G >0, E < 0, K < 1
↓Rxn Non Spontaneous
7102.2 cK
-1.58 V
Cu2+
I-Rxn bet Cu2+ +I-
Will it happen ?
2I- ↔ I2 + 2e Eθ = -0.54Cu2+ + 2e ↔ Cu Eθ = +0.342I- + Cu2+→ Cu + I2
Eθ = -0.20V
Pt(s) | I-, I2 || Cu2+(aq) | Cu (s)
Favour reactants
Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19Zn2+ + 2e- ↔ Zn -0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15
Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.52I2 + 2e- ↔ I- +0.54
Rxn not feasible
О
О - 0.20 V
Predicting will rxn occur with ΔG, E cell and Kc
Will I- oxidize Cu 2+ to Cu
Click here to view free energy
Predicting Spontaneity of Rxn
Thermodynamic, ΔG Equilibrium, Kc
1cK
1cK
KRTG lnG
veG
cK
1cK
Energetically favourable
0G
Predicting rxn will occur?
N2(g) + 3H2(g) ↔ 2NH3(g)
H2O(l) ↔ H+(aq)+ OH-
(aq)
Shift toward reactants
Energeticallyunfavourable Non spontaneous
Mixturereactant/productEquilibrium
veG Spontaneous Shift toward product
79G
33G610G
14101 cK
5105cK
Fe(s) + 3O2(g) ↔ 2Fe2O3(s)
261101cK
Shift toward reactants
Energeticallyunfavourable
Shift toward product
Energetically favourable
Energetically favourable
Kinetically unfavourable/(stable)Rate too slow due to HIGH activation energy
Rusting Process
Energy barrier
Shift toward product
Reaction too slow
Click here for notes
cellnFEG
Cell Potential
cellE
0cellE
0cellE
0cellE
0cellE
0cellE
0cellE
Eθ = +0.44V
IB Questions
Esterification produce ethyl ethanoate. ΔG = -4.38kJmol-1 Cal Kc
CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l)
Kc = 5.9
cKRTG lnRTGK c
ln
29831.84380ln
cK
2
?cK
NO oxidized to NO2. Kc = 1.7 x 1012. Cal ∆G at 298K
1
3 4
2NO + O2 ↔ NO2
?G
cKRTG ln
11
12
7.6969772
)107.1ln(298314.8
kJmolJmolG
G
Predict if iron react with HCI in absence air. Cal E cell , ∆G and Kc
Oxidized sp ↔ Reduced sp Eθ/VFe2+ + 2e- ↔ Fe -0.442H+ + 2e- ↔ H2 0.00O2 +2H2O+4e ↔ 4OH- +0.40
Fe2+ + 2e- ↔ Fe -0.442H+ + 2e- ↔ H2 0.00
ОО
Fe ↔ Fe2+ + 2e Eθ = +0.442H+ + 2e ↔ H2 Eθ = 0.00VFe + 2H+ → Fe2+ + H2
Eθ = +0.44V
cellnFEG
kJJG
G
8584900
44.0965002
cKRTG ln
298314.8)85000(ln
RTGK c
14108.7 cK
∆G –ve, E +ve, K > 1∆G <0, E > 0, K > 1
↓Rxn Spontaneous
Fe2+ + 2e- ↔ Fe -0.44O2 +2H2O+4e ↔ 4OH- +0.40
2Fe ↔ 2Fe2+ + 4e Eθ = +0.44O2+2H2O+4e ↔ 4OH- Eθ = +0.402Fe +O2
+2H2O→2Fe2++4OH- Eθ = +0.84V
Eθ = +0.84V
Oxidized sp ↔ Reduced sp Eθ/VFe2+ + 2e- ↔ Fe -0.442H+ + 2e- ↔ H2 0.00O2 +2H2O+4e ↔ 4OH- +0.40
Predict iron react HCI in presence of air. Cal E cell , ∆G and KcО
О cellnFEG
kJJG
G
324324000
84.0965004
cKRTG ln
298314.8)324000(ln
RTGK c
56108.2 cK
∆G –ve, E +ve, K > 1∆G <0, E > 0, K > 1
↓Rxn SpontaneousRusting is spontaneous
x 2
ОО ОО
Predict if manganate will oxidize chloride ion?MnO2 + 4H+ + 2CI- → Mn2+ + 2H2O + CI2
5
MnO2 +4H+ + 2e- ↔ Mn2+ + 2H2O +1.23
1/2CI2 + e- ↔ CI- +1.36
2CI- ↔ CI2 + 2e Eθ = -1.36MnO2 + 4H+ + 2e ↔ Mn2+ + 2H2O Eθ = +1.23MnO2 + 4H++2CI- → Mn2++2H2O+CI2
Eθ= -0.13V Eθ = -0.13V
Oxidized sp ↔ Reduced sp Eθ/VCr2O7
2-+ 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.33MnO2
+4H+ + 2e- ↔ Mn2+ + 2H2O +1.231/2CI2 + e- ↔ CI- +1.36MnO4
- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51
Predict if MnO4- able to oxidize aq CI- to CI2
2MnO4 + 16H+ + 10CI- → 2Mn2+ + 8H2O + 5CI2
ОО Oxidized sp ↔ Reduced sp Eθ/VCr2O7
2-+ 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.33MnO2
+4H+ + 2e- ↔ Mn2+ + 2H2O +1.231/2CI2 + e- ↔ CI- +1.36MnO4
- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51
О О2CI- ↔ CI2 + 2e Eθ = -1.36MnO4 - + 8H+ + 5e ↔ Mn2+ + 4H2O Eθ = +1.512MnO4 + 16H++10CI- → 2Mn2++8H2O+5CI2 Eθ= +0.15V
1/2CI2 + e- ↔ CI- +1.36MnO4
- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51
Eθ = +0.15V
IB Questions
cellnFEG
kJJG
G
2525000
13.0965002
cKRTG ln
298314.8)25000(ln
RTGK c
5105.4 cK∆G +ve, E -ve, K < 1∆G >0, E < 0, K < 1
↓Rxn Non Spontaneous
6
cellnFEG
kJJG
G
144144750
15.09650010
cKRTG ln
298314.8)144000(ln
RTGK c
25105.1 cK
∆G –ve, E +ve, K > 1∆G <0, E > 0, K > 1
↓Rxn Spontaneous
x 5x 2
О О О О
Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com
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