ib chemistry on gibbs free energy and equilibrium constant, kc
TRANSCRIPT
http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
Tutorial on Gibbs Free Energy Change and Equilibrium
Dynamic Equilibrium
Reversible (closed system)
Forward Rate, K1 Reverse Rate, K-1
Conc of product and reactant at equilibrium
At Equilibrium
Forward rate = Backward rateConc reactants and products remain CONSTANT/UNCHANGE
Equilibrium Constant Kc
aA(aq) + bB(aq) cC(aq) + dD(aq)
coefficient
Solid/liq not included in Kc Conc represented by [ ]
kf
Kr
ba
dc
c BADCK
r
fc k
kK
Equilibrium Constant Kc
express in
Conc vs time Rate vs time
A + B
C + D
Conc
Time
Catalyst
Factors affecting equilibrium (closed system)
TemperaturePressureConcentration
Equilibrium constant Kc ≠ Position equilibrium
forward rate constant
reverse rate constant
Effect of Temperature on position of equilibrium
Decrease Temp ↓ • Favour exo rxn• Equi shift to right → to increase ↑ Temp• Formation Co(H2O)6
2+ (pink)
Increase Temp ↑• Favour endo rxn• Equi shift to left ← to
reduce ↓ Temp• Formation of CoCl4
2- (blue)
CoCl42- + 6H2O ↔ Co(H2O)6
2+ + 4CI – ΔH = -ve (exo) (blue) (pink)
Increase Temp ↑ – Favour endo rxn – Absorb heat to reduce Temp again ↓Decrease Temp ↓ – Favour exo rxn – Release heat to increase Temp again ↑
Increase Temperature• Rate rxn increase ↑• Rate constant also change• Rate constant forward/reverse increase but to
diff extend• Position equi shift to endo to decrease ↓ Temp • Kc, equilibrium constant change
Click to view video
Le Chatelier’s Principle • System in dynamic equilibrium is disturbed, position of equilibrium will
shift so to cancel out the effect of change and new equilibrium can established again
Effect of Temperature on equilibrium constant, Kc
Le Chatelier’s Principle • System in dynamic equilibrium is disturbed, position of equilibrium will
shift so to cancel out the effect of change and new equilibrium can established again
Decrease Temp ↓• Favour exo rxn• Equi shift to left ← to
increase ↑ Temp• Formation N2O4 (colourless)
Increase Temp ↑• Favour endo rxn• Equi shift to right → to reduce
↓ Temp• Formation NO2
(brown)
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
(colourless) (brown)
Click to view video
Effect of Temperature on position of equilibrium
Effect of Temperature on equilibrium constant, Kc
Increase Temp ↑ – Favour endo rxn – Absorb heat to reduce Temp again ↓Decrease Temp ↓ – Favour exo rxn – Release heat to increase Temp again ↑
Increase Temperature• Rate rxn increase ↑• Rate constant also change• Rate constant forward/reverse increase but to
diff extend• Position equi shift to endo to decrease ↓ Temp • Kc, equilibrium constant change
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
Temp increase ↑ – Kc increase ↑ Why?
A ↔ B ΔH = +ve
Reverse rate constant = k r
Forward rate constant = kf
Kc
ABK c
r
fc k
kK
Temp affect rate constant
Temp change
cK
Increase Temp ↑
Position equilibrium shift to right Endo side – Absorb heat Temp decrease ↓
More product , less reactant treacproductK c tan
cK
Forward rate constant, kf > reverse rate, kr
r
fc k
kK
Decrease Temp ↓
Position equilibrium shift to left Exo side – Release heat Temp increase ↑
More reactant , less product treacproductK c tan
Forward rate constant, kf < reverse rate, kr
r
fc k
kK
cK
Conclusion : Endo rxn – Temp ↑ – Kc ↑ – Product ↑
Effect of Temperature on equilibrium constant, Kc
forward rate constant
reverse rate constant
Temp increase ↑ – Kc decrease ↓ Why?
A ↔ B ΔH = -ve
Increase Temp ↑
Position equilibrium shift to left Endo side – Absorb heat Temp decrease ↓
More Reactant , less product treacproductK c tan
cK
Forward rate constant, kf < Reverse rate, kr
r
fc k
kK
Decrease Temp ↓
Position equilibrium shift to right Exo side – Release heat Temp increase ↑
More Product , less reactant treacproductK c tan
Forward rate constant, kf > Reverse rate, kr
r
fc k
kK
cK
Conclusion : Exo rxn – Temp ↑ – Kc ↓ – Product ↓
H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1
Forward rate constant = kf
Reverse rate constant = k r
Effect of Temperature on equilibrium constant, Kc
Kc
ABK c
r
fc k
kK
forward rate constant
reverse rate constant
Temp affect rate constant
Temp change
cK
Equilibrium Constant Kc
express in
At equilibrium Rate of forward = Rate of reverse
DCkBAk rf
Forward Rate
Reverse Rate
Rate forward = kf [A] [B] Rate reverse = kr [A] [B]
BADC
kk
r
f
Change in Temp
=
r
fc k
kK
BADCK c
Equilibrium and KineticsForward Rate
Reverse Rate
At equilibrium Rate of forward = Rate of reverse
kf = forward rate constant
kr = reverse rate constant
same
BADCK c
r
fc k
kK
Ratio of product /reactant conc
Ratio of rate constant
Change rate constant, kf/kr Change in Kc
Temp increase ↑ – Kc increase ↑ Temp increase ↑ – Kc decrease ↓
A ↔ B ΔH = +ve
A ↔ B ΔH = -ve
Temp changes Kc with diff rxn?
Endothermic rxn
Exothermic rxn
kf
kr
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
Predicting rxn will 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
products
reactants
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
shift to left
shift to right
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 = 0Sys 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
Predicting rxn will 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, Q and Kc
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
∆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
Relationship bet ∆G and Kc
G, Gibbs free energy
A
B
100% A
100% B
∆G decreases ↓
30 % A70 % B
Equilibrium mixture 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 mixture 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 mixture 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 mixture 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
products
reactants
shift to left
shift to right
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 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
Gibbs Free Energy Change, ∆G
∆G - Temp/Pressure remain constantAssume ∆S/∆H constant with temp
Using ∆Gsys to predict spontaneity
syssyssys STHG
Easier
Unit ∆G - kJ mol-1
Only ∆S sys involved∆S surr, ∆S uni not needed
Using ∆Gsys to predict spontaneity
Easier
Method 1 Method 2
)()( reactfprofsys GGG At std condition/states
Temp - 298KPress - 1 atm
Gibbs Free Energy change formation, ∆Gf
0
At High Temp ↑
Temp dependent
syssyssys STHG
At low Temp ↓
veGSTG
HST sys
syssyssys STHG
veGHGSTH
spontaneous spontaneous
surrsysuni SSS
TH
S syssurr
syssysuni STHST
Deriving Gibbs Free Energy Change, ∆G
TH
SS syssysuni
∆S sys / ∆H sys
multi by -T
syssyssys STHG
∆Hsys ∆Ssys ∆Gsys Description
- + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at all Temp
+ - ∆G = ∆H - T ∆S ∆G = + ve Non spontaneous, all Temp
unisys STG syssyssys STHG
Only ∆H sys/∆S sys involved ∆S surr, ∆S uni not needed
Non standard condition
Gibbs Free Energy Change, ∆G
syssyssys STHG unisys STG
veGsys ∆S uni = +ve
Spontaneous SpontaneousveGsys
∆H = - ve∆S sys = +ve
∆Hsys ∆Ssys ∆Gsys Description
+ + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp
- - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp
Gibbs Free Energy change formation, ∆Gf
0
At High Temp ↑
Temp dependent
syssyssys STHG
At low Temp ↓
veGSTG
HST sys
syssyssys STHG
veGHGSTH
spontaneous spontaneous
∆Hsys ∆Ssys ∆Gsys Description
- + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at all Temp
+ - ∆G = ∆H - T ∆S ∆G = + ve Non spontaneous, all Temp
syssyssys STHG
∆Hsys ∆Ssys ∆Gsys Description
+ + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp
- - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp
Relationship Equilibrium and Energetics
At equilibrium∆G = 0
SHT
HST
CaCO3 (s) → CaO(s) + CO2(g)
CaCO3 (s) → CaO (s) + CO2(g) ∆Hf
0 - 1206 - 635 - 393S0 + 93 + 40 + 213
At what temp will decomposition
CaCO3 be spontaneous?
Reactant Product∆Hsys
θ = ∑∆Hfθ(pro) - ∑∆Hf
θ(react) ∆Ssys
θ = ∑Sfθ(pro) - ∑Sf
θ(react)
kJH sys 178)1206(1028
kJS
JS
S
SSS
sys
sys
sys
treacproductsys
16.0
160
93253)tan()(
STHG SHT
HST KT 111216.0178
Unit ∆H – kJUnit ∆S - JK-1
At equilibrium∆G = 0
Click here notes from chemwiki
∆H = +ve, ∆S = +ve → Temp ↑ High → Spontaneous∆H = -ve, ∆S = -ve → Temp ↓ Low → Spontaneous
∆Hsys ∆Ssys ∆Gsys Description
+ + ∆G = ∆H - T∆S ∆G = 0 Equilibrium at Temp, T
- - ∆G = ∆H - T∆S ∆G = 0 Equilibrium at Temp, T
∆G = 0
kJGG
STHG
130)16.0(298178
Predict what happen at diff Temp
Reactant Product
∆Hsysθ = ∑∆Hf
θ(pro) - ∑∆Hf
θ(react) ∆Ssys
θ = ∑Sfθ(pro) - ∑Sf
θ(react)
kJH sys 178)1206(1028
∆G > 0 Decomposition at
298K Non Spontaneous
CaCO3 (s) → CaO(s) + CO2(g)
CaCO3 (s) → CaO (s) + CO2(g) ∆Hf
0 - 1206 - 635 - 393S0 + 93 + 40 + 213
kJS
JS
sys
sys
16.0
16093253
Decomposition at 298K Decomposition at 1500K
Decomposition limestone CaCO3 spontaneous?
Gibbs Free Energy Change, ∆G
kJGG
STHG
62)16.0(1500178
∆H = +ve ∆S = +ve
Temp dependent
∆Hsys ∆Ssys ∆Gsys Description
+ + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp
- - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp
At Low Temp At High Temp
Unit ∆H – kJUnit ∆S - JK-1
Equilibrium, at what Temp?∆G = 0
∆G < 0 Decomposition at
1500K Spontaneous
STHG
HST
SHT
KT 111216.0178
Temp > 1112Krxn spontaneous
Temp dependentSpontaneous at
High ↑ temp
Predict what happen at diff Temp
Reactant Product
∆Hsysθ = ∑∆Hf
θ(pro) - ∑∆Hf
θ(react) ∆Ssys
θ = ∑Sfθ(pro) - ∑Sf
θ(react)
∆G > 0 Freezing at 298K Non Spontaneous
Freezing at 298K (25C) Freezing at 263K (-10C)
Is freezing water spontaneous?
Gibbs Free Energy Change, ∆G
∆H = -ve ∆S = -ve
Temp dependent
At High Temp At Low Temp
Unit ∆H – kJUnit ∆S - JK-1
Equilibrium, at what Temp?∆G = 0
∆G < 0 Freezing at 263K (-
10C) Spontaneous
STHG
HST
SHT
)0(273022.0010.6 CKT
H2O (l) → H2O(s)
Is Freezing spontaneous?
H2O (l) → H2O(s) ∆Hf
0 - 286 - 292S0 + 70 + 48
kJH sys 010.6)286(292 kJS
JS
sys
sys
022.0
227048
kJGG
STHG
55.0)022.0(298010.6
∆Hsys ∆Ssys ∆Gsys Description
+ + ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at high ↑ Temp
- - ∆G = ∆H - T ∆S ∆G = - ve Spontaneous at low ↓ Temp
Water start to freezeTemp < 0 , Spontaneous
kJGG
STHG
22.0)022.0(263010.6
Relationship betweenEquilibrium and Energetics
KRTG ln
HG cKG
EnergeticallyThermodynamicall
yFavourable/
feasible
Sign predict spontaneity of rxn
Negative (-ve)
spontaneous
Positive (+ve) NOT spontaneous
veG veG
NOT
favourable
Energetically
favourable
Product formation
NO product
KRTG ln
STHG
H2(g) + O2(g) → H2O2(l)
Energetically feasible ΔG / ΔH = -ve
Predicting if rxn will occur?veG veH f
Energetic favourable (-ve)
Product H2O2 more stableΔG and ΔH = -
negativeReaction wont happen!!!!!!
Kinetically unfavourable/stable
due to HIGH activation energy
H2(g) + O2(g) H2O2(l)
Energy barrier
Will rxn occur?
depends
Kinetically feasible
low activation energy
+To occur
ΔG < 0 (-ve) Activation energy LOW
Energetically favourable
Kinetically favourable
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
Energetic favourable (-ve) Graphite more stable
Relationship betweenEquilibrium and Energetics
KRTG ln
HG cKG
EnergeticallyThermodynamicall
yFavourable/
feasible
Sign predict spontaneity of rxn
Negative (-ve)
spontaneous
Positive (+ve) NOT spontaneous
veG veG
NOT
favourable
Energetically
favourable
Product formation
NO product
KRTG ln
STHG
Energetically feasible ΔG / ΔH = -ve
Predicting if rxn will occur?veG veH f
ΔG and ΔH = -negative
Reaction wont happen!!!!!!
Kinetically unfavourable/stable
due to HIGH activation energy
Energy barrier
Will rxn occur?
depends
Kinetically feasible
low activation energy
+To occur
ΔG < 0 (-ve) Activation energy LOW
Energetically favourable
Kinetically favourable
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
Diamond(s) → Graphite(s)
Diamond forever
Diamond(s) Graphite(s)
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 reactantsEnergeticallyunfavourable Non spontaneous
Mixturereactant/product
Equilibrium
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
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
Kc at 1338K is 0.0118. Cal Kc at 1473K
A + B ↔ C + D kJH 3.177
Qualitative(Le Chatelier Principle)
Quantitatively Formula
14731
13381
31.8177300
0118.0ln 2K
K2 = 0.051Temp increase ↑ – Kc increase ↑
Endothermic rxn
A + B ↔ C + D
Kc at 1000K and 1200K is 2.44 and 3.74. Cal ΔH.
?H
211
2 11lnTTR
HKK
12001
10001
31.844.274.3ln H
ΔH = 21.3kJmol-1
2
?cK
?cK
Temp decrease ↓ again
Temp increase ↑
Shift to right → - absorb heat
211
2 11lnTTR
HKK
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
Van’t Hoff Equation
cKRTG ln
Relationship bet Temp and Kc
STHG
STHKRT ln
RS
RTHK c
ln
cRTHK c
ln
Heat absorbed, ΔH +veTemp increase ↑ – Kc increase ↑
Heat released, ΔH –veTemp increase ↑ – Kc decrease ↓
Gibbs free energy change Equilibrium constant
Enthalpy change
Entropy change
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
Temp increase ↑ – Kc increase ↑
H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1
Temp increase ↑ – Kc decrease ↓
cTR
HK
1ln
Endothermic rxn
cRTHK
ln
Temp increase ↑ – Kc increase ↑
Exothermic rxn
cRTHK
ln
cTR
HK
1ln
Temp increase ↑ – Kc decrease ↓
Temp/K
250 400 650 1000
Kc 800 160 50 24
ΔH= +ve ΔH= -ve
Temp/K
350 400 507 550
Kc 3.89 47.9 1700 6030
Gibbs free energy change
Van’t Hoff Equation
cKRTG ln
Relationship bet Temp and Kc
RS
RTHK
ln
cRTHK
ln
Gibbs free energy changeEquilibrium constant
Enthalpy change
Entropy change
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
Temp increase ↑ – Kc increase ↑
Endothermic rxn
cTR
HK
1ln
Plot Kc against Temp
Temp/K
350 400 507 550
Kc 3.89 47.9 1700 6030
ln Kc 1.36 3.87 7.44 8.7
1/T(x 10-
3)2.86 2.50 1.97 1.82
RTH
c eK
Plot ln Kc against 1/T
N2O4 (g) ↔ 2NO2(g) ΔH = + 54kJmol-1
Endothermic rxn
Using Kc and Temp to find ΔH
Conclusion
RHGradient
31.8007.0 H
JH 58170
Temp increase ↑ Kc increase ↑ Endo rxn =+ΔH Relationship bet Temp, Kc and ΔH
ΔH=+ve
-0.007
STHG
Gibbs free energy change
STHKRT ln
Van’t Hoff Equation
KRTG ln
Relationship bet Temp and Kc
RS
RTHK
ln
cRTHK
ln
Gibbs free energy changeEquilibrium constant
Enthalpy change
Entropy change
Exothermic rxn
cTR
HK
1ln
Plot Kc against Temp
ln Kc 6.9 3.45 -3.3 -9.5
1/T(x 10-
3)2.9 2.6 2 1.4
RTH
c eK
Plot ln Kc against 1/T
Exothermic rxn
Using Kc and Temp to find ΔH
Conclusion
RHGradient
31.811316 H
JH 94000
Temp increase ↑ Kc decrease ↓ Exo rxn =-ΔH Relationship bet Temp, Kc and ΔH
H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1
H2(g) + I2(g) ↔ 2HI(g) ΔH = -9.6kJmol-1
Temp/K
345 385 500 700
Kc 1000 31.6 0.035 0.00007
Temp increase ↑ – Kc decrease ↓
ΔH=-ve
+11316
STHKRT lnSTHG
Gibbs free energy change
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