Dynamic Equilibrium

Reversible (closed system)

Forward Rate, K1 Reverse Rate, K-1

Conc of product and reactant at equilibrium

At Equilibrium

Forward rate = Backward rate Conc 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

cBA

DCK

r

f

ck

kK

Equilibrium Constant Kc

express in

Conc vs time Rate vs time

A + B

C + D

Conc

Time

Catalyst

Factors affecting equilibrium (closed system)

Temperature Pressure Concentration

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 ↑

A ↔ B ΔH = +ve Reverse rate constant = k r

Forward rate constant = kf

Kc

A

BK c

r

f

ck

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 treac

productK c

tan

cKForward rate constant, kf > reverse rate, kr

r

f

ck

kK

Decrease Temp ↓

Position equilibrium shift to left Exo side – Release heat Temp increase ↑

More reactant , less product treac

productKc

tan

Forward rate constant, kf < reverse rate, kr

r

f

ck

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 ↓

A ↔ B ΔH = -ve

Increase Temp ↑

Position equilibrium shift to left Endo side – Absorb heat Temp decrease ↓

More Reactant , less product treac

productK c

tan

cKForward rate constant, kf < Reverse rate, kr

r

f

ck

kK

Decrease Temp ↓

Position equilibrium shift to right Exo side – Release heat Temp increase ↑

More Product , less reactant treac

productKc

tan

Forward rate constant, kf > Reverse rate, kr

r

f

ck

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

A

BK c

r

f

ck

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]

BA

DC

k

k

r

f

Change in Temp

=

r

f

ck

kK

BA

DCK c

Equilibrium and Kinetics

Forward Rate

Reverse Rate

At equilibrium Rate of forward = Rate of reverse

kf = forward rate constant

kr = reverse rate constant

BA

DCKc

r

f

ck

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

cK

Temp dependent

Extend of rxn

Not how fast

Shift to left/ favour reactant

Shift to right/ favour product

cK

Relationship between Equilibrium and Energetics

cKRTG ln STHG

Enthalpy

change Entropy

change

Equilibrium

constant Gibbs free energy change

HG cK

G

Energetically 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

cK

Temp dependent

Extend of rxn

Not how fast

Shift to left/ favour reactant

Shift to right/ favour product

cK

Relationship 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 % A 70 % 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 = 0

Sys seek lowest possible free energy Product have lower free energy than reactant

∆G < 0 product reactant

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 % A 70 % B

Equilibrium mixture

∆G < 0

∆G = 0 (Equilibrium) ↓

Free energy minimum

∆G < 0

∆G < 0

∆G = 0

∆G < 0 product reactant

G, Gibbs free energy

reactant product ∆G < 0 A

B

∆G decreases ↓

100% A 100% B 30 % A 70 % 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 % A 70 % 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 ↔ B

G, Gibbs free energy

A

B

∆G decreases ↓

∆G < -100

100% A

100% B

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc > 1 Equilibrium mixture close to product

10 % A 90 % 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 ↔ B

G, 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 % A 30 % B

Equilibrium mixture close to reactant

∆G < 0

∆G = 0

A ↔ B

G, Gibbs free energy

∆G more +ve → Kc < 1 → (All reactant / no product at all)

A

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc < 1 100% A

100% B

Equilibrium mixture close to reactant/ No reaction.

∆G > +100 B

90 % A 10 % B

∆G increases ↑

∆G = 0

∆G < 0

reactant

reactant

reactant

reactant

product product

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 % A 70 % 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 = 0

System seek lowest possible free energy Product have lower free energy than reactant

∆G < -10

Kc > 1

A ↔ B A ↔ B

G, Gibbs free energy

A

B

∆G decreases ↓

∆G < -100

100% A

100% B

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc > 1 Equilibrium mixture

10 % A 90 % 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 reactants Kc <1

Complete rxn/Most products Kc > 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 constant Assume ∆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 - 298K Press - 1 atm

Gibbs Free Energy change formation, ∆Gf0

At High Temp ↑

Temp dependent

syssyssys STHG

At low Temp ↓

veG

STG

HST sys

syssyssys STHG

veG

HG

STH

spontaneous spontaneous

surrsysuni SSS

T

HS

sys

surr

syssysuni STHST

Deriving Gibbs Free Energy Change, ∆G

T

HSS

sys

sysuni

∆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 Spontaneous

veGsys ∆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, ∆Gf0

At High Temp ↑

Temp dependent

syssyssys STHG

At low Temp ↓

veG

STG

HST sys

syssyssys STHG

veG

HG

STH

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

S

HT

HST

CaCO3 (s) → CaO(s) + CO2(g)

CaCO3 (s) → CaO (s) + CO2(g) ∆Hf

0 - 1206 - 635 - 393 S0 + 93 + 40 + 213

At what temp will decomposition CaCO3 be spontaneous?

Reactant Product

∆Hsysθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)

kJHsys 178)1206(1028

kJS

JS

S

SSS

sys

sys

sys

treacproductsys

16.0

160

93253

)tan()(

STHG S

HT

HST KT 1112

16.0

178

Unit ∆H – kJ Unit ∆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

kJG

G

STHG

130

)16.0(298178

Predict what happen at diff Temp

Reactant Product

∆Hsysθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)

kJHsys 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 - 393 S0 + 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

kJG

G

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 – kJ Unit ∆S - JK-1

Equilibrium, at what Temp? ∆G = 0

∆G < 0 Decomposition at 1500K

Spontaneous

STHG

HST

S

HT

KT 111216.0

178

Temp > 1112K Decomposition spontaneous

Temp dependent Spontaneous at

High ↑ temp

Predict what happen at diff Temp

Reactant Product

∆Hsysθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Ssysθ = ∑Sf

θ(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 – kJ Unit ∆S - JK-1

Equilibrium, at what Temp? ∆G = 0

∆G < 0 Freezing at 263K (-10C)

Spontaneous

STHG

HST

S

HT

)0(273022.0

010.6CKT

H2O (l) → H2O(s)

Is Freezing spontaneous?

H2O (l) → H2O(s) ∆Hf

0 - 286 - 292 S0 + 70 + 48

kJH sys 010.6)286(292 kJS

JS

sys

sys

022.0

227048

kJG

G

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 freeze Temp < 0 , Spontaneous

kJG

G

STHG

22.0

)022.0(263010.6

Relationship between Equilibrium and Energetics

KRTG ln

HGcK

GEnergetically

Thermodynamically Favourable/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

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 stable

ΔG and ΔH = -negative

Reaction 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

STHG

Energetic favourable (-ve) Graphite more stable

Relationship between Equilibrium and Energetics

KRTG ln

HGcK

GEnergetically

Thermodynamically Favourable/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

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)

STHG

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 Energetically unfavourable

Non spontaneous

Mixture reactant/product

Equilibrium

veG Spontaneous Shift toward product

79G

33G

610G

14101 cK

5105cK

Fe(s) + 3O2(g) ↔ 2Fe2O3(s) 261101cK

Shift toward reactants

Energetically unfavourable

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

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 lnRT

GK c

ln

29831.8

4380ln

cK

Kc at 1338K is 0.0118. Cal Kc at 1473K

A + B ↔ C + D kJH 3.177

Qualitative (Le Chatelier Principle)

Quantitatively Formula

1473

1

1338

1

31.8

177300

0118.0ln 2K

K2 = 0.051

Temp 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 11ln

TTR

H

K

K

1200

1

1000

1

31.844.2

74.3ln

H

ΔH = 21.3kJmol-1

2

?cK

?cK

Temp decrease ↓ again

Temp increase ↑

Shift to right → - absorb heat

211

2 11ln

TTR

H

K

K

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

R

S

RT

HK c

ln

cRT

HK c

ln

Heat absorbed, ΔH +ve Temp increase ↑ – Kc increase ↑

Heat released, ΔH –ve Temp 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

cRT

HK

ln

Temp increase ↑ – Kc increase ↑

Exothermic

rxn

cRT

HK

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

R

S

RT

HK

ln

cRT

HK

ln

Gibbs free energy change Equilibrium

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

RT

H

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

R

HGradient

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

R

S

RT

HK

ln

cRT

HK

ln

Gibbs free energy change Equilibrium

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

RT

H

c eK

Plot ln Kc against 1/T

Exothermic rxn

Using Kc and Temp to find ΔH

Conclusion

R

HGradient

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 ln

STHG

Gibbs free energy change