IB Chemistry on Gibbs Free Energy and Entropy

http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

Tutorial on Gibbs Free Energy and Spontaneity .

http://lawrencekok.blogspot.com/

E = sum kinetic energy/motion of molecule, and potential energy represented by chemical bond bet atom

∆E = q + w

∆E = Change internal energy

q = heat transfer

w = work done by/on system

Thermodynamics Study of work, heat and energy on a system

∆E universe = ∆E sys + ∆E surrounding = 0

1st Law Thermodynamics

Entropy - Measure of disorder↓

∆S uni = ∆S sys + ∆S surr > 0 (irreversible rxn)↓

All spontaneous rxn produce increase in entropy of universe

2nd Law Thermodynamics

∆S uni = ∆S sys + ∆S surr

Isolated system - Entropy change of universe always increase

Click here thermodynamics entropyEntropy

Measure molecular disorder/randomness

↓More disorder - More dispersion of matter/energy

↓More random - Rxn toward right- Entropy Increases ↑

Direction to right- Spontaneous to right →

2nd Law Thermodynamics

Embrace the chaos

Over time - Entropy increase ↑

Direction to left ← Never happen !

Click here thermodynamics

Energy cannot be created or destroyed

> 0

https://www.youtube.com/watch?v=MALZTPsHSoo

https://www.youtube.com/watch?v=GOrWy_yNBvY

∆S = Entropy change

Entropy

Dispersal/DistributionMatter Energ

y

Matter more disperse ↑

Entropy increases ↑

solid liquid gas

spontaneous - entropy ↑

Over time - Entropy increase ↑

Phase change - sol → liq → gas

↓ Entropy increase ↑ Every energy transfer - increase entropy universe

Entropy universe can only go up - never go down Entropy increase - many ways energy spread out

Dispersion energy as heat - increase entropy

Stoichiometry- more gas/liq in product

↓Entropy increase ↑

TQS

Heat added ↑

Phase change

Stoichiometry

Embrace the chaos

N2O4(g) → 2NO2(g)

1 2

2H2O(l) → 2H2 (g) + O2 (g)

1 23

3

More gas in product - Entropy ↑

Heat added ↑

Entropy

Measure molecular disorder/randomness↓

More disorder - More dispersion of matter/energy↓

More randon - Rxn towards right- Entropy Increases ↑

Liq more disorder than solidGas more disorder than liq

kinetic energy distributed over wide range

Q = heat

transferT = Temp/K

Distribution matter in space Distribution energy bet particles

Direction to left ← Never happen !Direction to right- Spontaneous to right →

Statistical

Entropy

Entropy

Measure molecular disorder/randomness↓


More random - Entropy Increases ↑

1st Law Thermodynamics - Doesn't help explain direction of rxn

∆S uni > 0 (+ve) → More disorder - spontaneous∆S uni < 0 (-ve) → More order - non spontaneous

Change sol → liq → gas - Higher entropyGreater number particles in product - Higher entropyMore complex molecule - More atoms bonded - Higher

entropyHigher temp - Vibrate faster - More random - Higher

entropy

Why gas mixes and not unmix?

Why heat flow from hot to cold?

Entropy

Notes on Entropy

1st Law Thermodynamics 2nd Law Thermodynamics

Energy cannot be created or destroyedTransfer from one form to another


Isolated system ↓

∆S uni always increase

∆E = q + w

Method to calculate entropy

Number microstates

Thermodynamic

Entropy

Heat + Temp involved

Gas mixesSolution diffuse Heat flow hot →cold

X X X

∆E = internal energy

q = heat transfer

w = work done ∆S = Entropy universe

∆S = Entropy system

∆S = Entropy surrounding


Law Thermodynamics

1 2

∆S = Entropy uni

WkS ln

∆S = Entropy change

k = boltzmann constant

W = Microstate

Click here statistical entropy

Click here thermodynamics entropy

Why solution diffuse and not undiffuse?

Unit - J mol -1

K-1

surrsysuni SSS

∆S = Entropy sys and surr

High chaos factor

https://www.youtube.com/watch?v=bE7Z6Zkst1I&list=PLX2gX-ftPVXUeVHmfT2BkhcEibTeReWKg

https://www.youtube.com/watch?v=bE7Z6Zkst1I&list=PLX2gX-ftPVXUeVHmfT2BkhcEibTeReWKg

1st Law Thermodynamics - Doesn't help explain direction of rxn

∆S uni > 0 (+ve) → More disorder - spontaneous∆S uni < 0 (-ve) → More order - non spontaneous

Change sol → liq → gas - Higher entropyGreater number particles in product - Higher entropyMore complex molecule - More atoms bonded - Higher

entropyHigher temp - Vibrate faster - More random - Higher

entropyMeasure molecular disorder/randomness↓


More random - Entropy Increases ↑

Isolated system ↓

∆S uni always increase

Entropy



Notes on Entropy

1st Law Thermodynamics 2nd Law Thermodynamics

Energy cannot be created or destroyedTransfer from one form to another


∆E = q + w


X X X

∆E = internal energy

q = heat transfer

w = work done ∆S = Entropy universe

∆S = Entropy system

∆S = Entropy surrounding


Law Thermodynamics

3rd Law Thermodynamics

Unit - J mol -1

K-1

Standard Molar Entropy, S0

Entropy perfectly crystal at 0K = 0Std molar entropy, S0 (absolute

value)↓

S0 when substance heated from 0K to 298K

Std state - 1 atm / 1M sol

Temp = 298K

Std Molar Entropy/S0

S0 at 298 /JK-1 mol-1

Fe (s) + 27

H2O (s) + 48

Na (s) + 52

H2O (l) + 69

CH3OH (l) + 127

H2 (g) + 130

H2O (g) + 188

CO2 (g) + 218

Solid - Order↓

Entropy Lowest

Liq - Less order↓

Entropy Higher

Gas - Disorder↓

Entropy Highest

Entropy highest


High chaos factor

Entropy

Why gas mix and not unmix?




X X X

Unit - J mol -1

K-1


Entropy perfectly crystal at 0K = 0 (Absolute value)

↓S0 when substance heated from 0K to

298K


Temp = 298K


S0 at 298 /JK-1 mol-1

Fe (s) + 27

H2O (s) + 48

Na (s) + 52

H2O (l) + 69

CH3OH (l) + 127

H2 (g) + 130

H2O (g) + 188

CO2 (g) + 218

Solid - Order↓

Entropy Lowest

Liq - Less order↓

Entropy Higher

Gas - Disorder↓

Entropy Highest

Entropy highest

Entropy


Depend on

Temp increase ↑ - Entropy increase ↑

Physical/phase state

Dissolving solid Molecular mass

Click here thermodynamics entropy Ba(OH)2

Temp

Temp/K 273 295 298

S0 for H2 + 31 + 32 + 33.2

Sol → Liq → Gas - Entropy increase ↑

State solid liquid gas

S0 for H2O + 48 + 69 + 188

entropy increase ↑ entropy increase ↑

Depend on

Substance NaCI NH4NO3

S0 for solid + 72 + 151

S0 for aq + 115 + 260

More motion - entropy increase ↑ Higher mass - entropy increase ↑

Substance HF HCI HBr

Molar mass 20 36 81

S0 + 173 + 186 + 198

S0 = 0 at 0KAll sub > 0K, have +ve S0

https://www.youtube.com/watch?v=ZsY4WcQOrfk

https://www.youtube.com/watch?v=MALZTPsHSoo

Entropy perfectly crystal at 0K = 0 (Absolute value)

↓S0 when substance heated from 0K to

298K

Entropy

Why gas mix and not unmix?




X X X

Unit - J mol -1

K-1



Temp = 298K


S0 at 298 /JK-1 mol-1

H2O (s) + 48

Na (s) + 52

H2O (l) + 69

CH3OH (l) + 127

H2O (g) + 188

CO2 (g) + 218

Solid - Order↓

Entropy Lowest

Liq - Less order↓

Entropy Higher

Gas - Disorder↓

Entropy Highest

Entropy highest

Entropy


Depend on

Temp increase ↑ - Entropy increase ↑

Physical/phase state

Dissolving solid Molecular mass

Temp

Temp/K 273 295 298

S0 for H2 + 31 + 32 + 33.2

Sol → Liq → Gas - Entropy increase ↑

State solid liquid gas

S0 for H2O + 48 + 69 + 188

entropy increase ↑ entropy increase ↑

Depend on

More motion - entropy increase ↑

Click here entropy notes

Click here entropy, enthalpy free energy data

Click here entropy CRC data booklet

Higher mass - entropy increase ↑

S0 = 0 at 0KAll sub > 0K, have +ve S0

Substance NaCI NH4NO3

S0 for solid + 72 + 151

S0 for aq + 115 + 260

Substance HF HCI HBr

Molar mass 20 36 81

S0 + 173 + 186 + 198

http://www2.sunysuffolk.edu/sambass/index_files/ch34old/lectures/pdfs/ch19.pdf

http://www.chemistry-reference.com/Standard%20Thermodynamic%20Values.pdf



http://www.chemistry-reference.com/Standard%20Thermodynamic%20Values.pdf

http://www.fptl.ru/biblioteka/spravo4niki/handbook-of-Chemistry-and-Physics.pdf

∆Hf θ

(reactant)∆Hf

θ

(product)

Using Std ∆Hf θ formation to find ∆H rxn

∆H when 1 mol form from its element under std condition

Na(s) + ½ CI2(g) → NaCI (s) ∆Hf θ = - 411 kJ

mol -1

Std Enthalpy Changes ∆Hθ

Std condition

Pressure 100kPa

Temp 298K

Conc 1M All substance at std states

Std ∆Hf θ formation

Mg(s) + ½ O2(g) → MgO(s) ∆Hf θ =- 602 kJ

mol -1

Reactants Products

O2(g) → O2 (g) ∆Hf θ = 0 kJ

mol -1

∆Hrxnθ = ∑∆Hf

θ(products) - ∑∆Hf

θ(reactants)

∆Hf θ

(products)∆Hf

θ

(reactants)

∆Hrxnθ

Elements

Std state solid gas

2C(s) + 3H2(g)+ ½O2(g) → C2H5OH(I) ∆Hf θ =- 275

kJ mol -1

1 mole formed

H2(g) + ½O2(g) → H2O(I) ∆Hf θ =- 286

kJ mol -1

Std state solid gas 1 mol liquid

For element Std ∆Hf θ formation = 0

Mg(s)→ Mg(s) ∆Hf θ = 0 kJ

mol -1

No product form

Using Std ∆Hf θ formation to find ∆H rxn

PDF version

Click here chem database (std formation enthalpy)

Online version

Click here chem database (std formation enthalpy)

C2H4 + H2 C2H6

Find ΔHθ rxn using std ∆H formation

Reactants Products

2C + 3H2

ElementsC2H4 + H2 → C2H6

∆Hrxnθ

∆Hrxnθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react)

∆Hrxnθ = Hf

θ C2H6 - ∆Hf

θ C2H4+ H2

= - 84.6 – ( + 52.3 + 0 ) = - 136.9 kJ mol -1

Enthalpy Formation, ∆Hf

Std ∆Gfθ formation

∆Grxnθ = ∑∆Gf

θ(pro) - ∑∆Gf

θ(react)

∆Grxnθ = Gf

θ C2H6 - ∆Gf

θ C2H4+ H2

= - 33 – ( + 68 + 0 ) = - 101 kJ mol -1

∆Gf θ

(reactant)∆Gf

θ

(product)

Using Std ∆Gf θ formation to find ∆G rxn o

∆Gf when 1 mol form from its element under std condition

Na(s) + ½ CI2(g) → NaCI (s) ∆Gf θ = - 384 kJ

mol -1

Std Free Energy Change ∆Gθ

Std condition

Pressure 100kPa

Temp 298K

Conc 1M All substanceat std states

Gibbs Free Energy change formation, ∆Gf

Mg(s) + ½ O2(g) → MgO(s) ∆Gf θ =- 560 kJ

mol -1

Reactants Products

O2(g) → O2 (g) ∆Gf θ = 0 kJ

mol -1

∆Grxnθ = ∑∆Gf

θ(prod) - ∑∆Gf

θ(react)

∆Gf θ

(product)∆Gf

θ

(reactant)

∆Grxnθ

Elements

Std state solid gas

2C(s) + 3H2(g)+ ½O2(g) → C2H5OH(I) ∆Gf θ =- 175

kJ mol -1

1 mole formed

H2(g) + ½O2(g) → H2O(I) ∆Gf θ =- 237

kJ mol -1

Std state solid gas 1 mol liquid

For element Std ∆Gf θ formation = 0

Mg(s)→ Mg(s) ∆Gf θ = 0 kJ

mol -1

No product form

Using Std ∆Gf θ formation to find ∆G rxn

PDF version

Click here chem database (std ∆G formation)

Online version

Click here chem database (std ∆G formation)

C2H4 + H2 C2H6

Find ΔGθ rxn using std ∆G0 formation

Reactants Products

2C + 3H2

ElementsC2H4 + H2 → C2H6

∆Grxnθ

∆S sys + ve , ∆S surr - ve↓

∆S uni > 0 (+ve)(Rxn Spontaneous)

∆S sys - ve , ∆S surr + ve↓

∆S uni < 0 (-ve)(Rxn Non spontaneous)

spontaneous+ve

-ve

=

S /JK-1

∆Ssys = + ve

∆Ssurr = + ve∆Suni = + ve

+

∆Ssys = - ve

+

∆Ssurr = + ve∆Suni = + ve

= spontaneous

S /JK-1 S /JK-1

∆Ssys = + ve

+

∆Ssurr = - ve

=

∆Suni = + ve spontaneous

C6H12O6(s) + 6O2 (g) → 6CO2(g) + 6H2O(l)

Using ∆Hsys , ∆Suni , ∆S sys , ∆S surr to predict spontaneity

2NO(g) + O2(g) → 2NO2(g) CaCO3 (s) → CaO(s) + CO2(g)

∆H = -ve (Heat released)

Difficult !!



∆Ssys = + ve

∆Ssurr = - ve

+=∆Suni = - ve

Nonspontaneous

∆H = -ve (Heat released) ∆H = +ve (Heat absorb)

CaCO3 (s) → CaO(s) + CO2(g)

∆H = +ve (Heat absorb)∆Ssys = + ve

+

∆Ssurr = - ve∆Suni = - ve

Nonspontaneous

=

H2(g) → 2 H(g)

∆H = +ve (Heat absorb)

H2O (l) → H2O(s)

∆H = -ve (Heat released)

∆Ssys = - ve

+

∆Suni = - ve

∆Ssurr = + ve

=



∆S sys + ve , ∆S surr + ve↓


∆S sys - ve , ∆S surr + ve↓


∆Hsys ∆Ssys ∆Suni Description

- + > 0 (+) Spontaneous, All Temp

+ - < 0 (-) Non spontaneous, All Temp

+ + > 0 (+) Spontaneous, High ↑ Temp

- - > 0 (+) Spontaneous, Low ↓ Temp

Predicting Spontaneity rxn

∆G - Temp/Pressure remain constantAssume ∆S/∆H constant with temp

Using ∆Hsys , ∆Suni , ∆S sys , ∆S surr to predict spontaneity

Using ∆Gsys to predict spontaneity

syssyssys STHG

Difficult !!

surrsysuni SSS THSsurr

)()( reactfprofsys HHH

CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l)

CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(l) ∆Hf

0 - 74 0 - 393 - 286 x 2 S0 + 186 +205 x 2 + 213 + 171 x 2

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react) ∆Hsys

θ = ∑∆Hfθ(pro) - ∑∆Hf

θ(react)

1

)tan()(

41

596555

JKS

S

SSS

sys

sys

treacproductsys kJH sys 891)74(965

12990298

)891000(

JKS

S

THS

surr

surr

surr

12949299041

JKS

SSS

uni

surrsysuni

∆S uni > 0spontaneous

Easier

Unit ∆G - kJ

Unit ∆S - JK-1

Unit ∆H - kJ

CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l)

Only ∆S sys involved∆S surr, ∆S uni not needed

∆Hsys ∆Ssys ∆Gsys Description

- +∆G = ∆H - T∆S

∆G = - veSpontaneous, All Temp

+ -∆G = ∆H - T∆S

∆G = + veNon spontaneous, All Temp

+ +∆G = ∆H - T∆S

∆G = - veSpontaneous, High ↑ Temp

- -∆G = ∆H - T∆S

∆G = - veSpontaneous, Low ↓ Temp

CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2Reactant (+596) Product (+589)

kJGG

STHG syssyssys

888)007.0(298890

∆Hsys = - 890 kJ

kJS

JKS

S

SSS

sys

sys

sys

reactprodsys

007.0

7

5965891

)()(

∆G < 0spontaneous

Entropy change ∆S greater at low temp

∆Gsysθ = ∑∆Gf

θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -868 - (-51) = - 817 kJ

Predicting Spontaneity rxn



syssyssys STHG

CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l)

Reactant (-51) Product (-868)

∆G < 0spontaneous

Easier

Unit ∆G - kJ mol-1 CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l)



- + ∆G = ∆H - T ∆S ∆G = - ve

Spontaneous at all Temp

+ - ∆G = ∆H - T ∆S ∆G = + ve

Non spontaneous, all Temp

+ + ∆G = ∆H - T ∆S ∆G = - ve

Spontaneous at high ↑ Temp

- - ∆G = ∆H - T ∆S ∆G = - ve

Spontaneous at low ↓ Temp

CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2Reactant (+ 596) Product (+ 589)

kJGG

STHG syssyssys

888)007.0(298890

∆Hsys = - 890 kJ kJS

JKS

S

SSS

sys

sys

sys

reactprodsys

007.0

7

5965891

)()(

∆G < 0spontaneous


Easier

CH4(g) + 2 O2 (g) → CO2(g)

+ 2 H2O(l) ∆G0 - 51 0 - 394 - 237 x 2

Method 1 Method 2

)()( reactfprofsys GGG

CH4(g) + 2 O2 (g) CO2(g)

+ 2H2O(g)

C + 2O2 + 2H2

Reactants Products∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product) Elements

• Neither ∆H or ∆S can predict feasibility of spontaneous rxn• Gibbs Free Energy (∆G) – measure spontaneity and useful energy available• Gibbs Free Energy (∆G) - max amt useful work at constant Temp/Pressure• Involve ∆H sys and ∆S sys

• ∆G involve only sys while ∆S uni involve sys and surr• Easier to find ∆H and ∆S for system


0

At std condition/statesTemp - 298KPress - 1 atm



syssyssys STHG

Easier

Unit ∆G - kJ

CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l)



- +∆G = ∆H - T∆S

∆G = - veSpontaneous, All Temp

+ -∆G = ∆H - T∆S

∆G = + veNon spontaneous, All Temp

+ +∆G = ∆H - T∆S

∆G = - veSpontaneous, High ↑ Temp

- -∆G = ∆H - T∆S

∆G = - veSpontaneous, Low ↓ Temp

CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2Reactant (+ 596) Product (+ 589)

kJGGG

STHG syssyssys

8882890

)007.0(298890

∆Hsys = - 890 kJ

kJS

JKS

S

SSS

sys

sys

sys

reactprodsys

007.0

7

5965891

)()(

∆G < 0spontaneous

Gibbs Free Energy Change, ∆G

∆G sys

T∆S sys

Total energy change, ∆H

Measure spontaneity and useful energy availableMax amt useful work at constant Temp/Pressure

Free Energy

syssyssys STHG

Free energy available to do work not available

for work

syssyssys STHG

Free Energy

Total energy change, ∆H

∆G sys

T∆S sys

-890kJ

Free energy available to do work

not available for work

-888kJ +2 kJ




syssyssys STHG

Easier

Unit ∆G - kJ mol-1



Easier

Method 1 Method 2

)()( reactfprofsys GGG At std condition/states

Temp - 298KPress - 1 atm


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


- +∆G = ∆H - T ∆S

∆G = - veSpontaneous at all Temp

+ -∆G = ∆H - T ∆S

∆G = + veNon spontaneous, all Temp

unisys STG syssyssys STHG

Only ∆H sys/∆S sys involved ∆S surr, ∆S uni not needed

syssyssys STHGNon standard conditionStandard condition

or


syssyssys STHG unisys STG

veGsys ∆S uni = +ve

Spontaneous SpontaneousveGsys

∆H = - ve∆S sys = +ve


+ +∆G = ∆H - T ∆S

∆G = - veSpontaneous at high ↑ Temp

- -∆G = ∆H - T ∆S

∆G = - veSpontaneous at low ↓ Temp

kJGG

STHG

130)16.0(298178

Predict entropy change - quatitatively

Reactant Product

∆Hsysθ = ∑∆Hf

θ(pro) - ∑∆Hf

θ(react) ∆Ssys

θ = ∑Sfθ(pro) - ∑Sf

θ(react)

kJH sys 178)1206(1028

∆G uni > 0 - Decomposition at 298K - Non Spontaneous

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)∆Hsys


θ(react)


CaCO3 (s) → CaO (s) + CO2(g) ∆Hf

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

kJS

S

S

SSS

sys

sys

sys

treacproductsys

16.0

160

93253)tan()(

Decomposition at 298K Decomposition at 1500K



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

kJH sys 178)1206(1028

Rxn Temp dependentSpontaneous at High ↑ temp

1500K298K

Decomposition limestone CaCO3 spontaneous?


kJS

S

S

SSS

sys

sys

sys

treacproductsys

16.0

160

93253)tan()(

kJGG

STHG

62)16.0(1500178

∆G uni < 0 - Decomposition at 1500K - Spontaneous

∆H = +ve ∆S = +ve

Temp dependent


+ +∆G = ∆H - T ∆S


- -∆G = ∆H - T ∆S


At Low Temp At High Temp


Reactant Product


θ(pro) - ∑∆Hf

θ(react) ∆Ssys


θ(react)

∆G uni > 0 - Decomposition at 298K - Non Spontaneous

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)∆Hsys


θ(react)

Rxn Temp dependentSpontaneous at Low ↓ temp

298K (25C)


∆G uni < 0 - Decomposition at 1500K - Spontaneous

∆H = - ve ∆S = - ve

Temp dependent


+ +∆G = ∆H - T ∆S


- -∆G = ∆H - T ∆S


H2O (l) → H2O(s)

H2O (l) → H2O(s) ∆Hf

0 - 286 - 292S0 + 70 + 48

Freezing at 298K (25C)

Is Freezing spontaneous?

kJH sys 6)286(292

kJS

S

S

SSS

sys

sys

sys

treacproductsys

02.0

22

7048)tan()(

kJGG

STHG

55.0)022.0(2986

Freezing at 263K (-10C)

H2O (l) → H2O(s)


0 - 286 - 292S0 + 70 + 48

kJH sys 6)286(292

kJS

S

S

SSS

sys

sys

sys

treacproductsys

02.0

22

7048)tan()(

263K (-10C) kJGG

STHG

21.0)022.0(2636

At High Temp At Low Temp

C3H8(g) + 5 O2 (g) 3CO2(g)

+ 4H2O(l)

Entropy and Gibbs Free Energy


Why conc solution diffuse and not undiffuse?


Will rxn be spontaneous ?


X X X

1

syssyssys STHG

)tan()( treacprosys SSS

C3H8(g) + 5O2 (g) → 3CO2(g) + 4H2O(l) ∆H = -2220 kJ at 298K

C3H8(g) + 5 O2 (g) → 3 CO2(g) + 4 H2O(l) S0 +270 +205 x 5 +213 x 3 +70 x 4 1295 919

Reactant Product

kJGG

STHG

2108)376.0(2982220

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

376.0

376

12959191

)tan()(

∆H = -2220 kJ

Assume ∆S, ∆H at constant over Temp

∆G sys < 0 (-ve) → Spontaneous ∆G sys > 0 (+ve) → Non spontaneous

Is Combustion at 298K spontaneous?

Using Free Energy to predict spontaneity

Gibbs Free Energy, ∆G

syssyssys STHG

Unit ∆S - JK-

1

Unit ∆H - kJUnit ∆G - kJ

Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -2130 - (-23) = - 2153 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G < 0 - Combustion at 298K - Spontaneous

C3H8(g) + 5 O2 (g) → 3 CO2(g) + 4 H2O(l) ∆G0 - 23 0 - 394 x 3 - 237 x 4

Elements

3C + 5O2 + 4H2



CH4(g) + 2O2 (g) → CO2(g) + 2H2O(g) ∆H = - 890 kJ at 298K

CH4(g) + 2 O2 (g) CO2(g)

+ 2H2O(g)







X X X

2

syssyssys STHG


+ 596 + 589Reactant Product

kJGG

STHG

888)007.0(298890

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

007.0

7

5965891

)tan()(

∆H = - 890 kJ






syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -868 - (-51) = - 817 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)



CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(l) ∆G0 - 51 0 - 394 - 237 x 2

Elements

C + 2O2 + 2H2



CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(g) S0 + 186 +205 x 2 +213 + 188 x 2

H2O (g) → H2O(l) ∆H = - 44.1 kJ at 298K

H2O(g) H2O(l)







X X X

syssyssys STHG



kJGG

STHG

1.9)118.0(2981.44

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

118.0

118

188701

)tan()(

∆H = - 44.1 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -237 - (-228) = - 9 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)



H2O(g) → H2O(l) ∆G0 -228 - 237

Elements

H2 + O2



Condensation steam at 298K (25C) spontaneous?

H2O (g) → H2O(l) S0 + 188 + 70

3


H2(g) → 2 H(g) ∆H = + 436 kJ at 298K

H2(g) 2H(g)







X X X

syssyssys STHG



kJGG

STHG

406)1.0(298436

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

1.0

100

1302301

)tan()(

∆H = + 436 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = + 406 - (0) = +406 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G > 0 - Atomization at 298K - Non Spontaneous

H2(g) → 2H(g) ∆G0 0 + 203 x 2

Elements

H2

Reactant (0) Product ( + 406)

4Is Atomization of H2 at 298K spontaneous?

H2 (g) → 2 H(g) S0 + 130 + 115 x 2

∆G > 0 - Atomization at 298K - Non Spontaneous


H2O (l) → H2O(s) ∆H = - 6 kJ at 298K

H2O(l) H2O(s)







X X X

2

syssyssys STHG



kJGG

STHG

55.0)022.0(2986

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

022.0

22

70481

)tan()(

∆H = - 6 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -236.6 - (-237) = + 0.4kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G > 0 -Freezing at 298K - Non Spontaneous

H2O(l) → H2O(s) ∆G0 -237 - 236.6

Elements

H2 + O2

Reactant (-237) Product (-236.6)

5

H2O (l) → H2O(s) S0 + 70 + 48

∆G > 0 -Freezing at 298K - Non Spontaneous

Is Freezing water to ice at 298K (25C) spontaneous?


H2O (l) → H2O(s) ∆H = - 6 kJ at 263K

H2O(l) H2O(s)







X X X

2

syssyssys STHG



kJGG

STHG

21.0)022.0(2636

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

022.0

22

70481

)tan()(

∆H = - 6 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -237.2 - (-237) = - 0.2 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G < 0 -Freezing at 263K - Spontaneous

H2O(l) → H2O(s) ∆G0 -237 - 237.2

Elements

H2 + O2

Reactant (-237) Product (-237.2)

6

H2O (l) → H2O(s) S0 + 70 + 48

∆G < 0 -Freezing at 263K - Spontaneous

Is Freezing water to ice at 263K (-10C) spontaneous?

Assume std condition at 263K


CaCO3 (s) → CaO(s) + CO2(g) ∆H = + 178 kJ at 298K







X X X

syssyssys STHG



kJGG

STHG

130)16.0(298178

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

16.0

160

932531

)tan()(

∆H = + 178 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = - 999 - (- 1129) = + 130 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G > 0 - Decomposition at 298K - Non Spontaneous

CaCO3(s) → CaO + CO2(g) ∆G0 -1129 - 604 - 395

Elements

Ca + C + O2

Reactant ( -1129) Product (- 999)

7 Decomposition CaCO3 at 298K (25C) spontaneous?

CaCO3 (s) → CaO (s) + CO2(g) S0 + 93 + 40 + 213

∆G > 0 - Decomposition at 298K - Non Spontaneous

CaCO3 (s) CaO (s) + CO2(g)


CaCO3 (s) → CaO(s) + CO2(g) ∆H = + 178 kJ at 1500K







X X X

syssyssys STHG



kJGG

STHG

62)16.0(1500178

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

16.0

160

932531

)tan()(

∆H = + 178 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = - 999 - (- 939) = - 60 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G < 0 - Decomposition at 1500K - Spontaneous

CaCO3(s) → CaO + CO2(g) ∆G0 -939 - 604 - 395

Elements

Ca + C + O2

Reactant (- 939) Product (- 999)

8

CaCO3 (s) → CaO (s) + CO2(g) S0 + 93 + 40 + 213

CaCO3 (s) CaO (s) + CO2(g)

Decomposition CaCO3 at 1500K (1227C) spontaneous?


Assume std condition at 1500K


2NO(g) + O2(g) → 2NO2(g) ∆H = - 114 kJ at 298K







X X X

syssyssys STHG



kJGG

STHG

101)042.0(298114

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

042.0

42

5224801

)tan()(

∆H = - 114 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = + 104 - (174) = - 70 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)



2 NO + O2 → 2NO2(g) ∆G0 + 87 x 2 0 + 52 x 2

Elements

N2 + O2

Reactant (+ 174) Product (+ 104)

9

2 NO(g) + O2 (g) 2NO2(g)


Is Oxidation of NO at 298K (25C) spontaneous?

2 NO(g) + O2 (g) → 2NO2(g) S0 + 210 x 2 + 102 + 240 x 2


N2(g) + 3H2(g) → 2NH3(g) ∆H = - 92 kJ at 298K







X X X

syssyssys STHG



kJGG

STHG

32)2.0(29892

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

2.0

201

5853841

)tan()(

∆H = - 92 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = - 34 - (0) = - 34 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G < 0 - NH3 production at 298K - Spontaneous

N2 + 3H2 → 2NH3(g) ∆G0 0 0 - 17 x 2

Elements

N2 + H2

Reactant (0) Product (- 34)

10

N2(g) + 3H2 (g) 2NH3(g)

Is Haber, NH3 production 298K (25C) spontaneous?

NH3

N2(g) + 3H2 (g) → 2NH3(g) S0 + 192 + 131 x 3 + 192 x 2

∆G < 0 - NH3 production at 298K - Spontaneous


Fe2O3(s) + 2AI(s) → 2Fe(s) + AI2O3(s) ∆H = - 851 kJ at 298K







X X X

syssyssys STHG



kJGG

STHG

840)038.0(298851

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

038.0

38

1431051

)tan()(

∆H = - 851 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -1576 - (-741) = - 835 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G < 0 - AI production at 298K - Spontaneous

Fe2O3 + 2AI → 2Fe + AI2O3∆G0 - 741 0 0 - 1576

Elements

Fe + AI + O2

Reactant (-741) Product (- 1576)

11 Is Thermite, AI production 298K (25C) spontaneous?

Fe2O3(s) + 2AI(s) → 2Fe(s) + AI2O3(s) S0 + 87 + 28 x 2 + 27 x 2 + 51

∆G < 0 - AI production at 298K - Spontaneous

Fe2O3(s) + 2AI(s) 2Fe(s) + AI2O3(s)


4KCIO3(s) → 3KCIO4(s) + KCI(s) ∆H = - 144 kJ at 298K







X X X

syssyssys STHG



kJGG

STHG

133)037.0(298144

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

037.0

37

5725351

)tan()(

∆H = - 144 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -1317 - (-1160) = - 157 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)



4KCIO3 → 3 KCIO4 + KCI∆G0 - 290 x 4 - 303 x 3 - 408

Elements

K + CI2 + O2


13


Is decomposition KCIO3

298K (25C) spontaneous?

4KCIO3(s) → 3KCIO4(s) + KCI(s) S0 + 143 x 4 + 151 x 3 + 82

4KCIO3(s) 3KCIO4(s) + KCI(s)








X X X

syssyssys STHG



kJGG

STHG

3071)877.0(2982810

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

877.0

877

82116981

)tan()(

∆H = - 2810 kJ




syssyssys STHG

Unit ∆S - JK-

1


Reactants Products


θ(pro) - ∑∆Gf

θ(react)

∆Gsysθ = -3792 - (-910) = - 2882 kJ

∆Gsysθ

∆Gf θ

(reactant)∆Gf

θ

(product)


∆G < 0 Combustion sugar at 298K - Spontaneous

Elements

C + H2 + O2


14 Is combustion sugar 298K (25C) spontaneous? C6H12O6(s) + 6O2 (g) → 6CO2(g) + 6H2O(l) ∆H = - 2810

kJ at 298K

C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(l) S0 + 209 +102 x 6 + 213 x 6 + 70 x 6


C6H12O6 + 6O2 6CO2 + 6H2O(l)

C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆G0 - 910 0 - 395 x 6 - 237 x 6


Entropy






X X X

Reactant Product

CH4(g) + 2O2 (g) → CO2(g) + 2H2O(l)

CH4(g) + 2 O2 (g) → CO2(g) + 2 H2O(l) ∆Hf

0 - 74 0 - 393 - 286 x 2 S0 + 186 +205 x 2 +213 + 171 x 2


θ(pro) - ∑∆Hf

θ(react) ∆Ssys


θ(react)

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

041.0

41

5965551

)tan()(

kJH sys 890)74(964


Unit for ∆S - JK-1 Unit for ∆H - kJ

C3H8(g) + 5O2 (g) → 3CO2(g) + 4H2O(l)

C3H8(g) + 5 O2 (g) → 3 CO2(g) + 4 H2O(l) ∆Hf

0 - 104 0 - 393 x 3 - 286 x 4S0 +270 +205 x 5 + 213 x 3 + 171 x 4

Reactant Product∆Ssys


θ(react)∆Hsys


θ(react)

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

028.0

28

129513231

)tan()(

kJH sys 2219)104(2323

1 2

kJGG

STHG

877)041.0(298890


kJGG

STHG

881)028.0(2982219


Entropy






X X X

Reactant Product


θ(pro) - ∑∆Hf

θ(react) ∆Ssys


θ(react)

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

118.0

118

188701

)tan()(

kJH sys 44)242(286

Is Condensation/Freezing at 298K spontaneous?

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)∆Hsys


θ(react)

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

022.0

22

70481

)tan()(

kJH sys 6)286(292

3 4 H2O (g) → H2O(l)

H2O (l) → H2O(s)

H2O (g) → H2O(l) ∆Hf

0 - 242 - 286S0 + 188 + 70


0 - 286 - 292S0 + 70 + 48

kJGG

STHG

1.9)118.0(2981.44

∆G < 0 Condensation at 298K - Spontaneous

kJGG

STHG

55.0)022.0(2986

∆G > 0 Freezing at 298K – Non Spontaneous

Entropy






X X X

Reactant Product


θ(pro) - ∑∆Hf

θ(react) ∆Ssys


θ(react)

kJH sys 92)0(92

Are these rxn at 298K spontaneous?

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)∆Hsys


θ(react)

kJH sys 168)1564(1732

5 6N2(g) + 3H2(g) → 2NH3(g)

N2(g) + 3H2 (g) → 2NH3(g) ∆Hf

0 0 0 - 46 x 2S0 + 192 + 131 x 3 + 192 x 2

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

201.0

201

5853841

)tan()(

4KCIO3(s) → 3KCIO4(s) + KCI(s)

4KCIO3(s) → 3KCIO4(s) + KCI(s) ∆Hf

0 - 391 x 4 - 432 x 3 - 436S0 + 143 x 4 + 151 x 3 + 82

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

037.0

37

5725351

)tan()(

kJGG

STHG

32)2.0(29892

∆G < 0 NH3 production at 298K - Spontaneous

kJGG

STHG

157)037.0(298168

∆G < 0 KCIO3 production at 298K - Spontaneous

Entropy






X X X

Reactant Product


θ(pro) - ∑∆Hf

θ(react) ∆Ssys


θ(react)

kJH sys 178)1206(1028

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)∆Hsys


θ(react)

7 8CaCO3 (s) → CaO(s) + CO2(g)


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

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

16.0

160

932531

)tan()(

Decomposition at 298K Decomposition at 1500K



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

kJH sys 178)1206(1028

Rxn Temp dependentSpontaneous at High ↑ temp

1500K298K (25C)

Decomposition limestone CaCO3 spontaneous?

kJGG

STHG

130)16.0(298178

∆G > 0 Decomposition at 298K – Non Spontaneous

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

16.0

160

932531

)tan()(

kJGG

STHG

62)16.0(1500178

∆G < 0 Decomposition at 1500 K - Spontaneous

At Low Temp At High Temp

Entropy






X X X

Reactant Product


θ(pro) - ∑∆Hf

θ(react) ∆Ssys


θ(react)

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

022.0

22

70481

)tan()(

kJH sys 6)286(292

Is Freezing spontaneous?

Reactant Product

∆Ssysθ = ∑Sf

θ(pro) - ∑Sf

θ(react)∆Hsys


θ(react)

kJS

JKS

S

SSS

sys

sys

sys

treacproductsys

022.0

22

70481

)tan()(

kJH sys 6)286(292

9 10 H2O (l) → H2O(s)

H2O (l) → H2O(s)


0 - 286 - 292S0 + 70 + 48


0 - 286 - 292S0 + 70 + 48

Freezing at 298K (25C) Freezing at 263K (-10C)

Rxn Temp dependentSpontaneous at Low ↓ temp

263K (-10C)298K (25C)kJG

GSTHG

55.0)022.0(2986

∆G > 0 Freezing at 298K – Non Spontaneous

kJGG

STHG

21.0)022.0(2636

∆G < 0 Freezing at 263K – Spontaneous

At High Temp At Low Temp

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

IB Chemistry on Gibbs Free Energy and Entropy

Education

Transcript of IB Chemistry on Gibbs Free Energy and Entropy