Entropy, Free Energy and Spontaneity

• How will a reaction go ?

• Why gas MIXES and NOT UNMIX ?

• Why concentrated solute DIFFUSE and NOT UNDIFFUSE?

Entropy

• Measures the degree of DISORDER for a system

• Measures the probability or chance for a system, both in distribution of particles in space

and distribution of energy

Entropy and Spontaneity

• Unit for entropy – JK-1mol-1

• Formula for entropy, S = k ln W k = Boltzmann constant

W = Number ways a particle can arrange in space

• Absolute entropy can be measured!

• Entropy of a perfectly order solid crystal at OK is ZERO

• Entropy of an element under standard condition is NOT ZERO

ΔSθ (H2) gas = 130.6JK-1mol-1

• Standard entropy change, ΔSθ = Entropy change per mole of a substance

heated from OK to standard Temp of 298K

Important concepts for Entropy

Entropy and Spontaneity

ENTROPY CHANGE, ΔSθ

• Change in the disorder of a system

• Higher ↑ disorder – Higher entropy ↑

• Change in state from SOLID → LIQUID → GAS → Higher entropy ↑

• Greater ↑ number of particles formed in products → Higher entropy ↑

• Complex molecules (more atoms bonded) → Higher entropy ↑

• Higher temperature ↑ → Particles vibrate faster → More random → Higher entropy ↑

Standard entropy, ΔSθ /298K Sθ / JK-1mol-1

Second Law of Thermodynamics • For all spontaneous reactions, TOTAL entropy of the universe ΔSθ

uni always INCREASES

• Formula for ΔSθsys = ΔSθ (products) - ΔSθ (reactants)

• Reaction will likely to happen if -

• Formula for ΔSθ(surrounding)

• Conclusion : For spontaneous reaction, ΔSθuni must > O (positive)

Gibbs Free Energy, ΔG

To predict if a reaction will likely to happen

• ΔG is used or preferred because it involves the system while ΔS involves system and surrounding

• Easier to determine ΔH and ΔS for a system

ΔG = ΔH -TΔS

• ΔSuni = ΔSsys + ΔSsurr > 0, (+ve) positive

• ΔGsys =ΔHsys –TΔSsys < 0, (-ve) negative

•ΔHsys = -ve (exothermic)

• ΔSsys = +ve (entropy increses ↑)

• Δssurr = +ve (entropy increases ↑ )

•ΔSuni = +ve (entropy increases ↑ )

Gibbs Free Energy, ΔG is used

Unit for ΔGsys = kJmol-1 ΔHsys = kJmol-1 ΔSsys = JK-1mol-1

Combination ΔHsys , ΔSsys and ΔGsys to predict if reaction is spontaneous

ΔHsys +ve ΔSsys +ve ↑

Spontaneous if Temp High ↑

ΔHsys -ve ΔSsys –ve ↓

Spontaneous if Temp low ↓

ΔHsys -ve ΔSsys +ve Always Spontaneous

ΔHsys +ve ΔSsys –ve Non Spontaneous

ΔSθsys ΔSθ

surr

ΔSuni = ΔSsys + ΔSsurr

= (-118.7) + 148.0 = +29.3 JK-1mol-1

Conclusion :

• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to the surrounding

INCREASES ↑ the entropy of surrounding, ΔSsurr

• Spontaneous - ΔGθ = -ve

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (70.0) – (188.7) = -118.7 JK-1mol-1

ΔSsurr = - ΔHsys /T

= -(-44100)/298 = +148.0JK-1mol-1

ΔGθ = ΔH -TΔS

= (-44.1) – (298 x -118.7/1000) = - 8.72kJmol-1

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K

AND

+

ΔSθsys ΔSθ

surr

Conclusion :

• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to the surrounding

INCREASES ↑ the entropy of surrounding, ΔSsurr

• Spontaneous - ΔGθ = -ve

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K

AND

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (213.6 + 2x171) – (186.0 + 2x205) = -40.4 JK-1mol-1

ΔSsurr = -ΔHsys /T

= -(-890000)/298 = +2986.5JK-1mol-1

+

ΔSuni = ΔSsys + ΔSsurr

= -40.4 + 2986.5 = +2946 JK-1mol-1

ΔGθ = ΔH -TΔS

= (-890) – (298 x -40.4/1000) = - 878 kJmol-1

ΔSθsys ΔSθ

surr

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K

AND

+ ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (2 x 115) – (130.6) = +99.4 JK-1mol-1

ΔSsurr = -ΔHsys /T

= -(+436000)/298 = -1463 JK-1mol-1

ΔSuni = ΔSsys + ΔSsurr

= +99.4 + (-1463) = -1363.6 JK-1mol-1

ΔGθ = ΔH -TΔS

= (436) – (298 x 99.4/1000) = +406 kJmol-1

Conclusion : • Non Spontaneous - Entropy of system ΔSθ

sys INCREASES ↑ BUT heat absorbed from surrounding

DECREASES ↓ the entropy of surrounding, ΔS surr

• NON spontaneous - ΔGθ = +ve

ΔSθsys ΔSθ

surr

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K

AND

+

Is it possible to freeze water at 298K ( 25oC)

At 25oC

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (48) – (70) = -22 JK-1mol-1

ΔSsurr = -ΔHsys /T

= -(-6010)/298 = +2016 JK-1mol-1

ΔSuni = ΔSsys + ΔSsurr

= -22 + (+2016) = -1.84 JK-1mol-1

ΔGθ = ΔH -TΔS

= (-6.01) – (298 x -22/1000) = +0.54 kJmol-1

Conclusion :

• Non Spontaneous - Entropy of system ΔSθsys DECREASES ↓ more than > the

INCREASE ↑ in entropy of surr, ΔSsurr due to heat released to the surrounding

• NON spontaneous - ΔGθ = +ve

ΔSθsys ΔSθ

surr

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 263K

AND

+

Is it possible to freeze water at 263K ( -10oC)

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (48) – (70) = -22 JK-1mol-1

At -10oC

ΔSsurr = -ΔHsys /T

= -(-6010)/263 = +22.85 JK-1mol-1

ΔSuni = ΔSsys + ΔSsurr

= -22 + (+22.85) = +0.85 JK-1mol-1

ΔGθ = ΔH -TΔS

= (-6.01) – (263 x -22/1000) = -0.23 kJmol-1

Conclusion :

• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to surrounding

INCREASES ↑ the entropy of surrounding, ΔSsurr

• Spontaneous - ΔGθ = -ve

ΔSθsys ΔSθ

surr

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K

AND

+

Is it possible to decompose CaCO3 (s) → CaO (s) + CO2 (g)

At 25oC

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (39.7 + 213.6) – (92.9) = + 160.4 JK-1mol-1

ΔSsurr = -ΔHsys /T

= -(+178300)/298 = - 598.3 JK-1mol-1

ΔSuni = ΔSsys + ΔSsurr

= +160.4 + (- 598.3) = - 438 JK-1mol-1

ΔGθ = ΔH -TΔS

= (+ 178.3) – (298 x +160.4/1000) = + 130.5 kJmol-1

Conclusion :

• Non Spontaneous - Entropy of system ΔSθsys INCREASES ↑ BUT heat absorbed from surrounding

DECREASES ↓ the entropy of surrounding, ΔSsurr

• NON spontaneous- ΔGθ = +ve

ΔSθsys ΔSθ

surr

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 1500K

AND

+

Is it possible to decompose CaCO3 (s) → CaO (s) + CO2 (g)

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (39.7 + 213.6) – (92.9) = + 160.4 JK-1mol-1

At 1227oC

ΔSsurr = -ΔHsys /T

= -(+178300)/1500 = - 118.8 JK-1mol-1

ΔSuni = ΔSsys + ΔSsurr

= +160.4 + (- 118.8) = + 41.6 JK-1mol-1

ΔGθ = ΔH -TΔS

= (+ 178.3) – (1500 x +160.4/1000) = - 62.3 kJmol-1

Conclusion :

• Spontaneous - Entropy of system ΔSθsys INCREASES ↑ more than >

DECREASES ↓ in entropy of surr, ΔSsurr due to heat absorbed from surrounding

• Spontaneous - ΔGθ = -ve

ΔSθsys ΔSθ

surr

Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K

AND

+

Is the reaction possible 2NO (g) + O2 (g) → 2NO2 (g)

ΔSsys = ΔSsys (product) – ΔSsys (reactant)

= (2 x 240) – (2 x 210.7+102.5) = - 43.9 JK-1mol-1

ΔSsurr = -ΔHsys /T

= -(-114000)/298 = +382.5 JK-1mol-1

ΔSuni = ΔSsys + ΔSsurr

= -43.9 + (+382.5) = +339 JK-1mol-1

ΔGθ = ΔH -TΔS

= (- 114) – (298 x -43.9/1000) = -100.9 kJmol-1

Conclusion :

• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ less than <

INCREASE ↑ in entropy of surr, ΔSsurr due to heat released to the surrounding

• Spontaneous - ΔGθ = -ve

Reaction which is Temperature Dependent whereby ΔHsys (+ve) and ΔSsys (+ve)

Finding the temperature which make reaction spontaneous

ΔGsys = (-ve) Spontaneous ΔGsys = (+ve) Non Spontaneous

ΔGsys = O (Equilibrium)

Equilibrium Temperature whereby reaction becomes spontaneous happen when ΔG = O

ΔG = ΔHsys - TΔSsys

0 = ΔHsys -TΔSsys

TΔS = ΔH

T = ΔH/ΔS

At what temperature will reaction becomes spontaneous

0 = ΔH -TΔS

TΔS = ΔH

T = ΔH/ΔS

T = 178300/160.4 = 1111.6K

ΔHsys = +178.3 kJmol-1

ΔSsys = +160.4 JK-1mol-1

Standard Enthalpy Change of Formation, ΔHf

• Energy released when ONE mole of compound form from its constituent elements in their std states

EX : C(s) + 2H2 → CH4 (g) ΔHf = -75 kJmol-1

• Using standard enthalpy of formation to calculate ΔH reaction

• Free energy released when ONE mole of compound form from its constituent elements in their std states

EX : C(s) + 2H2 → CH4 (g) ΔGf = -51 kJmol-1

• Using standard free energy of formation to calculate ΔG reaction

Standard Free Energy of Formation, ΔGf

• Standard free energy of formation, ΔGf of compound

C(s) + 2H2 → CH4(g) ΔGf = -51 kJmol-1

H2(g) + ½ O2(g) →H20(g) ΔGf = -228 kJmol-1

2C (s) + 3H2(g) → C2H6(g) ΔGf = -33 kJmol-1

• Standard free energy of formation, ΔGf of element is ZERO

C(s) → C(s) ΔGf = 0 kJmol-1

ΔGreaction = (Sum ΔGf (product) – Sum ΔGf(reactant) )

ΔGreaction = (-ve) Reaction is Spontaneous

Using ΔG to predict if reaction is Spontaneous

Ex 1

Ex 2

Ex 3

ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) )

= ( -394 + 2 x -237) – (-51 + 0)

= - 817 kJmol-1 (Spontaneous)

ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) )

= ( 3 x -303.2 + (-409.2) ) – ( 4 x -296.3)

= -134 kJmol-1 (Spontaneous)

ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) )

= (-305) – (-267.8 + 0 )

= -37.2 kJmol-1 (Spontaneous)

Combination ΔH, ΔS and ΔG to predict if reaction is Spontaneous

Ex 1

Ex 2

ΔHrxn = (Sum ΔHf (product) – Sum ΔHf (reactant) )

= ( -986.6) – (-635.5 + (-285.9) )

= - 65.2 kJmol-1

ΔHrxn = (Sum ΔHf (product) – Sum ΔHf (reactant) )

= ( 3 x -432.8 -436.7) – ( 4 x -397.7)

= - 144 kJmol-1

ΔSrxn = (Sum S(product) – Sum S(reactant) )

= ( +76.1) – (+39.7 + 70.0)

= - 33.6 JK-1mol-1

ΔSrxn = (Sum S(product) – Sum S(reactant) )

= ( 3 x 151 + 82.6) – ( 4 x 143.1)

= - 36.8 JK-1mol-1

ΔGθ = ΔH -TΔS

= (-65.2) – (298 x -33.6/1000) = -55.2 kJmol-1

ΔGθ = ΔH -TΔS

= (-144) – (298 x -36.8/1000) = -133 kJmol-1

(Spontaneous)

(Spontaneous)