Entropy changes in irreversible Processes

To obtain the change in entropy in an irreversible process we have to calculate S along a reversible path between the initial state and the final state.

Freezing of water below its freezing point

H2O( l , -10 °C) H2O( s , -10 °C)

H2O( l , 0°C) H2O( s , 0 °C)

Irrev

273

263

263

273ln ice

crysliq C

T

HCS

A sample of 1.00 mole of monoatomic perfect gas with Cv,m = 1.5 R, initially at 298 K and 10 L, is expanded, with the surroundings maintained at 298 K, to a final volume of 20 L, in three ways (a) isothermally and reversibly (b) isothermally against a constant external pressure of 0.5 atm (c) adiabatically against a constant external pressure of 0.5 atm . Calculate ΔS and ΔSsurr, ΔH and ΔT for every path.

Question?

• A 50 gm mass of Cu at a temperature of 393 K is placed in contact with a 100 g mass of Cu at a temperature of 303 K in a thermally insulated container. Calculate q and ΔStotal for the reversible process. Use a value of 0.4184 Jg-1K-1 for specific heat capacity of Cu.

Absolute entropy of a substance

)(

)(

)()0()(

0

T

T

p

b

vT

T

p

f

fT p

b

b

f

f

T

dTgC

T

H

T

dTlC

T

H

T

dTsCSTS

Third law of thermodynamics:

The entropy of each pure element or substance in a perfectly crystalline form is zero at absolute zero.

energy. free helmholtz called isA

0)(

0)(

)(

T and Vconstant At

0

or 0

0

0

workadditional no olume,constant vAt

0

0

,

,

,

,

TV

TV

VS

VU

v

sys

surrsys

surrsys

Ad

TSUd

TSddUTdSdU

dU

dS

TdSdU

TdSdq

TdSdqT

dqdS

dSdS

dSdS

Spontaneous process

energy. free sGibb' called isG

0)(

0)(

)(

T and Pconstant At

0

or 0

0

0

workadditional no pressure,constant At

0

0

,

,

,

,

TP

TP

pS

pH

p

sys

surrsys

surrsys

Gd

TSHd

TSddHTdSdH

dH

dS

TdSdH

TdSdq

TdSdqT

dqdS

dSdS

dSdS

Enthalpy/ Entropy driven process

A-Enthalpy drivenB-Entropy driven

Hydrophobic Effect in Protein Hydrophobic Effect in Protein FoldingFolding

More Hydrocarbon-Water Interfacial Area

Less Hydrocarbon-WaterInterfacial Area

HOH

HOH+

Folded Unfolded

S=+

Hydrophobic forcesHydrophobic forces

• Hydrophobic interactions are considered to make a major contribution to stabilizing the native structures of proteins in aqueous environment.

• The hydrophobic effect is a unique organizing force based on repulsion by the solvent instead of attractive forces at the site of organization.

• Thermodynamics of protein unfolding can be explained on the basis of transfer of non-polar groups from organic solvent to water.

Enthalpy and EntropyEnthalpy and Entropy

Go = Ho - TSo

•Typically, G and H are measurable and S calculated

•H and S Provide Mechanistic Insight

•dH = CpdT Gives Thermal Dependence of K

•In very rough generalities:H related to bond formation/breakingS related to configurational freedom and water ordering

O

H

H

O H

HWater molecules seek

favorable positions and partially freeze their

thermal motion.

Extreme ordering of waters caused by hydrophobic molecule

Hydrophobic bond

• Net effect is entropic rather than energetic in nature just because the energy of H-bonds is extremely high and waters would prefer to become frozen and sacrifice a part of their freedom than to lose the large energy of a hydrogen bond.

Reason for negative Reason for negative SS

A simple binding process

++

Binding can be enthalpy or entropy driven

Enthalpic contribution to Free energy of binding

• Ionic and hydrogen bonds.

• Electrostatic interaction.

• Van der Waal interaction

entropy of bindingentropy of binding

• Binding is favoured if it leads to a net increase in disorder or entropy.

• This includes entropy of– the system (interacting molecules and solvent)

• represented as change in entropy or S – the environment (everything else)

• as the system releases or absorbs heat it changes the entropy of the surroundings

• heat release is measured as change in enthalpy or H

Combining the First and Second law

ps

Sp

Vs

sV

S

V

p

T

Vp

HT

S

H

VdpTdSdH

VdppdVpdVTdSdH

VdppdVdUdH

S

p

V

T

pV

UT

S

U

VSfU

pdVTdSdU

),(

P

V

S

T

H

U G

A

Vs

ps

Tp

VT

S

p

V

T

S

V

p

T

p

S

T

V

T

p

V

S

Good people have studied under very able teachers.

Tp

VT

p

S

T

V

SdTVdpdG

SdTTdSVdpTdSdG

SdTTdSdHdG

T

p

V

S

pdVSdTdA

SdTTdSpdVTdSdA

SdTTdSdUdA

TSUddA

The Thermodynamic Square

P

V

S

T

H

U G

A

SdTVdpdG

pdVSdTdA

VdpTdSdH

pdVTdSdU

Thermodynamic equation of state

VT

VT

P

H

PT

pT

V

U

PT

VT

Prove it

PT

P

T

VT

P

C

2

2

Derive the expression for for a gas following the virial equation with Z=1+B/V

TV

U

TV

U

TV

U

True or False ??

No heat transfer occurs when liquid water is reversibly and isothermally compressed.

During an adiabatic, reversible process at constant volume the internal energy can never increase.

The internal energy of a system and its surroundings is never conserved during an irreversible process, but is conserved for reversible processes.

The work done by a closed system can exceed the decrease in the system’s internal energy

When extra (useful) work is involved.

Important Thermodynamic equations

e

e

e

e

dwSdTVdpdG

dwpdVSdTdA

dwVdpTdSdH

dwpdVTdSdU

0)(

0

,

TP

surrsys

Gd

dSdS

Helmholtz free energy

e)Temperaturconstant (At

0

process ReversibleFor

0

0

max dATdSdUdw

TdSdUdw

TdSdwdU

TdSdqT

dqdS

dSdS

dSdS

sys

surrsys

surrsys

Maximum work done (including expansion work) can be calculated by measuring the change in A.

Gibb’s Free Energy

revadddwdG ,

rev add,

pressure, and emperatureconstant tAt

dwVdP-SdTdG

work,extra of presence In the

E.M.F. work

EnFG

nFdEdwdG

dwpdVSdTdG

TP

eTP

e

,

,

0

0

0

,

,

,

TP

TP

TP

E

EnF

G

Displacement of metal from metal salts

• A more active metal can "displace" a less active one from a solution of its salt.

• “Active" metals are all "attacked by acids“.

• Zn(s) + Cu2+ → Zn2+ + Cu(s) .

VveE TP 09.1,

Pressure as a function of height

• An increase in height will correspond to a decrease in pressure.

• Boiling point of water is raised if the pressure above water is increased; it is lowered if the pressure is reduced. This explains why water boils at 70 °C up in the Himalaya.

)(ln

mequilibriuAt

211

2 hhRT

Mg

P

P

dhRT

Mg

P

dP

gdhRT

PMdP

gdhdP

mgdhVdP

mgdhVdPG

mgdhVdPSdTG

T

The Thermodynamics of a Rubber band

LppT T

fTf

L

H

fdLpdVTdSdU

,,

• Q.Experiments show that the retractive force f of polymeric elastomers as a function of temperature and expansion L is given by f(T,L) = aT(L-L0) where a and L0 are constants.

• (a)Use Maxwell relations to determine the entropy and enthalpy at constant T and p.

• (b) If you adiabatically stretch a rubber band by small amount, its temperature increases but volume does not change. Derive an expression for its temperature as a function of L, L0, a and C (heat capacity).

The variation of Gibb’s energy with temperature.

2

/

T

H

T

TG

Variation of the Gibb’s energy with pressure

p

pRTG

pp

p

pRTdp

p

RTG

ppVG

VdpdG

SdTVdpdG

fm

i

i

fm

m

ln

bar) 1at pressure (standard

ln

Gas Ideal

(b)Gases

constantV

solids and Liquids (a)

,emperatureconstant tAt

12

m

Question ?

• Calculate ΔG for the conversion of 3 mol of liquid benzene at 80 C (normal boiling point) to vapour at same temperature and a pressure of 0.66 bar? Consider the vapour as an ideal gas.

• In the transition CaCO3 (aragonite) to CaCO3 (calcite), ΔGm (298) = -800 J and Vm = 2.75 cm3. At what pressure would aragonite become the stable form at 298 K?.