lecture 4. the second law of thermodynamicsthe second law of thermodynamics. limitation of the first...

Lecture 4. The Second Law of Thermodynamics

LIMITATION OF THE FIRST LAW:-Does not address whether a particular process is spontaneous or not.

-Deals only with changes in energy.

Consider this examples:

-Drop a rock from waist-high height, rock will fall spontaneously

-Plunger of a spray is presses , gas comes out spontaneously

-Metallic sodium is placed in a jar with chlorine gas, reaction occurs

Spontaneous Processes:

Why does the color spread when placing a drop of dye in a glass of clean water?

SPONTANEOUS PROCESSES:

Spontaneous processes are those that can proceed without any outside intervention.

The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously go back to B.


Processes that are spontaneous in one direction are not spontaneous in the reverse direction.

SPONTANEOUS PROCESSES

Processes that are spontaneous at one temperature may be not spontaneous at other temperatures.Above 0C it is spontaneous for ice to melt.Below 0C the reverse process is spontaneous


Changes in the extent of disorder.

When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess?

natures’s

way to seek disorder. It is easy to create disorder; difficult to create order.

ENTROPY:

Entropy can be thought of as a measure of the randomness or disorder of a system.

It is related to the various modes of motion in molecules.

Spontaneity and the Sign of S

Why does a room with a fragrance bottle at the other end of the room is suddenly filled with the aroma?

Limonene (l) Limonene (g)

S(process) = S (final state) -

S(initial

state)

A spontaneous process is accompanied by S of positive sign.

Reaction Spontaneity by Inspection

Why do damp clothes become dry when hung outside?

S(g) >> S(l) > S(s)

By inspection alone, decide whether the sublimation of solid carbon dioxide is spontaneous or not. How about the condensation of water?

Spontaneity by inspection.

By inspection alone, decide whether the following reaction is spontaneous or not.

SOCl2

(l) + H2

O(g) 2HCl(g) + SO2

(g)

N2

(g) + 3H2

(g) 2NH3

(g)

ENTROPY: Statistical Definition

1

2

VVlnnRS

when an ideal gas expands isothermally from V1

to V2

where n is the number of moles of the gas present

Sample Problem:

Calculate the entropy change when 2.0 moles of an ideal gas are allowed to expand isothermally from an initial volume of 1.5 L to 2.4 L.

THERMODYNAMIC DEFINITION OF ENTROPY:

ST

qVVlnnR

Tq

VVnRTln q

process reversible a for

-wqprocess, isothermal In

rev

1

2rev

1

2rev

Although for the expansion of gases, they are

applicable to all types of processes at constant

temperature

THE CARNOT CYCLE:

1.

Reversible Isothermal Expansion.

1

2

1

222

VVlnRTq absorbed heat

VVln-RTw done work

0U

THE CARNOT CYCLE:

2. Reversible Adiabatic Expansion.

)T-(TC U done work

0q

21v

3. Reversible isothermal Compression

3

411

3

421

VVlnRTq released heat

VVln-RTw done work

0U

CARNOT CYCLE:

4. REVERSIBLE ADIABATIC COMPRESSION

)T-(TC U done work

0q

12v

CARNOT CYCLE SUMMARY:

3

4

1

22

12

VVlnRT

VVln-RTw(cycle) done work

qq)cycle(q0)cycle(U

THERMODYNAMIC EFFICIENCY:

2

1

2

12

2

TT1

TTT

qw

engine the by absorbed heatengine heat ny done work netEfficiency

REFRIGERATOR:

1.

CONDENSER (hot side heat exchange)

2. EXPANSION VALVE (gas expands, cools and liquifies)

3.

EVAPORATOR (cold side heat exchanger)

4. COMPRESSOR (gas compressed, heated) RED –

gas at high pressure and temperaturePINK –

gas at high pressure reduced temperatureBLUE –

liquid at low pressure and greatly reduced temperatureLIGHT BLUE –

gas at low pressure and warmer temperature

REFRIGERATORS

HEAT

ENGINE

Heat Source T2

Cold Reservoir T1

q2

q1

work w

12

1

1

12

T-TT

wq COP

COP e,Performanc of tCoefficien

wqq-

AIR CONDITIONERS; HEAT PUMPS

AIR CONDITIONERS: OPERATION IS SIMILAR TO REFRIGERATORS:

HEAT PUMPS: USE TO HEAT A ROOM RATHER THAN LOWER THE TEMPERATURE:

21

2

2

12

T-TT

wq

COP

COP e,Performanc of tCoefficien

wqq-

SAMPLE PROBLEM:

Calculate the coefficient of performance of a reversible refrigerator operating between and interior temperature of 4oC and an exterior temperature of 22oC.

The Second Law of Thermodynamics:

0SSS,Thus

0)1V/2Vln(nRSSS

processes leirreversib for

0T

)V/Vln(nRTT

)V/Vln(nRTT

qT

qSSS

processes; reversible for:universetheof entropy in Change

surrsysuniv

surrsysuniv

surrsys

surrsysuniv

The entropy of an isolated system increases in an irreversible processes and remain unchanged in a reversible process.

ENTROPY CHANGES DUE TO MIXING:

B and Agas offractions mole are x and x where)xlnnxlnn(RS

VVVlnRnS B;Gas For

VVVlnRnS A;Gas For

BA

BBAAmix

B

BAB

A

BAA

ENTROPY CHANGES PHASE CHANGE:

TH

S

THS

Hq so,process pressure constantOHOH

:Consider

vapvap

fusfus

fusrev

)l(2atm1,C0

)s(2

o

SAMPLE PROBLEM

The enthalpy of vaporization of methanol is 35.27 kJ/mol at its normal boiling point of 64.1oC. Calculate (a) the entropy of vaporization of methanol at this temperature and (b) the entropy change of the surrounding

ENTROPY and changes in Temperature:

1

2p

T

T

p

p

rev

TTlnnCdT

TC

S

) pressure constant system, (closed dTCndHdq

TdqdS

2

1

Sample Problem:

Calculate S and Ssurr

for reversibly heating 2.0 moles of liquid water from 0.00oC to 100oC at constant pressure of 1.0 atm. Specific heat capacity of liquid water is 4.184 J/K-g.

SAMPLE PROBLEM:

Calculate the change in entropy when 1.0 mole of ice at -10oC is heated until it is a superheated steam at 120oC.

mol-J/K 6.33O(g)H Cmol-J/K 3.75O(l)H Cmol-J/K 7.37O(s)H C

mol-J/K 4.109SmolK/J 0.22S

2p

2p

2p

vap

fus

THIRD LAW OF THERMODYNAMICS:

01lnkS1W K,0 at WlnkS

dTTC

S

zero;toetemperaturthelowerweSuppose

b

b

T

0

p

Every substance has a finite positive entropy, but at absolute zero the entropy maybe come zero, and it does in case of pure perfect crystalline

substance.


The entropy of a perfect crystal at 0 K is zero.

It is impossible to reach a temperature of absolute zero

It is impossible to have a (Carnot) efficiency equal to 100% (this would imply Tc = 0).


T = 0, S = 0 T > 0, S > 0

ENTROPY OF CHEMICAL REACTIONS:

)tstanreac(Sv)products(SvS

(B)Sb-(A)Sa-(D)Sd(C)ScS

:by givenis reaction of entropy ethdDcCbBaA

reaction, alhypothetic a For

ooor

ooooor

SAMPLE PROBLEM:

Calculate the value of the standard molar entropy changes for the following reactions at 298 K.

210.6 :NO 205, :O 191.5, :N69.9 :OH 205, :O 130.6, :H

213.6 :CO 39.8, :CaO 92.9, :CaCOmol) (J/K :1bar) K,(298 values S

2NO(g) (g)O (g)N c)

O(l)2H (g)O (g)2H b)

(g)CO CaO(s) (s)CaCO a)

22

222

23

o22

222

23

The GIBBS Energy:

ST-HG :EnergyGibbs TS-HG :function a define now we

0TdS-dHT by multiply

0T

dqdS

0dSdSdSnot? ors spontaneouIs

l)O(gH (g)O(g)Hreaction, the Consider

syssys

surrsys

surrsysuniv

2

22

EQUILIBRIUM and SPONTANEITY CRITERIA:

other. each reinforce canons contributi enthalpy and entropy- entropy. and enthalpy bothes incorporat-

:energyGIBBS of ceSignificanpressure) and etemperatur constant (at

2 to 1 fromprocess s spontaneou ;01G2GGmequilibriu atis system ;01G2GG

0dGsys

THE GIBBS ENERGY: Factors affectingH S G

+ + Positive at low Temp; negative at high Temp; Reaction is spontaneous at forward at high T and spontaneous in reverse direction at low temperature.

+ - Positive at all temperatures, Reaction is spontaneous in the reverse reaction at ll

temperatures- + Negative at all temperatures. Reaction is spontaneous in the

forward direction at all T.- - Negative at low temperatures; positive at high temperatures.

Reaction is spontaneous at Low temperatures. Tends to reverse at high temperatures.

The HELMHOLTZ Energy:

0A:mequilibriu and yspontaneit for criteria

processes. volume and etemperatur constant forEnergy HelmholtzTS-UA

sys

SAMPLE PROBLEM:

Calculate the value of G for the melting of ice at a) 0oC b) 10oC c) -10oC. The molar enthalpy and entropy of fusion of water are 6.01 and 22.0 J/K-mol.

STANDARD MOLAR GIBBS ENERGY OF FORMATION

)tstanreac(Gv)products(GvG

(B)Gb-(A)Ga-(D)Gd(C)GcG

:by givenis reaction of EnergyGibbs molar standard thedDcCbBaA

reaction, alhypothetic a For

ooor

ooooor

GIBBS FREE ENERGY, GGIBBS FREE ENERGY, G

∆∆GGoo

= = ∆∆HHoo

--

TT∆∆SSoo

Two methods of calculating Two methods of calculating ∆∆GGoo

A. Determine A. Determine ∆∆HHoo

rxnrxn

and and ∆∆SSoo

rxnrxn

and use and use GIbbsGIbbs

equation.equation.

B. Use tabulated values of B. Use tabulated values of free energies of formation, free energies of formation, ∆∆GGff

oo..

∆Go

rxn

=

∆Gf

o

(products) -

∆Gf

o

(reactants)∆∆GGoo

rxnrxn

= =

∆∆GGff

oo

(products) (products) --

∆∆GGff

oo

(reactants)(reactants)

FREE ENERGIES OF FORMATIONFREE ENERGIES OF FORMATION

Note that Note that ∆∆GG˚̊ff for an element = 0for an element = 0

SAMPLE CALCULATION,SAMPLE CALCULATION,

∆∆GGoo

rxnrxnFor the combustion of acetyleneFor the combustion of acetylene

CC22

HH22

(g) + 5/2 O(g) + 5/2 O22

(g) (g) ----> 2 CO> 2 CO22

(g) + H(g) + H22

O(g)O(g)a)a)

by inspection is the reaction spontaneous or not?by inspection is the reaction spontaneous or not?b)b)

Calculate the Calculate the ∆∆GGoo

rxnrxn

using standard molar enthalpies and using standard molar enthalpies and entropies.entropies.

c) Is the reaction spontaneous or not? Is it entropy or enthalpc) Is the reaction spontaneous or not? Is it entropy or enthalpy y driven?driven?

CALCULATINGCALCULATING

∆∆GGoo

rxnrxn

Is the dissolution of ammonium nitrate productIs the dissolution of ammonium nitrate product--favored? favored? If so, is it enthalpyIf so, is it enthalpy--

or entropyor entropy--driven?driven?

NHNH44

NONO33

(s) + heat (s) + heat ------> NH> NH44

NONO33

(aq)(aq)

CALCULATING CALCULATING ∆∆GGoo

rxnrxn

From tables of thermodynamic data we find:From tables of thermodynamic data we find:∆∆HHoo

rxnrxn

= = --25.7 kJ25.7 kJ∆∆SSoo

rxnrxn

= +108.7 J/K or +0.1087 kJ/K= +108.7 J/K or +0.1087 kJ/K∆∆GGoo

rxnrxn

= = --25.7 kJ 25.7 kJ --

(298 K)(+0.1087 J/K)(298 K)(+0.1087 J/K)= = --6.7 kJ6.7 kJ

Reaction is Reaction is spontaneousspontaneous

in spite of negative in spite of negative ∆∆HHoo

rxnrxn

. . Reaction is Reaction is ““entropy drivenentropy driven””

NHNH44

NONO33

(s) + heat (s) + heat ------> NH> NH44

NONO33

(aq)(aq)

Gibbs Free Energy, GGibbs Free Energy, G

∆∆GGoo

= = ∆∆HHoo

--

TT∆∆SSoo

Two methods of calculating Two methods of calculating ∆∆GGoo

a)a)

Determine Determine ∆∆HHoo

rxnrxn

and and ∆∆SSoo

rxnrxn

and use Gibbs and use Gibbs equation.equation.

b)b)

Use tabulated values of Use tabulated values of free energies of free energies of formation, formation, ∆∆GGff

oo..∆Go

rxn

=

∆Gf

o

(products) -

∆Gf

o


rxnrxn

= =

∆∆GGff

oo


∆∆GGff

oo


Calculating Calculating ∆∆GGoo

rxnrxn

Combustion of carbonCombustion of carbonC(graphiteC(graphite) + O) + O22

(g) (g) ----> CO> CO22

(g) (g) ∆∆GGoo

rxnrxn

= = ∆∆GGff

oo(CO(CO22

) ) --

[[∆∆GGff

oo(graph(graph) + ) + ∆∆GGff

oo(O(O22

)])]∆∆GGoo

rxnrxn

= = --394.4 kJ 394.4 kJ --

[ 0 + 0][ 0 + 0]Note that free energy of formation of an element in its standardNote that free energy of formation of an element in its standard

state is 0.state is 0.∆∆GGoo

rxnrxn

= = --394.4 kJ394.4 kJReaction is Reaction is spontaneousspontaneous

..

∆Go

rxn

=

∆Gf

o

(products) -

∆Gf

o


rxnrxn

= =

∆∆GGff

oo


∆∆GGff

oo


FREE ENERGY AND TEMPERATUREFREE ENERGY AND TEMPERATUREIron metal can be produced by reducing its ore (Iron metal can be produced by reducing its ore (Iron(IIIIron(III) )

oxide with graphite:oxide with graphite:2 Fe2 Fe22

OO33

(s) + 3 (s) + 3 C(sC(s) ) ------> 4 > 4 Fe(sFe(s) + 3 CO) + 3 CO22

(g)(g)∆∆HHoo

rxnrxn

= +467.9 kJ= +467.9 kJ

∆∆SSoo

rxnrxn

= +560.3 J/K= +560.3 J/K∆∆GGoo

rxnrxn

= +300.8 kJ= +300.8 kJA) Is the reaction spontaneous or not?A) Is the reaction spontaneous or not?B) At what temperature will the reaction become B) At what temperature will the reaction become

spontaneous?spontaneous?At what T does At what T does ∆∆GGoo

rxnrxn

just change from being (+) to being (just change from being (+) to being (--)? )? When When ∆∆GGoo

rxnrxn

= 0 = = 0 = ∆∆HHoo

rxnrxn

--

TT∆∆SSoo

rxnrxn

FACT: FACT: ∆∆GGoo

rxnrxn

is the change in free energy is the change in free energy when pure reactants convert COMPLETELY to when pure reactants convert COMPLETELY to pure products.pure products.

FACT: ProductFACT: Product--favored systems have favored systems have KKeqeq

> 1.> 1.

Therefore, both Therefore, both ∆∆GG˚̊rxnrxn

and and KKeqeq

are related to are related to reaction favorability.reaction favorability.

Thermodynamics and Thermodynamics and KKeqeq

KKeqeq

is related to reaction favorability and so to is related to reaction favorability and so to ∆∆GGoo

rxnrxn

..The larger the value of K the more negative the value The larger the value of K the more negative the value

of of ∆∆GGoo

rxnrxn

∆∆GGoo

rxnrxn

= = -- RT RT lnKlnK

where R = 8.31 J/where R = 8.31 J/KK••molmol

THERMODYNAMICS AND THERMODYNAMICS AND KKeqeq

Calculate K for the reactionCalculate K for the reaction

NN22

OO44

------>2 NO>2 NO22

∆∆GGoo

rxnrxn

= +4.8 kJ= +4.8 kJ

∆∆GGoo

rxnrxn

= +4800 J = = +4800 J = --

(8.31 J/K)(298 K) (8.31 J/K)(298 K) lnln

KK

∆∆GGoo

rxnrxn

= = --

RT RT lnKlnK

ln K = - 4800 J(8.31 J/K)(298 K)

= - 1.94

THERMODYNAMICS and THERMODYNAMICS and KKeqeq

K = 0.14K = 0.14When When ∆∆GGoo

rxnrxn

> 0, then K < 1> 0, then K < 1

Free Energy and Chemical Equilibrium

•

The sign of G°

tells the direction of spontaneousreaction when both reactants and products arepresent at standard state conditions.

•

Under nonstandard conditions, G°

becomes G.G = G° + RT lnQ

•

The reaction quotient is obtained in the same way as an equilibrium expression

∆∆G, G, ∆∆GG˚̊, and , and KKeqeq

Figure 19.10Figure 19.10

••

Product favored reaction Product favored reaction (spontaneous)(spontaneous)

••

––∆∆GGoo

and K > 1and K > 1

••

In this case In this case ∆∆GGrxnrxn

is is < < ∆∆GGoo

rxnrxn

, so state with both , so state with both reactants and products reactants and products present is MORE STABLE present is MORE STABLE than complete conversion.than complete conversion.


Spontaneous reaction. Spontaneous reaction. 2 NO2 NO22

------> N> N22

OO44

∆∆GGoo

rxnrxn

= = ––

4.8 kJ4.8 kJHere Here ∆∆GGrxnrxn

is less than is less than ∆∆GGoo

rxnrxn

, so the state with , so the state with both reactants and both reactants and products present is more products present is more stable than complete stable than complete conversion.conversion.


Non spontaneous reaction. Non spontaneous reaction. NN22

OO44

------>2 NO>2 NO22

∆∆GGoo

rxnrxn

= +4.8 kJ= +4.8 kJHere Here ∆∆GGoo

rxnrxn

is greater than is greater than ∆∆GGrxnrxn

, so the state with , so the state with both reactants and both reactants and products present is more products present is more stable than complete stable than complete conversion.conversion.


KKeqeq

is related to reaction favorability.is related to reaction favorability.

When When ∆∆GGoo

rxnrxn

< 0, reaction moves energetically < 0, reaction moves energetically ““downhilldownhill””

∆∆GGoo

rxnrxn

is the change in free energy when is the change in free energy when reactants convert COMPLETELY to products.reactants convert COMPLETELY to products.

Thermodynamics and Thermodynamics and KKeqeq

END OF CHAPTEREND OF CHAPTER

lecture 4. the second law of thermodynamicsthe second law of thermodynamics. limitation of the first...

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