2nd Law ThermodynamicK1-K6_2nd Law - Entropy (14,15) - Edaran

8/10/2019 2nd Law ThermodynamicK1-K6_2nd Law - Entropy (14,15) - Edaran

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Physical Chemistry II

Prof. Dr. Ishak AhmadPPSKTM, FST, UKM

G150 (Chemistry Building)[email protected]


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Part II Physical transformations of pure substances The stabilities of phases Phase boundaries Supercritical fluids

Three typical phase diagram Thermodynamic criterion of equilibrium The dependence of stability on the condition The location of phase boundaries

Solid-liquid boundaryLiquid-vapour boundaryClausius-Clapeyron equation

Ehrenfest Calssification of phase transitions


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SECOND LAWS OFTHERMODYNAMICSMany processes are reversible and yet tendto proceed in a direction in which they aresaid to be spon taneous .

What makes a reaction spontaneous? Whatdrives the reaction in one direction and notthe other?

2.1 Spontaneous ProcessesSpontaneous change : Occurs by itself,without continuous outside assistance underspecific condition (temperature, pressure,and concentration).

Non spontaneous : Occurs with continuousoutside assistance . As soon as supplyenergy cut off, the process stop.


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Some spontaneous changes occur veryrapidly i.e. biochemical reactions.

Touch something very hot, spontaneous jerk hand quickly.

Others occur slowly and many years suchas erosion of mountain.

Spontaneous process cannot occur inboth direction under the same conditions.


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Disp ers al of Energy Spontaneous changes are accompaniedby dispersal of energy into a moredisordered form.

Bo unc ing b al l:

Ball does not return to original height , asenergy is redistributed to molecules in thefloor and ball as heat - inelastic energylosses.

Ball eventually comes to rest, losing allenergy (dispersed) into thermal motion ofatoms in the floor

the reverse process will never occur!


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Heated f lo o r an d b al l : Molecules and atoms in the floor (and

ball) undergo thermal (random) motion In order for the ball to spontaneously

bounce, it would require aspontaneous localization of motion ,to create an upwards motion of all ofthe atoms - a virtually impossibleprocess**


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Normally, spontaneous processes occur soas to decrease the energy of the system,e.g. spring in a clock unwind.

Examples of spontaneous chemicalreactions that are exothermic (most cases):

2 Al( s) + 3 Br 2 (l) 2 AlBr 3 (s) H o = -511 kJ/mol

2 H 2 ( g) + O 2 (g) 2 H 2 O(g) H o = -241.82 kJ/mol

P4 ( s) + 5 O 2 (g) P 4O10 (s) H o = -2984 kJ/mol


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However endothermic reaction can alsospontaneous e.g. ice melts spontaneousabove 0 oC.

E.g.:

1. Decomposition of mercury(II) oxide:2HgO(s) 2Hg(l) + O 2(g)

H= +ve

2. Dissolution of ammonium nitrate inwaterNH4NO 3(s) NH 4+(aq) + NO 3- (aq)

H= +ve


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Conclusion:

Exothermic favors the spontaneity ofreaction but does not guarantee it.

Just as it possible for an endothermicreaction to be spontaneous.

It is also possible for an exothermicreaction to be nonspontaneous.

We cannot confirm a chemical reactionwill occur spontaneously solely on thebasis of energy changes ( H) in the

system.

Another thermodynamic quantity must betake into consideration which turns out to

be en t ropy .


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2.2 ENTROPY

A measure of the randomness ordisorder of a system.

Symbol S .

The greater the disorder of the system,the greater its entropy.

Conversely, the more ordered a system,the smaller its entropy.

Particles in the solid state are moreordered than those in the liquid state,which in turn are more ordered thanthose in gaseous state.

S solid < S liquid


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Figure shows several processes thatlead to an increase in entropy.

In each case more ordered stated

to less ordered.

Factors that increase the entropyof the system


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Melting and vaporization

Melting and vaporization processes

have S > 0.S solid < S liquid


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Polymorphism and Molar mass

For different substances in the samephase, the molecular complexity andmolar mass determine which oneshave higher entropy.

Case 1:Diamond and graphite. Diamond hasa smaller entropy because itsstructure is more ordered.

Case 2:Neon and Helium. Neon has greaterentropy because its molar mass isgreater.


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2nd law and entropy

First LawUses internal

energy, U, toidentify permissiblechanges

Second LawUses entropy, S, to

identify thespontaneous changesamong the permissiblechanges

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The law used to identify spontaneouschange, and can be quantified in terms of astate function* known as en t ropy

*Value depends only on the current state of the systemand is independent of how the state has been prepared


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The change in a disorderly mannerdepends on the quantity of energytransferred as heat.

The thermodynamic definition ofentropy is based on the expression:

For a measurable change between two

state i and f this expression integratesto

f i

T

dq s


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Entropy change: ReversibleIsothermal Expansion of

Perfect GasU = 0 so q = -w ( U = q + w)

V = V f - V i so w < 0

ORq = nR T ln V 2

V1

Hence,S = nR ln V 2

V1or

S = - nR ln P 2P 1

1

2

2

1

lnV

V nR T w

V

dV nR T w

V

V

Work exp ansion = -ve


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The Variation of Entropywith Temperature

From the measurable changebetween two state i and f;

f i

T dq

s

S(T f ) = S(T i ) + f i

T

dqor

From the definition of constant-pressureheat capacity;

q p = C p T

Consequently, at constant pressure;

f i

pi f T

dT C )T ( S )T ( S


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Variation of Entropy withTemperature for Expansion of

Perfect Gas By combining the entropy at different

temperature:

i

f

p T T lnnC ) sys( S

and the entropy of the isothermalexpansion of perfect gas :

1

2

V V

lnnRS

Hence we get (factor T and V areincluded):

1

2

1

2

V V

lnnRT T

lnC ) sys( S p


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Quiz

Write the equation for each types ofwork:a. Free expansionb. Expansion against constant pressurec. Isothermal reversible expansion

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Answer

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Entropy at phase transition

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Melting of a solid and vaporization of aliquid correspond to accept thermal energy Energy flow into a system, at a constant

temperature

These inflows of thermal energycorrespond to the heats of fusion andvaporization.The entropy increase associated withmelting, for example, is just H fusion /T m. Heat cap acity, C of a phase expressesthe quantity of heat required to change thetemperature by a small amount T.

At 1 atm : standard entropy of a substance At 298K, and are expressed in units of J K 1 mol 1


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Entropy at a Phase Transition Consider the phase transitions of water, at transition

temperatures T trs . For ice water, T trs = 273 K, ice in equilibrium with liquid

water at 1 atm. The external pressure is constant for a glass of ice

water, and in order to match attractive forces betweenice molecules, energy must come from kinetic energy

of the water molecules or the surroundings. At T trs , any transfer of heat between the system and

surroundings is reversible since the two phases in thesystem are in equilibrium (the forces pushing the icetowards melting are equal to those pushing the watertowards freezing) - so a phase transition isreversible.

It does not matter how the ice melts (what path ittakes) since entropy is a state function. What doesmatter for this particular expression is that the systembe isothermal . If it was not isothermal, one would

have a problem examining the process in steps aswe shall see.

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Entropy of PhaseTransition Temperature

Because at constant pressure,q = trans H

The change in molar entropy of thesystem is;

trs

trstrs

T

H S


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Troutons Rule

Troutons Rule:This empirical observation(see Table 4.2) states that most liquids have

approximately the same standard entropy ofvaporization, vap S o 85 J K -1mol -1: So,vap H o = T b 85 J K -1 mol -1

See Tab le 3.1 A tk ins 9 th ed )

xceptions :

In water , molecules are more organised inthe liquid phase (due to hydrogen bonding ),so a greater change of disorder occurs uponvapourizationIn methane , the entropy of the gas is slightlylow (73.2 J K -1mol -1 at 298 K) and in lightmolecules very few rotational states areaccessible at room temperature -associated disorder is low

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Examples

Calculate S when argon at 25 o

C and 1.00 atmin a container of volume 500 cm 3 expands to1000 cm 3 and is simultaneously heated to100 oC. C p,m (Ar) = 20.786 J K -1 mol -1

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Methodo logy :Since S is a state function, we can choose a convenientpath from initial to final state: (1) isothermal expansionto final volume, then (2) reversible heating at constantvolume to final temperatureAmount of Ar present is n = pV/RT = 0.0204 mol

C p,m (Ar) = 20.786 J K -1 mol -1

(1) Expansion from 500 cm 3 to 1000 cm 3 at constant T:S = nR ln 2.00 = +0.118 J K -1

(2) Reversible heating from 25 oC to 100 oC at cons tan tV :

S =(0.0204 mol) (12.47JK -1 )ln(373 K/298 K)= +0.057 J K -1

Overall entropy change:S = +0.118 J K -1 + 0.057 J K -1 = +0.173 J K -1


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Entropy Change of Melting Ice Consider the entropy changes for putting

an ice cube in a glass of warm water andletting it melt ( adiabatic container ).

Start at Ti (1) Calculate S to cool the water to 0 oC by

reversibly removing heat, q 1, from the system (2) At T fus , calculate the amount of heat, q 2 , to be

added to the system to melt the ice cube (3) Calculate the difference between the two

amounts of heat and add back remaining heat sothat the total heat lost or gained is zero(adiabatic system) - determine the entropy changein this process

At each step, the infinitesimal entropy change forthe system, dS, is just dq divided by the T . For thecooling and heating of water, integrate over thetemperature range , since the temperature is not

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Is the process spontaneous? (We knowintuitively that it is - sticking an ice cubeinto warm water melts the ice cube!!).

How do we prove this?Show that the total entropy change (the

system plus the surroundings) is positive.

The entropy change of the system weknow is:S 1 + S 2 + S 3

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Adiabatic Changes

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Carnot Cycle The Carnot cycle, named after Sadi Carnot, has

four reversible stages1. Reversible isothermal expansion from A to B atT h, S = q h /T h, q h is heat supplied from a hot source,and is positive2. Reversible adiabatic expansion from B to C, noheat leaves system ( S = 0), temperature falls fromT h to T c , where T c is the temperature of the cold sink3. Reversible isothermal compression from C to D,S = q c /T c , qc is heat released into a cold sink, and isnegative

4. Reversible adiabatic compression from D to A,no heat enters the system ( S = 0), temperature risesfrom T c to T h

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Total entropy change in Carnot cycle:

This S zero since

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c

c

h

h

T

q

T

qdS

c

h

c

h

T

T

q

q


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S = 0 fo r Carn o t Cyc le

for reversible isothermal expansion of aperfect gas:

And for reversible adiabatic processes

Multiplying the expressions

and simplifying:

Finally we get:

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A

Bhh V

V nRT q ln

C

Dcc V

V nRT q ln

cc D

ch A T V T V

ch B

ccC T V T V

c

iic

f f T V T V

R

C c mV ,

cc

ch B D

cc

chC A T T V V T T V V

C

D

B

A

V V

V V

B

Acc V

V nRT q ln


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Thermodynamic in Carnot Engine

Step 1 : Isothermal expansion,

Step 2 : Adiabatic expansion, U 2 = w 2

Step 3 : Isothermal compression, U 3 = q 2 + w 3

Step 4 : Adiabatic compression, U 4 = w 4

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1

2

VV

nRTw ln

cii

c f f T V T V RC c

mV ,

3

4

VV

nRTw ln


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If we combine the net work ofthe engine:

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1

2c-h

VV

)TnR(Tw ln


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Efficiency The same calculation just completed for

gases applies to all types of materials andsystems. We define the efficiency , , of aheat engine:

The greater the work output froma given supply of heat, thegreater the efficiency of theengineWork performed by the engine isthe difference between heat supplied fromthe heat source and returned to the cold sink:

Since q c < 0,

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hq

w

absorbedheat performedwork

h

c

h

ch

q

q

q

qq 1

h

c

T T

1


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Carnot cycles and Engines

2nd Law : all reversible engines have thesame efficiency regardless ofconstruction - Two engines A and B,assume A more efficient than B, coupled

together using the same reservoirs.

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A: takes heat q h, releases heat q cB: takes heat q c releases q h

Since A more efficient than B, not

all work A produces is needed forthis process, and difference canbe used to do work


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2nd Law ThermodynamicK1-K6_2nd Law - Entropy (14,15) - Edaran