Lecture 1 ElectroChem and Corr

7/30/2019 Lecture 1 ElectroChem and Corr

1/65

Electrochemistry and Corrosion

Professor Brian Kinsella


2/65

Corrosion

Corrosion can be defined as the deterioration

of a substance by reaction with its

environment.

Strictly speaking this definition includes non-

metals as well as metals, for example, plastics

and ceramics. However, for this course

corrosion will focus on the corrosion of metalswhich is mainly by electrochemical reactions.


3/65

Assignment 1 Corrosion of

Reinforcing Bar in Concrete

Often referred to as concrete cancer this area ofcorrosion amounts to tremendous costs togovernment, industry and society.

It affects concrete structures of wharves, bridges, tallbuildings and pathways.

The problem is caused by oxygen corrosion of themild steel reinforcing bar. The iron oxide corrosion

product occupies a larger volume then the iron andthis results in high internal pressure causing theconcrete to crack and spall (break in chips or flakes).


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Under normal circumstances the steel does notcorrode due to the high pH of concrete and theformation of a passive (protective oxide) film whichcan form under this condition.

Destruction of the passive film can be attributed toingression of chloride ions and carbon dioxide.

There are a number of techniques used to preventcorrosion taking place, including cathodic protection.

There are also a number of techniques used to treatthe problem once it has occurred.

In your assignment you will need to cover thecorrosion mechanism, the methods of preventionand the methods of treatment.


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Key topics to look for:

1. Pourbaix diagram

2. Cathodic protection methods

3. Inhibitors

4. Quality of concrete

Comment on the effectiveness of preventionand cure.


6/65

Basic Thermodynamics

Gas Laws


7/65

Basic Thermodynamics Gas Laws and Constant

1-1-

32-5

1-1-7

32-6

1-1-

moledegjoules8.314

deg16273mole1

meter02241360meternew tons100131

moledegergs108.314

deg16273mole1

cm641322cmdyne100131

unitssecond)-kilograms-(metermksandcgsinRconstantGas

moledegatm-l08205490

deg160273mole1

l413622atm1

.

..

.

.,.

.

.

.

nT

PVR

nT

PVR

nT

PVR

nRTPV


8/65

Gas Constant

The calorie is equal to 4.1840 joules

The equation is used to determine the number of

moles of a gas if P, V and T are known or P, V or T if

the other parameters are known.

It is important to remember that 1 mole of gas willoccupy 22.4 liters at 1 atmosphere and 0oC.

1-1-

1

-1-1

moledegcal987.1caljoules184.4

moledegjoules314.8

R


9/65

Gas Constant

Note: The value of R = 8.314 joules deg-1 mole-1

is used in the Nernst Equation

r

p

r

p

a

a

nEE

n

EE

aa

nFRTEE

log0591.0

500,96

1.298314.8303.2

ln

0

0

0


10/65

Daltons Law The total pressure of a gas in a

liquid is equal to the sum of the partial pressures

of the gases in the mixture

The total number of moles of a gas in a mixture is equal to

the sum of the numbers of moles of the different gases

Where lower case p is used to represent the partial

pressure

cba

cbatotal

cbatotal

ppp

V

RTn

V

RTn

V

RTn

V

RTn

nnnn

totalP


11/65

Raoults Law

The mole fraction of a component in the

vapor is equal to the pressure fraction in the

vapor

E.g. for two components

componentspuretheofpressurepartialtheareandandfractionsmoletheareandw here

2

121

222

111

ppXX

pXp

pXp


12/65

Raoults Law

2211

11

21

1,1

pXpX

pX

pp

pX vap


13/65

Raoults Law Ideal Solutions

Example: The vapor pressure of benzene and toluene at 60o

are 385 and 139 mm. Calculate the partial pressure of

benzene and toluene, the total vapor pressure of the solution,

the mole fraction of toluene in the vapor above a solution

with 0.60 mole fraction toluene.

pbenzene = 0.4 x 385 = 154

ptoluene = 0.6 x 139 = 83.4

Ptotal = 237.4

35104237

483vaptoluene

..

.X

Note that the more

volatile component is

more concentrated in the

vapor. In this case

benzene is more volatile

and the mole fraction of

benzene in the vapor will

be 154/237.4 = 0.649


14/65

Raoults Law Ideal Solutions

The law enables the determination of volatilecomponents such as pentane and hexane in the gasphase for top of the line corrosion mechanisms.

The law predicts that the vapor pressure of a purecomponent will decrease due to the presence of anon volatile soluble material, e.g. salt in water willdecrease the mole fraction of water and its vaporpressure.

Does the corrosivity of CO2 gas decrease under supersaline conditions?


15/65

Henry's Law - Non Ideal Solutions

For dilute solutions, the partial pressure of the

component present at lower concentration is

directly proportional to its mole X2

For ideal solutions K2 = p2o and Henrys law

becomes identical to Raoults law

constantlawsHenry'

(solute)conc.lowatcomponenttheisw here

2

2

222

K

X

KXp


16/65

Henry's Law - Non Ideal Solutions

Gas K2 Water 25oCH2 5.34 x 10

7

N2 6.51 x 107

O2 3.30 x 107

CO 4.34 x 107

CO2 1.25 x 106

CH4 31.4 x 106

C2H6 23.0 x 106

K2 = P2/X2 The partial pressure of

the gas is in mm and the conc.

units are mole fraction


17/65

Henry's Law - Non Ideal Solutions Calculate the solubility of CO2 in water at 25

oC at a CO2 pp of

760 mm

1-2

6

2

2

6

2

2

litermole1038.31025.1

55.55760

1000/18.02OHofmolesofnumbertheto

comparednegligibleisandg1000mL1000assume

solutionof

litreperCOofmolesofnumbertheiswhere

02.18

10007601025.1

2

2

2

22

CO

CO

CO

CO

CO

n

n

n

n

n

X

pK


18/65


19/65

Consequence of Daltons Raoults and Henrys Law

It is important to remember Daltons, Raoults andHenrys laws when preparing corrosive gas mixtures.

E.g., If you need to determine the corrosive effect of1 ppm dissolved oxygen. Knowing that seawatercontains about 10 ppm dissolved oxygen at 20oC and1 bar (760 mm), i.e., 80% N2 and 20% O2. A gasmixture of 98% N2 and 2% O2 should result in 1 ppmdissolved oxygen.

The same result can be obtained by bubbling

nitrogen and oxygen into your test solution at therates of 98 and 2 mL s-1 respectively (or 196 and 4 mLminute-1,respectively etc).


20/65

Consequence of Daltons Raoults and Henrys Law

Alternatively, to study the corrosivity of CO2 at 0.2bar, you can achieve the required concentration bybubbling N2 and CO2 gas into your test solution at 20and 80 mL minutes-1 respectively.

You are required to compare the corrosivity of aceticacid with CO2 at 0.2 bar to steel. Determine themolar concentration of acetic acid required since forcomparison, both acids need to be at the same molar

concentration. CO2 + H2O H2CO3


21/65

First Law of Thermodynamics

Energy can be transformed (changed from one form

to another), but it can neither be created nor

destroyed."

The increase in the internal energy of a system isequal to the amount of energy added by heating the

system, minus the amount lost as a result of the

work done by the system on its surroundings.


22/65

Reversible Expansion of a Gas

2112

2

1

1

2

2

1

2

1

2

1

sinceP

Plog3032

V

Vlog3032

PPVV

nRTnRTwV

dVnRTdV

V

nRTPdVw

wqdE

wqE

rev

v

v

v

v

v

v

rev

..

The work performed during the expansion of a gas

can be related to changes in pressure or volume.


23/65

Dependence of Internal Energy on Temperature

and Volume

VV

V

V

V

TV

TV

T

E

T

qC

CdTq

dTT

Eq

dVV

EPdT

T

Ed

PdVqdE

dVVEdT

TEdE

volume.constantatsystemtheofcapacityheattheasknow nisandvolumeconstantat

allyexperimentmeasuredbemayquantitythe

constantheldissystemtheofvolumetheif


24/65

Energy

For an ideal gas:

2

1

21 istofromheatedgasof

moleaforenergyinternalinchangeThe

T

TV

V

V

V

dTC

TT

dTCdTTEdE

PdVdTCq

E


25/65

Enthalpy

Constant pressure process

dpPHdT

THdH

Hq

PVEH

PVEPVEq

VVPqEE

VPqE

TP

enthalpyfunctionstateNew

adsorbedheatthe-

---

1122

1212


26/65

Constant pressure and Relationship between CP and

CV

The ratio is the heat capacity at constant pressure, Cp

dTT

Hq

pp

dTqP

RCC

CC

dPT

HdTCdH

T

H

dT

q

C

VP

VP

P

p

P

p

p

pressureconstantatheatedissubstanceawhendoneisworkvolume

pressurebecausethanlargeralwaysis


27/65

Reversible Adiabatic Expansion of a Gas

1

2

1

2

V

VR

T

TC

V

dVR

T

dTC

V

V

lnln

No loss or gain of heat to the environment

the system is thermally isolated so that q = 0


28/65

Reversible Adiabatic Expansion of a Gas

Calculate the temperature increase and final pressure of He ifa mole is compressed adiabatically and reversibly from 44.8

liters at 0oC to 22.4 liters. The molar heat capacity of He = 3.00

cal deg-1 mole-1 . R must be expressed in the same units.

o

o

V

KT

T

T

VVRTTC

159.3etemperaturinIncrease

4432

1273210039871

844422303298711273303203

2

2

2

1212

.

.loglog./.log

..log...log..

lnln


29/65

Molar Heat Capacity of Gases cal deg-1 mole-1 at 25oC

Gas Cp Cv Cp / Cv

Argon 4.97 2.98 1.67

Helium 4.97 2.98 1.67

Hydrogen 6.90 4.91 1.41

Oxygen 7.05 5.05 1.40

Nitrogen 6.94 4.95 1.40

Carbon Dioxide 8.96 6.92 1.29

Methane 8.60 6.59 1.31

Ethane 12.71 10.65 1.19


30/65

Heat Capacity of Solids

Debye showed that at sufficiently low

temperature

The quantity , the characteristic temperature

of a substance, is usually of the order of 100-

400oK. The equation is suitable for

extrapolating heat capacity data below 15oK

3

4

5

12

T

RCV

S d d S


31/65

Standard State The change in enthalpy H or internal energy E for a

reaction depends on the states of the reactants and

products. E.g. the heat of combustion of graphite isdifferent from that of diamond. To facilitate thetabulation of thermodynamic data certain standard stateare adopted and thermodynamic properties aretabulated to these standard states. The standard state of

a gas is the ideal gas at 1 atm at the temperatureconcerned; for a liquid it is the pure liquid at 1 atm at thetemperature concerned; for a solid it is a specifiedcrystalline state at 1 atm at the temperature concerned.The standard form of carbon is graphite and the standard

form of sulfur is rhombic sulfur. In writing equations,solids, liquids, and gases are designated by (s), (l), and (g)respectively, since the enthalpy change depends on thephysical state of the reactants and products. Thestandard state refers to temperature at 25oC unless

otherwise specified.


32/65

Calculation of the Heat of a Reaction at Constant Pressure, from

the Heat of Reaction at Constant Volume

The standard heat of combustion of n-heptane at constant volume and

25oC is 1148.93 kcal

C7H16(l) + 11O2 (g) = 7CO2 (g) + 8 H2O(l)

Eo = -1,148.93 kcal mole-1

Estimate the heat absorbed at constant pressure

qp = qv + RTn = -1,148,930 (1.987)(298.1)(4)

= 1,151,300 cal mole-1 or1,151.30 kcal mole-1

nRTqq

nRTEH

VPEH

VP

Pp

Pp


33/65

Application of First Law to Thermodynamics

Calc. of H when carbon burns to carbon monoxide It is difficult to burn C only to CO

kcal4157.26sC

kcal6361.67

kcal0518.94sC

kcal6361.67

kcal0518.94)()(

221

221

2

22

2221

22

HgCOgOHgOgCOgCO

HgCOgO

HgCOgOgCO

HgCOgOsC

Note the change in polarity

for the reverse reaction


34/65

Enthalpy of Formation

Standard enthalpy change for a reaction is equal to the sum of

the enthalpies of the products minus the sum of the

enthalpies of the reactants at the temperature concerned.

vi and vj are the stochiometric coefficients in the balanced

chemical equations for reactants and products, respectively.

The molar enthalpy for each element in its standard state is

given the value zero.

000reactiprodj HvHvH


35/65


For example the enthalpy of formation of CO2 is the enthalpy

change for the following reaction.

Since the enthalpy of combustion of hydrogen to form liquidwater at 298oK is68,317.4 cal mole-1, the enthalpy offormation of H2O(l) is68,317.4 cal mole

-1. The enthalpy offormation of gaseous water is less negative by the molar heatof vaporisation of water at 25oC, which is 10,519.5 cal mole-1.Therefore the enthalpy of formation of H2O(g) @ 25

oC is-57,797.9 cal mole-1

1-

298,

-1

22

molecal8.051,94soand

molecal8.051,94H

o

f

o

oH

gCOgOgraphiteC


36/65


When a substance cannot be formed directly in a

rapid reaction from its elements, the enthalpy

change for a series of suitable reactions may be

utilized in calculating the enthalpy of formation. Example: calculate the enthalpy change of H2SO4(l)

from the enthalpy change for the combustion of

sulfur to SO2, the oxidation of SO2 to SO3 using a

platinum catalyst, and the heat of solution of SO3 inwater to give H2SO4 at 25

oC


37/65


kcal193.91-2

kcal-68.32

kcal-31.14

kcal-23.49

kcal-70.96

0

4222

0

2221

2

04223

0

3221

2

0

22

fHlSOHgHgOsS

HlOHgOgH

HlSOHlOHgSO

HgSOgOgSO

HgSOgOsS


38/65


Consider the following

0

21

00

0

21

00000

0

,

0

,0

2221

22

222

2

OCOCO

OCCOOCCO

COfCOf

HHH

HHHHHH

HHH

COOCO


39/65


The enthalpy of formation of many compounds, ions,

and atoms are accurately know. From these

enthalpies of formation the enthalpy change for

many reactions may be calculated by use of theequation given.

The standard state of a solute in aqueous solution is

taken as the hypothetical ideal state of unit molality,

in which the enthalpy of the solute is the same as inthe infinitely dilute solution.


40/65


Enthalpy of formations at 25o

C, elements andinorganic compounds in kcal mole-1

H2O(l) -68.3174 CO2(g) -94.0158

H2O(g) -57.7979 H2S(g) -4.815

Br2(g) 7.34 HCl(g) -22.063

Fe2O3(s) -196.5 Fe3O4(s) -267.0

S

(monoclinic0.071

C(s,

diamond)0.4532


41/65


Enthalpy of formations at 25oC, in kcal mole-1, organic

compounds

Methane,

CH4(g)-17.889 Ethane C2H6(g) -20.236

Benzene,

C6H6(g)19.820

Benzene,

C6H6(l)11.718

Acetic acid,

CH3COOH(l)-116.4

Formic acid,

HCOOH(l)-97.8

Oxalic acid,

(COOH)2(s)-197.6

Acetylene,

C2H2(g)54.194


42/65



C, in kcal mole-1

, Ions in Water

H+ 0.000 OH- -54.957

Cl

-

-40.023 ClO4-

-31.41

CO32- -161.63 NO3

- -49.372

SO42- -216.90 PO4

3- -306.9

S2- 10.0 HS- -4.22

Na+ -57.279 Ca2+ -129.77


43/65



C, in kcal mole-1

, Solutes and Solution

NaOH(s)

in 100 H2O

in 200 H2O

in H2O

-101.99

-112.108

-112.1

-112.236

NaCl(s)

in 100 H2O

in 200 H2O

in H2O

-98.232

-97.250

-97.216

-97.302NaAc(s)

in 100 H2O

in 200 H2O

in H2O

-169.8

-173.827

-173.890

-174.122

HAc(l)

in 100 H2O

in 200 H2O

in H2O

-116.4

-116.705

-116.724

-116.743HCl(g)

in 100 H2O

in 200 H2O

in H2O

-22.063

-39.731

-39.798

-40.023

NaNO3(s)

in 100 H2O

in 200 H2O

in H2O

-111.54

-106.83

-106.73

-106.651

Heat

absorbed

solution

becomes

colder

Heat

evolved

solution

heats up


44/65

Heat of Solutions

Substance dissolving in water

Calculate the integral heat of solution of 1 mole of HCl(g) in

200H2O(l)

The reaction involves HCl gas dissolving in two hundred moles

of water to produce a solution of HCl (1 mole in 200 moles of

water)

kcal3517

0632279839

200in200

0

0

2200in0298

22

...HHH

OHHCllOHgHCl

gHClfOHHClf

,,


45/65

Heat of Dilution

Calculate the integral heat of dilution for the addition

of 195 moles of water to 1 mole of HCl in 5 moles of

water. The enthalpy of formation of HCl in 5 moles of

H2O is -37.37 kcal and the enthalpy of formation in

200 moles of water is -39.798 kcal.

kcal43.2

37.3779839

200in1955in

0

5

0

2000

298

222

22

.

HHH

OHHCllOHOHHCl

OHf, HCl inOHf, HCl in


46/65

Enthalpies of Formation of Ions

Since for strong electrolytes in dilute solution thethermal properties of the ions are essentially

independent of the accompanying ions, it is

convenient to use relative enthalpies of formation of

individual ions. The sum of the enthalpies offormation of H+ and OH- ions may be calculated.

kcal957.54

kcal317.68

kcal360.13

0

221

2

0

22212

0

2

HaqOHaqHgOgH

HlOHgOgH

HaqOHaqHlOH


47/65


The separate enthalpies of formation of H+ and OH-

cannot be calculated, but if the enthalpy offormation of H+(aq) is arbitrarily assigned the valuezero, it is possible to calculate relative enthalpies offormation for other ions. Denoting the electron by e,

the convention is that.

Therefore the enthalpy of formation of OH- is givenby

0)( 0221 HeaqHaqgH

kcal957.54 0221

221 HaqOHeaqgOgH


48/65


kcal023.40

kcal023.400

221

0

221

221

fHaqClaqgCl

HaqClaqHaqgClgH


49/65

Bond Energies

The bond energy E(A-B) is the contribution of the bond between a

particular pair of atoms A-B in a molecule to the total bindingenergy of the molecule. The total binding energy is required to

dissociate the molecule into atoms, each in its ground state. Bond

energies are usually given for a reference temperature of 298oC.

For a diatomic molecule the bond energy is the heat of dissociation

of the molecule.

In molecule of the type, A-Bn where all n bonds are identical, the

bond energy of the A-B bond is 1/n times the heat of dissociation of

into A + nB.

For polyatomic molecules having several types of bonds, it is notpossible to make precise assignment of bond energies, but in a

series of similar compounds bond energies for different bonds may

be assigned in such a way as to reproduce as much as possible the

heat of atomization of all compounds in the series.


50/65

Bond Energies

E(C-H) = 394/4 = 98 kcal. This may be compared with the energy of

dissociation of successive hydrogen atoms of methane.

kcal3944

kcal170

kcal20642

kcal182

04

0

0

2

0

24

HgHgCgCH

HgCgraphiteC

HgHgH

HgHgraphiteCgCH


51/65

Bond Energies

These bond-dissociation energies are all different because of the different

structure of the parent molecules. The sum of the four bond-dissociationenergies is equal to four times the bond energy.

Since the initial and final states are the same in the two cases.

The bond dissociation energies D are known only for a few polyatomicmolecules. When they are known they are more useful in kinetic andother calculations than the bond energies calculated from thermodynamicdata.

kcal81kcal125kcal87

kcal103

2

223

334

HCDgHgCgCH

HCHDgHgCHgCHHCHDgHgCHgCH

HCHDgHgCHgCH

HCEHCDHCHDHCHDHCHD 423


52/65

Bond Energies

Bond Energies E(A-B) (in kcal mole-1)

CC 80.5 OO 34

C

C 145 O

H 109.4CC 198 HH 103.2

CH 98.2 NN 37

CCl 78 NH 92.2

CO 79 ClCl 57.1


53/65

Bond Energies

The heat of a reaction may be estimated by adding the bond

energies for the bonds that are broken and subtracting fromthis total the bond energies for the new bonds formed in thereaction.

C2H4 + H2 = C2H6

HH bond is broken Ho = 103.2 kcal CC bond is broken Ho = 145 kcal

CC bond is formed Ho = -80.5 kcal

2(CH) bonds are formed Ho = -196.4 kcal

Ho = -29 kcal

Compared with experimental value of 32.73 kcal mole-1


54/65

Heat Adsorbed in Heating a Gas

Molar heat capacity of gases at constant pressure (in cal deg-1

mole-1

2cTbTaCp Gas a b x 103 C x 107

H2 6.9469 -0.1999 4.808

N2 6.4492 1.4125 -0.807

O2 6.0954 3.2533 -10.171

CO2 6.3957 10.1933 -35.333

H2O 7.1873 2.3733 2.084

NH3 6.189 7.887 -7.28

CH4 3.422 17.845 -41.65

Heat

Capacity is a

function of

temperature


55/65

Heat Absorbed in Heating a Gas

The molar heat capacities of simple gases can be

calculated on the basis of spectroscopic data more

accurately than can be determined directly,

especially at high temperature.

To calculate the heat adsorbed per mole at constantpressure when the temperature of a substance is

raised, the equation dH = CpdT

31323

2

1

2

2212

22

1

2

11

2

TTTTTTa

dTcTbTadTCHHH

cb

T

T

T

TPTT


56/65

Heat Absorbed in Heating a Gas

When the temperature change is small or the

variation of the heat capacity with temperature is

small, it is satisfactory to use the average heat

capacity in this temperature range.

12avg TTCH p

D d f th H t f R ti


57/65

Dependence of the Heat of Reaction on

Temperature

If we know the enthalpy of reaction at temperatureT1 we can calculate the temperature of reaction attemperature T2, provided the heat capacities of Aand B are known

The enthalpy change of reaction T2 is equal to thesum of three terms.

dTCT

TAP

1

2,

2

1,

T

TBP dTC

A B

A B

H2T2

H1

T1



58/65


Temperature

Since the enthalpy change is independent of the path

chosen between two states. It is convenient to write

this equation

2

1

1

2

,1,2

T

TBP

T

TAP dTCHdTCH

APBPP

T

TP

CCC

dTCHH

,,

12

where

2

1



59/65


Temperature

If the heat capacities of reactant and product are independentof temperature between T1 and T2, may be placed in

front of the integration sign and

Calculate the heat evolved in the freezing of water at constant

pressure and a temperature to -10oC.

H2O(l) = H2O(s)

Given H273 =-79.7 cal g-1, Cp,H2O.l = 1.00 cal deg-1 g-1, Cp,H2O,S =0.49 cal deg-1 g-1

PC

1212

TTCHH P

D d f h H f R i


60/65


Temperature

1-

273263

273263

22273263

gcal67415779

10510

1001490

273263

...

.

..

,,

HH

HH

CCHHlOHPsOHP

G l T t t f th Eff t f T t th


61/65

General Treatment of the Effect of Temperature on the

Enthalpy of Reaction

For a general chemical reaction

aA + bB = cC + dD

the enthalpy change is

The rate of change of H with temperature is obtained by

differentiating with respect to temperature at constant pressure

BADC HbHaHdHcH

PBPAPDPCP

P

PP

P

B

P

A

P

D

P

C

P

CCbCaCdCcdT

Hd

CdTHd

dT

Hdb

dT

Hda

dT

Hdd

dT

Hdc

dT

Hd

,,,,

thatseewe,thatgRememberin

G l T t t f th Eff t f T t


62/65

General Treatment of the Effect of Temperature

on the Enthalpy of Reaction

The previous equation may be stated in words as follows: the

change in enthalpy of reaction at constant pressure per

degree rise in temperature is equal to the change in heat

capacity at constant pressure of the system as the result of

the reaction. The equation may be integrated between two temperatures

T1 and T2 to obtain the relation between the enthalpy changes

at these two temperatures.

2

112

2

1

T

TP

H

H

dTCHHHd



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By use of this equation it is possible to calculate H for areaction at another temperature if it is known at one

temperature and if the values of CP are known in the

intervening temperature range. The empirical equations for

expressing as a function of temperature are very good for

this purpose.

Calculate the heat of combustion of hydrogen at 1500oK

PC

Heat Capacity of Gases


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Heat Capacity of Gases

@ Const. P in cal deg-1 mole-1

Gas a b x 103 c x 107

H2 6.9469 -0.1999 4.808

O2 6.0954 3.2533 -10.171

CO2 6.3957 10.1933 -35.333

H2O 7.1873 2.3733 2.084

NH3 6.189 7.877 -7.28CH4 3.422 17.845 -138.27

C2H6 1.375 41.852 -225.82

2CTbTaCP



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767,119

10723.4108931.16146.58.598,115

obtainto22

:equation

followingtheintodsubstitutearetablesfromcapacitiesheatmolarThecal8.598,115H22

0

1500

1500

298

2730

1500

1500

298

,,,0

298

0

1500

o

298222

222

H

dTTTH

dTCCCHH

gOHgOgH

OPOPOHP

The equation is only applicable if there is no change in

phase in going from T1 to T2.

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Lecture 1 ElectroChem and Corr