Chemical Thermodynamics- Lectures 1 and 2

1

Thermodynamics (classical)

Description of chemical and physical bahaviour of matter and transformation between different forms of energies on a macroscopic scale (in terms of the bulk properties)

Statistical thermodynamics Based on Microscopic properties of speicies

2

What is What is thermodynamics ?thermodynamics ?

Will a particular chemical reaction take place at all (of course, at what speed, not in scope of thermodynamics)?

If thermodynamics predicts that a reaction cannot take place, can we some how make it to occur ?

How much energy we can extract from a reaction ? Or

How much energy we have to spend for carrying out a reaction ?

Why do chemical reactions or processes take place ?

Description of chemical and physical bahaviour of matter and transformation between different forms of energies on a macroscopic scale (in terms of the bulk properties)

3

Four Laws of ThermodynamicsFour Laws of Thermodynamics

Generalisations from experience Generalisations from experience (not mathematical theorems) (not mathematical theorems) So far no violationsSo far no violations

Zeroth lawZeroth law – – defines thermometrydefines thermometry First law First law – – Conservation of energyConservation of energy Second lawSecond law –– provides a concept of entropyprovides a concept of entropy Third LawThird Law – – gives a numerical value for entropygives a numerical value for entropy

4

System and SurroundingSystem and Surrounding – thermodynamic definition

5

Equilibrium between Equilibrium between (closed)(closed) System and Surrounding System and Surrounding

Psystem = Psurrounding

Mechanical Equilibrium Thermal Equilibrium

Insulated Insulated wallwall

Diathermic wallDiathermic wall

6

Hot ColdQ

Warm

Thermal Equilibrium

Zeroth Law of ThermodynamicsZeroth Law of Thermodynamics

Two systems that are separately in thermal Two systems that are separately in thermal equilibrium with a third system are also in equilibrium with a third system are also in thermal equilibrium with each other.thermal equilibrium with each other.

Examination of thermal Examination of thermal equilibrium between two equilibrium between two bodies possible without bodies possible without bringing them into contact.bringing them into contact.

HOW ?HOW ?

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Equilibrium within the systemEquilibrium within the system

P, T, and equal chemical potential of all P, T, and equal chemical potential of all species within the system is impliedspecies within the system is implied

When the system is under When the system is under (thermodynamic) (thermodynamic) equilibrium, all its properties equilibrium, all its properties are well defined.are well defined.

No fluxes.No fluxes.

How many parameters or variables are needed to define the system ?

8

Equilibrium within the systemEquilibrium within the system

P, T, and equal chemical potential of all P, T, and equal chemical potential of all species within the system is impliedspecies within the system is implied

Mass, position, velocity, etc of individual molecules are needed to define the microscopic state of a system

The macroscopic The macroscopic description of a system description of a system of ~10of ~1023 23 particles may particles may involve only a few involve only a few variables! variables!

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Macroscopic state of the system Macroscopic state of the system under equilibriumunder equilibrium

For simple systems of fixed quantity, defining For simple systems of fixed quantity, defining two of the three independent variables viz, T, P two of the three independent variables viz, T, P and V defines all the properties.and V defines all the properties.

Equation of stateEquation of state Relation between the Relation between the variablesvariables

Macroscopic state of a system can be specified only if it is in internal equilibrium

Ideal gases: Ideal gases: PV = n RTPV = n RTFirst “real-gas law” : (P+a/Vm

2) (Vm-b) = RT

defined when a minimum defined when a minimum number of variables are fixednumber of variables are fixed

““Simple systems”:Simple systems”:

Macroscopically homogeneous, Macroscopically homogeneous, isotropic, uncharged, large isotropic, uncharged, large enough that surface effects can enough that surface effects can be neglected, not acted upon by be neglected, not acted upon by (external) electric, magnetic, or (external) electric, magnetic, or gravitational fields.gravitational fields.

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How about equations of states for condensed phases ?How about equations of states for condensed phases ?

Molar volume, Vm = C1 + C2T + C3 T2 – C4P – C5 P T

At P ~ 1 bar, 4th and 5th terms are neglected.

The constants C1, C2 etc are obtained by fitting the experimental data to the expression.

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12

Atoms and molecules in a substance have energy in different forms:

Internal Internal energyenergy

1. Translational energies

2. Vibrational energies

3. Rotational energies

4. Electronic energies5. Attractive forces

between molecules (Lattice energies in case of solid, cohesive energies in case of liquids)

6. Rest energies (E = mc2)

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Internal energy, E = EInternal energy, E = Etran tran + E+ Erot rot + E + Evibvib + E + Eelcelc + E + E interinter + E + Erestrest

EEtran tran = = 3/2 RT for 3/2 RT for (ideal)(ideal) gases; gases; liquids also have translational motion; solidsliquids also have translational motion; solids

EErot rot = = RT (linearRT (linear ideal ideal molecules), 3/2 RT (non-linear molecules), 3/2 RT (non-linear idealideal molecules)molecules)

EEvibvib == nR nR 3N-63N-6∑∑s=1s=1 ((vib,svib,s) / [exp() / [exp(vib,svib,s/T) -1] /T) -1] wherewhere vib,svib,s = h = h ss /k /k

s s isis the frequency of normal mode of vibrationthe frequency of normal mode of vibration

A normal mode is a collective motion of all the atoms in the molecule where each atom in the molecule moves in phase with each other at a particular frequency. To a first approximation, the motion in a normal vibration can be described as a kind of simple harmonic motion

EEelcelc == n ( n ( eqeq - - ))

where where eq eq = energy of the molecule with nuclei at rest (no translation, rotation or vibration) = energy of the molecule with nuclei at rest (no translation, rotation or vibration) and and energy when all the atoms are at rest at infinite distance from each other energy when all the atoms are at rest at infinite distance from each other

E inter consists of cohesive forces for liquids, lattice energies for solidsconsists of cohesive forces for liquids, lattice energies for solids

EErestrest == molar rest mass of nuclei and electronsmolar rest mass of nuclei and electrons

Is it possible to determine absolute energy of the substance ? NO.

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Eproducts - Ereactants = E

Work done by the system taken to be +

Heat absorbed by system taken to be +

q = E + w

First law

It is not possible to determine absolute energy of the substance.But we can find the changes in internal energy changes in a process.

Heat flowing into the system =

WW

qq

E + Work done by the system

Energy can neither be created nor destroyed(in an isolated system)

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Types of WorkTypes of Work

Volume ExpansionVolume Expansionw = w = P Pextext dV dV (P = pressure)(P = pressure)

StretchingStretchingw = w = d d ll ( ( = tension, l = length) = tension, l = length)

Surface ExpansionSurface Expansionw = w = d d (( = surface tension, = surface tension, = surface area = surface area))

ElectricalElectricalw = w = dq dq (( = electrical potential) = electrical potential)

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Enthalpy, HEnthalpy, H

q = H + w’’

When PV work is separated out from other forms of work done,

q = E + w’ + w’’Where w’’ is other forms of work.

q = E + PV + w’’

Most of the chemical processes conducted under const. P conditions.

w’ = PV

q = (E + PV) + w’’

Volume Expansion:Volume Expansion:

w = w = P Pextext dV dVw = w = d (P d (PextextV)V)

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Enthalpy, HEnthalpy, H

Most reactions take place in an open vessel at constant pressure, pex. Volume can change during the reaction i.e. V 0 (expansion work).

Definition: H = qp i.e. heat supplied to the system at constant pressure

Enthalpy is the sum of internal energy and the product of PV of that substance.i.e H = E + PV (P = Pex)

Hi = Ei + PVi

Hf = Ef + PVf

Hf – Hi = Ef – Ei + P(Vf – Vi)or

H = E + P V

Some properties of H

work done by the system = Pex V H = (q - Pex V ) + P V

(Pex= P) H = ( - P V + q) + P V = q H = qp

(Constant P)q = E + w

Like internal energy, E, absolute value of H cannot be determined. Only H can be determined.

when pV work is the only work done

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Modes of Transferring Modes of Transferring EnergyEnergy

By doing work– Requires a macroscopic

displacement of the point of application of a force

By heat– Occurs by random molecular

collisions Results of both

– Change in internal energy of the system

– Generally accompanied by measurable macroscopic variables

• Pressure• Temperature• Volume

Similar processes when system receives energy

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When work is done, orderly motion is stimulated.

e.g. Atoms shown may be part of weight that is being raised.

Energy transfer as Work

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Heat is the transfer of energy that makes use of disorderly molecular motion.

Energy transfer as Heat

Molecular difference between heat and workMolecular difference between heat and work

21

Specific heats: Cp & Cv

dTCH 2

1

T

T P

dTCE 2

1

T

T V

Substance at T1 Substance at T2

Constant volume process

Constant pressure process

or

T

Vv

CTE

PP

CTH

Slope at a

given temp.

22

CCPP is always higher than C is always higher than CVV

Consider the slopes at a temp. T

What is the value of CWhat is the value of Cpp – C – Cvv ? ?

CCpp – C – Cvv = ( = (22 TV / TV / TT))

where where

= expansion coefficient of a material= expansion coefficient of a material

TT = isothermal compressibility= isothermal compressibility

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Pex pex

P

h

Pressure-volume workPressure-volume work

Work done by the system taken to be +

Heat absorbed by system taken to be +Work done by the system = pex × ∆V

Work done by the system = opposing force x distance moved Pressure = Force / areaw = h × (pex × A) = pex × h A

h x A = ∆V = (Vf – Vi)A = area of piston

pressure (P)

h is distance moved

h

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When the system is compressed ΔV is negative The work done by the system is negative

When the system is allowed to expand ΔV is positive The work done by the system is positive

When the volume remains constant No work is done on the system

W = P ΔV

If the pressure remains constant during the expansion or compression, the process is called an isobaric process

Work = Area under the curve

Work done by the system

W = P ΔV

Pressure-volume workPressure-volume work

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Work = Area under the curve

W = P ΔV

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PV PV DiagramsDiagrams

Used when the pressure and volume are known at each step of the process

The work done by a gas that takes it from initial state to final state is the area under the curve on the PV diagram

–This is true whether or not the pressure stays constant

The curve on the diagram is called the path taken between the initial and final states

The work done depends on the particular path– Same initial and final states, but different

amounts of work are done

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ExampleExamplework done by a gas in this cycle. 

Note: work is equal to the area

1212 VVppW P2

P1

V1 V2

28

Reversible ProcessesReversible Processes

Isothermal expansion: Pressure decrease by P at each step Each state of the system is in an equilibrium (Infinite time allowed for heat to flow to surroundings)

State variables reach equilibrium

29

Isothermal expansion of a Van der Waals gasIsothermal expansion of a Van der Waals gas

30

State Functions and Path Functions:State Functions and Path Functions:

P T isochore

Initia

l St

ate

Final

State

Common PathsCommon Paths

• Isochore:Isochore: V = constant (dV=0)V = constant (dV=0)

• Isobar:Isobar: p = constant (dp=0)p = constant (dp=0)

• Isotherm:Isotherm: T = constant (dT=0)T = constant (dT=0)

• Adiabat:Adiabat: q = 0q = 0

Combination of the above common pathsCombination of the above common paths

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q = E + w

Minimum energy required is E.

Any extra energy given to the system will be used to perform work (when energy is supplied as heat)

or reject as heat (when energy is supplied as work).

Efinal - Einitial = E

System at initial state System at final state

WW

qq

32

State Functions and Path Functions:State Functions and Path Functions:

q = E + w

– Same initial and final states (on an isotherm)but different amounts of work are done

• A A state functionstate function is a property of a system that is a property of a system that depends only on its current state and depends only on its current state and notnot on how that on how that state was reached.state was reached.P T

isochore

Initia

l St

ate

Final

State

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P T isochore

Initia

l St

ate

Final

State

Isothermal Processes for gases:Isothermal Processes for gases:

E is constant at a constant temperature for ideal gases ( PV = nRT = constant; ( PV) = 0)

H = E = 0 gasesrealforVE

T0

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Example:

Given:

n = 1 moleTi = 96.2 KTf = 144.3 KVi = 0.2 m3

Vf = 0.3 m3 P = const

Find:

W=?

J

mVVPVPW if

400

2.00.3mPa4000 33

1. Isobaric expansion:

Calculate work done by expanding gas of 1 mole if initial pressure is 4000 Pa, initial volume is 0.2 m3, and initial temperature is 96.2 K. Assume a two processes: (1) isobaric expansion to 0.3 m3, Tf=144.3 K (2) isothermal reversible expansion to 0.3 m3.

Also:

5.12.03.0

3

3

mm

VV

nRVP

nRVP

TT

i

f

ii

ff

i

f

A 50% increase in temperature!

i f

Vi Vf

35

Example:

Given:

n = 1 moleTi = 96.2 KVi = 0.2 m3

Vf = 0.3 m3 T = constFind: W=?

J3240.2m0.3mln0.2mPa4000

VVlnVP

VVnRTlnW

3

33

i

fii

i

f

2. Isothermal reversible expansion:

Calculate work done by expanding gas of 1 mole if initial pressure is 4000 Pa, initial volume is 0.2 m3, and initial temperature is 96.2 K. Assume a two processes: (1) isobaric expansion to 0.3 m3, Tf=144.3 K (2) isothermal expansion to 0.3 m3.

Also:

PammPa

VVPP

f

iif 2667

3.02.04000 3

3

A ~ 67% decrease in pressure!

i

f

V i V f

P f

P i

Vf

36

q = 0 dE= - w

for an ideal gas, CVdT = - pexdV

if reversible, pex = p = nRT/V

CV(dT/T) = - nR(dV/V)

CV ln (Tf/Ti) = - R ln (Vf/Vi)

qqw

Reversible Adiabatic ExpansionReversible Adiabatic Expansion

q = dE + w

f

iRC

i

f

VV

TT

v

iR

C

iif

RC

ff VVPn

RVR

VPnvv

37

v

P

v

v

CC

CRC

c1cγwhere

11

iVipfVfp

RCcwherec

iVcipc

fVcfp v

iR

C

iif

RC

ff VVPn

RVR

VPnvv

Reversible Adiabatic ExpansionReversible Adiabatic Expansion

TV TV (( -1)-1) = constant= constant

38

Adiabats vs. IsothermsAdiabats vs. Isotherms

39

State vs. Path FunctionState vs. Path Function

• A A state functionstate function is a property of a system that depends only on its is a property of a system that depends only on its current state and current state and notnot on how that state was reached. on how that state was reached.

• A A path functionpath function depends on how the state was reached.depends on how the state was reached.• Heat Heat and and work work are path functions.are path functions.

state A

state B

path 1

path 2

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0dE

0dH

A state function can be expressed as a function A state function can be expressed as a function of the two variables of the two variables

(in case of a fixed amount of one component)(in case of a fixed amount of one component)

e.g. e.g. P = (RT/V)P = (RT/V)

Its differential is also an exact differential.Its differential is also an exact differential.

41

In literature, E and U are interchangeably used for internal energy.

When the property Z of a system is determined by two independent variables x and y

[Fixed amount of one component ]

dyyZdx

xZdZ

xy

42

43

Exact and inexact differentialsExact and inexact differentials

q = dE + w

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Is there a function, Is there a function, ff = = ff((xx,,yy) such that ) such that

In other words, does a function In other words, does a function ff((xx,,yy) exist ) exist such that such that

Euler's test:Euler's test: A way to check whether such a function, A way to check whether such a function, ff((xx,,yy) exists. ) exists.

Based on the fact that for "nice" functions the Based on the fact that for "nice" functions the mixed second derivatives must be equal. That ismixed second derivatives must be equal. That is

df = g(x,y) dx + h(x,y) dydf = g(x,y) dx + h(x,y) dy

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So there exists a function, f(x,y) (actually x2y3) such that the following equations are equal:

The differential The differential dfdf given by this equation is called an "exact given by this equation is called an "exact differential" for the very reason that a function, differential" for the very reason that a function, f ( = xf ( = x22yy33)), exists such , exists such that this equation can be used to calculate it.that this equation can be used to calculate it.

46

Let's consider the differential:Let's consider the differential:

Is Is dgdg an exact differential ? an exact differential ? Use Euler's test to find out.Use Euler's test to find out.

The differential, The differential, dgdg, is not exact , is not exact and there does not exist a and there does not exist a function, function, gg((xx,,yy), such that ), such that dg(x,y) obeys the dg(x,y) obeys the Equation. Equation.

Both of the differentials, Both of the differentials, dfdf and and dgdg can be integrated from, say, can be integrated from, say, xx11, , yy11 to to xx22, , yy22

47

The differential The differential dgdg also can be also can be integrated, but there is no equivalent to integrated, but there is no equivalent to the above Equation for the integral of the above Equation for the integral of dgdg because there is no function, because there is no function, gg((xx,,yy).).

The integral of The integral of dgdg would have to be would have to be carried out along some path and we carried out along some path and we would find that the value of the would find that the value of the integral depends on the path as well integral depends on the path as well as on the initial and final points. as on the initial and final points.

Note: Sum of two inexact differentials can result in an exact differential.

aa bb dZ = dZ =

aa bb y dx y dx Depends on path a to bDepends on path a to b

yy

XX

II

IIIIaa

bb

aa bb dZ = dZ =

aa bb y dx + y dx +

aa bb x x

dydy II IIIIZ = XYZ = XY

dZ is dZ is exactexact

InexactInexact

48

When the differential of a physical quantity is inexact, When the differential of a physical quantity is inexact, it may be possibleit may be possible to define to define another physical quantity that has an exact differential by multiplying it by an a factoranother physical quantity that has an exact differential by multiplying it by an a factor ((integrating factorintegrating factor).).

dq = dE + dw dq = dE + dw = C= Cvv dT + PdV dT + PdV C Cv v dT + (RT/V) dV for an ideal gasdT + (RT/V) dV for an ideal gas

Is Is

Result:Result: dq is inexact.dq is inexact.

How about (dq / T)?How about (dq / T)?dq / T is exact differentialdq / T is exact differential

49

Cyclic ruleCyclic rule If Z is a function that depends on x and y (for example If Z is a function that depends on x and y (for example x, y and z may correspond to P, V and T)x, y and z may correspond to P, V and T)

Note how the variables in this product of partial derivatives appear in a cyclic order:

PT

V

50

0dE

0dH

A state function can be expressed as a function A state function can be expressed as a function of the two variables of the two variables

(in case of a fixed amount of one component)(in case of a fixed amount of one component)

e.g. e.g. P = (RT/V)P = (RT/V)

Its differential is also an exact differential.Its differential is also an exact differential.

51

Reference StatesReference StatesAll thermodynamic measurements are of differences between states; there is no absolute value for any property.

(ExceptionException:: entropy does have an absolute value from theory, but it’s the only one.)

In order to quantify thermodynamic properties, we choose by convention a reference state.

“Standard Ambient Temperature and Pressure (SATP)”.Pressure = 1 bar (105 Pa)Temperature = Any chosen T

Reference state of an elementReference state of an element is its most stable state at the specified temperature and 1 bar pressure.

Reference state for carbon at 25oC : Graphite

52

Standard Enthalpy of formation of a compoundStandard Enthalpy of formation of a compound

Standard state of a substanceStandard state of a substance (element or compound) is the pure substance at 1 bar and at the specified temperature.

Reference state of an elementReference state of an element is its most stable state at the specified temperature and 1 bar pressure.

Reference state for carbon at 25oC : Graphite

Standard enthalpy of formation of a substanceStandard enthalpy of formation of a substance, is the enthalpy change (per mole of the substance) for its formation at its standard state from the constituent elements from their reference states.

Standard state for carbon at 25oC : Graphite at 1 bar: Fullerene at 1 bar: Diamond at 1 bar

)(reactantsH(products)HHof

mm C (s, graphite)) + ½ O2(g) CO(g)

C (s, graphite)) C (s, diamond)

53

Standard enthalpy of formation of a substanceStandard enthalpy of formation of a substance, is the standard enthalpy change (per mole of the substance) for its formation from the constituent elements from their reference states.

)(reactantsH(products)HHof

mm

)(reactantsH(products)HHof

mm

Standard Enthalpy of formation of a compoundStandard Enthalpy of formation of a compound

Definition: Standard enthalpy of formation of the elements in their reference state is equal zero.

Stan

dard

Ent

halp

y of

form

atio

n,

ofH

CO(g) + ½ O2(g) CO2 (g)

C (s, graphite)) C (s, diamond)

C (s, graphite)) C (s, graphite)

54

Like internal energy, E, absolute value of H cannot be determined. Only H can be determined.

Kirchoff’s law”:Kirchoff’s law”: dTCP 2

112

T

T

oTf

oTf HH

55

Two conventions of defining enthalpy of elements:Two conventions of defining enthalpy of elements:

a.a. Enthalpy formation ofEnthalpy formation of an elementan element at its reference state is equal to zero.at its reference state is equal to zero.

HHoof, T f, T = 0= 0

b.b. Enthalpy ofEnthalpy of an elementan element at its reference state at 298K is zero:at its reference state at 298K is zero: H Hoo 298298 = 0 = 0

They are not contradictory:They are not contradictory:

Enthalpy formation ofEnthalpy formation of an elementan element at its reference state is equal to zero.at its reference state is equal to zero.

C (s) + OC (s) + O22(g) (g) CO CO22(g)(g)

C (s) + OC (s) + O22(g) (g) CO CO22(g)(g)

T = 298 K and P = 1 barT = 298 K and P = 1 bar

C (s) + OC (s) + O22(g) (g) CO CO22(g)(g)

T = T K and P = 1 barT = T K and P = 1 bar

How How HHoof,Tf,T (CO (CO22) and ) and HHoo

f,298 f,298 (CO(CO22) are related ?) are related ?

56

Enthalpy formation ofEnthalpy formation of an elementan element at its reference state is equal to zero.at its reference state is equal to zero.

C (s) + OC (s) + O22(g)(g)

T = 298 K and P = 1 barT = 298 K and P = 1 bar

C (s) + OC (s) + O22(g)(g)

T = T K and P = 1 barT = T K and P = 1 bar

C (s) + OC (s) + O22(g)(g)

COCO22(g)(g) COCO22(g)(g)

C (s) + OC (s) + O22(g)(g)

(1)(1)(2)(2)

(3)(3)

(4)(4)

(5)(5)

(6)(6)

RxnRxn

HHoof, 298 f, 298 (CO(CO22))

H(4) =H(4) = HHoof, 298 f, 298 <graphite> + <graphite> + HHoo

f,298f,298 (oxygen) (oxygen) = 0= 0

H(5) =H(5) =

H(6) = CH(6) = CPPCO2 CO2 * (T-298)* (T-298)

- - HHoof,298f,298<graphite> - <graphite> - HHoo

f,298f,298 (oxygen) (oxygen)

H(2) =H(2) = HHoof, T f, T <graphite> + <graphite> +

HHoof,Tf,T (oxygen) (oxygen) = 0= 0

H(3) = H(3) = HHoof, T f, T (CO(CO22) )

- - HHoof,Tf,T<graphite> <graphite>

- - HHoof,Tf,T(oxygen)(oxygen)

H(1) = CH(1) = CPPgrapgrap * (T-298) + C * (T-298) + CPP

O2O2 * (T-298) * (T-298)

H(Rxn) = CH(Rxn) = CPPgrapgrap * (T-298) + C * (T-298) + CPP

O2O2 * (T-298) * (T-298) + HHoo

f,Tf,T (CO (CO22) )

H(Rxn) = H(Rxn) = HHoof,298f,298 (CO (CO22)) + + CCPP

CO2 CO2 * (T-298)* (T-298)

HHoof,T f,T (CO(CO22)) = = HHoo

f,298f,298 (CO (CO22)) + CCPPCO2CO2 * (T-298) * (T-298)

- C- CPPgrapgrap * (T-298) * (T-298)

- C- CPPO2O2 * (T-298) * (T-298)

57

Enthalpy ofEnthalpy of an elementan element at its reference state at 298K is equal to zero.at its reference state at 298K is equal to zero.

C (s) + OC (s) + O22(g)(g)

T = 298 K and P = 1 barT = 298 K and P = 1 bar

C (s) + OC (s) + O22(g)(g)

T = T K and P = 1 barT = T K and P = 1 bar

COCO22(g)(g) COCO22(g)(g)

(1)(1)(2)(2)

(4)(4)

RxnRxn

H(3) = H(3) =

H(4) = CH(4) = CPPCO2 CO2 * (T-298)* (T-298)

H(2) =H(2) =

H(1) = CH(1) = CPPgrapgrap * (T-298) + C * (T-298) + CPP

O2O2 * (T-298) * (T-298)

HHooTT(Rxn) = C(Rxn) = CPP

grapgrap * (T-298) + C * (T-298) + CPPO2O2 * (T-298) * (T-298)

+ HHoof,Tf,T (CO (CO22) )

HHoof,T f,T (CO(CO22)) = = HHoo

f,298f,298 (CO (CO22)) + CCPPCO2CO2 * (T-298) * (T-298)

- C- CPPgrapgrap * (T-298) * (T-298)

- C- CPPO2O2 * (T-298) * (T-298)

(3)(3) HHoof, T f, T (CO(CO22) )

HHoof, 298 f, 298 (CO(CO22))

= H= Hoo298298(CO(CO22) – H) – Hoo

298298<C> - H<C> - Hoo298298(O(O22))

HHooTT(Rxn) = (Rxn) = HHoo

f,298f,298 (CO (CO22)) + + CCPPCO2 CO2 * (T-298)* (T-298)