TOPIC 4: Thermodynamics

Chemistry for Biomedical Engineering. TOPIC 4: Thermodynamics Open Course Ware Universidad Carlos III de Madrid 2012/2013

Autors: Juan Baselga & María González

TOPIC 4: Thermodynamics

Key concepts Heat and work First law of thermodynamics Enthalpy Heat capacity Second law of thermodynamics: Entropy Third law of thermodynamics Free energy



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AIMS: 1. To learn how to calculate the absorbed or released heat of a chemical reaction at a reference temperature. 2. To learn how to calculate the absorbed or released heat of a chemical reaction at any temperature. 3. To learn how to predict if a given reaction takes place as it is written or in the opposite way. 4. To learn quantitative criteria to handle equilibrium reactions

Key concepts: 1. Thermodynamic state 2. Thermodynamic function 3. Reversible and irreversible processes 4. Chemical equilibrium



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Energy: the ability to do work Heat : is the form energy uses to be transferred from a system to other at lower temperature. Dynamics: time evolution of a physical system

System: A system is the specific portion of the universe that is being studied Environment: Everything outside the system

Universe = System + Surroundings (environment)

Thermochemistry is a part of thermodynamics: Thermochemistry is the scientific study of heat that is released or absorbed during chemical changes .

Thermodynamics is the part of physics that studies heat and its transformations

Hint: heat is not a form of energy, is an energy transference method

Key concepts



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Systems

Open systems

Closed systems

Isolated systems

Mass and energy (generally in the form of heat) can be exchanged with environment.

Energy can be transferred but not mass

Neither mass nor energy can be transferred

Isolated system

Closed system heat heat

Closed system heat heat

matter http://es.wikipedia.org/wiki/Agua_embotellada

http://creepypasta.wikia.com/wiki/File:Glass-‐of-‐water.jpg



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State variables

♣ These macroscopic properties are called thermodynamic variables, for example P, T, V, ...

Is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state. When you fix a small number of these, the rest becomes automatically fixed.

State Function:

♣ A system is in thermodynamic equilibrium when a set of macroscopic properties are fixed. Equilibrium State

♣ Thermodynamic variables can be classified as: •  intensive: independent of amount of matter: pressure, temperature •  extensive: depend on the amount of matter: mass, volume (dividing the system→ divides the variable)

♣ Some thermodynamic variables are called state functions (functions of state, state quantities or state variables)



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Thermodynamics distinguishes between two modes energy uses to be transferred:

• heat (q) • work (w)

The term work includes all the forms of energy except heat (mechanic, electric, magnetic, etc.).

Work is the way energy is transferred due to a mechanical link between system and surroundings. Heat is the way energy is transferred due to a temperature difference between the system and the surroundings

A system can do two kinds of work: expansion work (associated to a change of volume of the system) and nonexpansion (does not involve change in volume, i.e. a battery)

Heat and work



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r2

r1

fex

Assume a gas contained in a piston that expands against an external force fex DEFINITIONS w: work DONE BY THE ENVIRONMENT OVER THE SYSTEM

Expansion: the SYSTEM (GAS) MAKES WORK OVER THE SURROUNDINGS

VPVVAfrrA

Afrrfw ex

exexex Δ−=−−=−×−=−−= )()()( 121212

A

Let us consider the transformation from (P1,V1) to (P2,V2). There are several ways: a, b…

P1

P2

V1 V2

P1

P2

V1 V2

a) b)

VPwa Δ−= 1 VPwb Δ−= 2

wa ≠ wb so work is NOT a state function

For a general process where pressure P is not constant dVPw 21

VV ex∫−=

isobaric

isoc

horic



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Reversible process: TD functions differ infinitesimally in two successive instants Consider a gas inside a piston experiencing an expansion or a compression. The pressure inside will be Pint and the pressure exerted by surroundings Pext.

Reversible Expansion

dPPPdPPP

ex

ex

+=

−=

int

int

Reversible Compression

Irreversible Expansion

ex

ex

PPPP

<

>

int

int

Irreversible Compression

revintexirrev

intintexrev

wdVPdVPw

dVPdV)dPP(dVPw

=∫−<∫−=

∫∫ −≅±∫ −=−=

irrevrev ww >

The product between two infinitesimals is insignificant and can be disregarded

Reversible and irreversible processes

Let us calculate the expansion work done OVER the system for the two situations

More work is done in a reversible process



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In mechanics, total energy is the sum of kinetic and potential energy. These two forms of energy can be exchanged in such a way that total energy

remains constant: law of conservation energy This definition can be extended including an additional term that depends on the

nature of the system. Therefore, total energy depends on three contributions:

a)  Rest or motion state of the system b)  Position state of the system c)  Nature of the system

Internal Energy, E, of a system is that energy that only depends on the nature of the system and is independent on its position in any force field or its rest or motion state.

Internal Energy is the sum of the kinetic energy of the molecules or atoms of the

system (associated to translations, rotations and vibrations) and the potential energy associated to intermolecular forces.

First Law of thermodynamics



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Energy can neither be created nor destroyed: it can only be transformed from one state to another

To increase internal energy of a system, the environment must transfer heat and make work on it.

wqE +=Δ q= heat transferred TO THE SYSTEM (+)

w= work DONE OVER THE SYSTEM (+)

¿Is E a state function?

State 1

State 2 ΔEb

ΔEa Assume two states 1 and 2 with different internal energy. We can imagine at least two trajectories and let us assume that the energy change depends on trajectory. If ΔEb≠ΔEa then, for example, ΔEb>ΔEa. This means that the system returns more energy than we communicate on going from 1 to 2. → law of conservation energy is violated → YES, E is a state function .



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Sign criteria:

How is measured ΔE?

dVPqwqEV

V∫−=+=Δ2

1

If the process happens in a closed container at constant volume

VqqE =+=Δ 0

+w -w

+q -q

Surroundings

System



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Is the absorbed or released heat in a constant pressure process. (H)

pVEH += H is a state function

Its absolute value can not be

determined, only its change during a process. (ΔH)

PqVPVPqVPwqpVEH =Δ+Δ−+=Δ++=Δ+Δ=Δ )()(

Enthalpy

Heat Heat

System System

Surroundings Surroundings

ΔH > 0 Endothermic process

ΔH < 0 Exothermic process



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In processes where liquids or solids take part, change of volume is very small due to its inherent incompressibility

pVEH +=

EpVEH Δ≈Δ+Δ=Δ )(

In processes in which (ideal) gases take part and at constant temperature, ΔE and ΔH are very different if gases are produced or consumed.

nRTEnRTEpVEH Δ+Δ=Δ+Δ=Δ+Δ=Δ )()(



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Physical quantity that characterizes the amount of heat, q, required to change a body's temperature by a given amount, ΔT, : q = C· ΔT

Molar Heat Capacity Heat capacity per mol of substance (Units J/mol·K), intensive property.

q = m·C·ΔT

q = n·C·ΔT

It can be used for identification of substances.

Specific Heat Capacity Heat capacity per unit mass (units J/Kg·K)

Substance c (J/gºC) Water (l) 4,18 Cu(s) 0.382 Cl2(g) 0.478 Benzene 1,72 NaCl 0.866

Heat capacity



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dTdH

dT

dqC pp == ∫ Δ==∫==Δ 2

121

TT PP

TT pp TnCdTnCdTCnqH

V

vv

q0q)VP(qwqE

dTdE

dTdq

C

=+=

=Δ−+=+=Δ

==TnCdTnCdTCnqE v

TTv

TT vv

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Δ=∫=∫==Δ

V = const. TnCE vΔ=Δ

P = const. TnCH pΔ=Δ

Heat capacity at constant pressure CP or constant volume CV

Assuming that CP and CV are independent of T



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Relation between Cv and CP

pVEH += dTpVd

dTdE

dTdH )(

+= dTpVdCC VP)(

+=

Gases pV= nRT

Cp – Cv = nR

Solids and liquids

ΔV ≅ 0 Cp ≅ Cv

nRCdTpVdCC VVP +=+=)(



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Enthalpy and chemical reactions

Standard state: (refer to transparency 14 Topic 0) the most stable state at P = 1 atm and an specific temperature (usually 298,15 K)

Standard Molar Enthalpy of reaction, ΔH0 : change of enthalpy for a chemical reaction when reactants and products are in standard state, per mole of reaction as it is written.

Standard Molar Enthalpy of formation, ΔH0 f: change of enthalpy when one mole of a pure compound is formed from its elementary antecedents.

Chemical reaction: a process in which reactants are transformed in products.

C(graphite) + 1/2O2(g) → CO2(g) ΔH0f = -393,51 kJ/mol

H2(g) + 1/2O2(g) → H2O(l) ΔH0f = -285,83 kJ/mol

By definition, the standard molar enthalpy of formation of a pure element is cero.



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Some rules of thermochemistry:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

ΔH = -890.4 kJ/mol

CO2(g) + 2H2O(g) → CH4(g) + 2O2(g) ΔH = +890.4 kJ/mol

1- Enthalpy of reaction ΔH is proportional to the amount of reactants or products:

2- The value of ΔH in a reaction is equal and opposite to the value of ΔH for the reverse reaction:

2CH4(g) + 4O2(g) → 2CO2(g) + 4H2O(g)

ΔH = -2x890.4 = 1780.8 kJ



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)(2)()(2)( 2224 lOHgCOgOgCH +→+ ΔH = -890.4 kJ/mol

)()( 22 gOHlOH → ΔH = 40.7 kJ

)(2)()(2)( 2224 gOHgCOgOgCH +→+ ΔH = ? kJ

)(2)()(2)( 2224 lOHgCOgOgCH +→+ ΔH = -890.4 kJ/mol )(2)(2 22 gOHlOH → ΔH = 2x40.7 kJ

)(2)()(2)( 2224 gOHgCOgOgCH +→+ ΔH = -890.4 + 81.4 = -809 kJ/mol

3- The value of ΔH for a reaction is the same whether it occurs in one stage or in a series of stages ⇒ Hess law

Consider the following reactions

What is the enthalpy of reaction for the following?



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∑∑ Δ−Δ=Δi

ifii

ifireaction RHnPHnH )()( 000

4- For a given reaction, the reaction enthalpy ΔHºreaction can be calculated from

the enthalpy of formation of both the reactants ΔHºf(Ri) and the products ΔHº

f(Pi) according to

Example

)(2)(2)(2

)(2)()(2)(2)(

)()(2)(

)()(2)(2)(

)(2)()(

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lgg

lgggs

ggs

lggs

lgl

OHOH

OHCOHOC

COOC

HCOOHHOC

OHCOHCOOH

→+

+→++

→+

→++

+→ Enthalpy of combustion of formic acid

Enthalpy of formation of formic acid, carbon monoxide and water



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Example (cont.)

)(2)(2)( ggs HOC ++

)(2)( lg OHCO +)(lHCOOH

ΔH1 = ΔHºf (CO) + ΔHº

f (H2O ) - ΔHºf (HCOOH) = ΔH2

ΔH3

According to Hess law, ΔH3 = ΔH2 + ΔH1 ΔH3 = - ΔHº

f (HCOOH) + ΔHºf (CO) + ΔHº

f (H2O ) ΔH3 = ΔHº

f (CO) + ΔHºf (H2O ) - ΔHº

f (HCOOH) ΔH3 = Σ ΔHº

f (P) - Σ ΔHºf (R)



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Temperature dependence of enthalpy

dDcCbBaA +→+

( ) ( ) ( ) ( )][][)( 0,

0,

0,

0,1

01111BHbAHaDHdCHcTH TfTfTfTf Δ+Δ−Δ+Δ=Δ

Assume the following reaction Its reaction enthalpy at T = T1 es ΔH0(T1)

We want to calculate the reaction enthalpy at T = T2.

(aA + bB) at T1

(aA + bB) at T2

(cC + dD) at T1

(cC + dD) at T2

ΔH0 (T1)

ΔH0 (T2)

ΔH’(T1→T2 ) ΔH’’(T2→T1 )

To go from 1 to 2 we have two paths:

a)  Direct reaction at T1 b)  Heating the reagents at T2,

reacting at T2, cooling the products at T1. 1 2

ΔH0 (T1) = ΔH’(T1→T2 ) + ΔH0 (T2) + ΔH’’(T2→T1 )



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ΔH’ is the heating enthalpy of reagents

)(' 2

1

RCHT

T p∫=Δ

ΔH’’ is the cooling enthalpy of products

)(' 1

2

PCHT

T p∫=Δ

)()()( BbCAaCRC PPP +=

)()()()( 00000 BbCAaCDdCCcCC PPPPP −−+=Δ

)(''')( 20

10 THHHTH Δ+Δ+Δ=ΔTherefore

∫∫∫∫∫

Δ+Δ=−+Δ=

=−−Δ=Δ−Δ−Δ=Δ

2

1

2

1

2

1

1

2

2

1

01

01

0

10

10

20

)()()()(

)()()(''')()(T

T P

T

T P

T

T P

T

T P

T

T P

dTCTHdTRCdTPCTH

dTPCdTRCTHHHTHTH

Where

If CP of reagents and products do not differ too much, reaction enthalpy does not change appreciably with temperature

)()()( DdCCcCPC PPP +=



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Is a measure of the microscopic disorder within the system

It is a state function. In a process, the entropy change is: initialfinal SSS −=Δ

0>−=Δ initialfinal SSSIn all cases

Second Law of thermodynamics. Entropy

Liquid Solid Gas Liquid

Solid+liquid Solution T1 T2 (T2>T1)



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Reversible process: TD functions differ infinitesimally in two successive instants

irrevirrev

revrev

wqEwqE+=Δ

+=Δ

irrevrev qq >

irrevrev ww >

As state function internal energy does no depend on the path. So for two processes: reversible and irreversible

For an expansion, the work done reversibly is more negative (the system decreases its internal energy doing work) than irreversibly.

The more negative the work is, the more positive the heat must be to keep constant the internal energy variation

The heat absorbed by a system as it does reversible expansion work is higher than the heat absorbed irreversibly



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Wording

In an irreversible or spontaneous process, the entropy of Universe increases but remains constant in a reversible process or in equilibrium

0>Δ+Δ=Δ gssurroundinsystemuniverse SSSSpontaneous process

0=Δ+Δ=Δ gssurroundinsystemuniverse SSS Equilibrium process

0<Δ universeS Impossible process

http://en.wikipedia.org/wiki/File:Skier-‐carving-‐a-‐turn.jpg



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Let us assume: dDcCbBaA +→+

Standard reaction entropy ][][ 00000BADCreactionsystem bSaSdScSSS +−+=Δ=Δ

∑∑ −=Δ 000reactivesproductsreaction mSnSS

∫=Δ2

1 TdqS revDefinition

Temperature dependence of entropy

• Constant P process dqrev = dqP = dH = nCPdT

• Constant V process dqrev = dqV = dE = nCVdT

2

12

1

2

12

1

ln

ln

2

1

2

1

TTnCdT

TnC

TdqS

TTnCdT

TnC

TdqS

V

T

TVrev

P

T

TPrev

≈==Δ

≈==Δ

∫∫

∫∫ (constant P)

(constant V)

Reaction entropy

Absolute entropy can not be calculated (unless third law TD), only entropy changes.



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As a system approaches absolute zero, all processes cease and the entropy of the system approaches a minimum value. This minimum value, the residual entropy, is not necessarily zero, although it is almost always zero in a perfect, pure crystal

This allows defining absolute entropies

dTTCSSS

T PTT ∫==−

0

000

00

dTTGC

SdTTLC

SdTTSCS

T

TP

v

T

TP

f

T PT

b

b

f

f

∫

∫

∫

+

+Δ++

+Δ+=

)(

)(

)(

0

0

0

00

Third Law of Thermodynamics

Temperature

Ent

ropy

(S)

Entropy of vaporization

Entropy of fusion

Boiling point Melting point



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Absolute entropies , S0 (J·mol-1·K-1) at 298.15 K GASES LIQUIDS SOLIDS

C 158 C(g) 5.74 C2 199 C(d) 2.38 S 238 Srhombic 31.8

Smonoclinic 32.3 H2O 189 H2O 69.9 SO3 257 SO3 114 CO 198 CO2 214

Fe 27 Fe2O3 87



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Gibbs energy is the capacity of a system to do non-mechanical work and ΔG measures the non-mechanical work done on it. The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a closed system; this maximum can be attained only in a completely reversible process. Therefore, ΔG is the available energy for doing work.

TSHG −=

TERMOQUÍMICA

rev

rev

dqdqdG

qTdSdqdH

dT

SdTTdSdHdG

−=

=

=

=

−−=

PconstantatTconstantat0 If the process is reversible

dq =dqrev → dG = 0 Equilibrium If the process is irreversible q<qrev → dG < 0 (Spontaneous)

Mahan, page 360

Free Energy



Spontaneity in chemical reactions

STHG Δ−Δ=Δ

0<ΔG Spontaneous reaction

0>ΔG Non spontaneous reaction

0=ΔG Equilibrium reaction



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Standard free energy of formation ΔGf0

dDcCbBaA +→+For a general reaction :

000reactivesproductsreaction GGG Δ−Δ=Δ

[ ] [ ])()()()( 00000 BGbAGaDGdCGcG ffffreaction Δ+Δ−Δ+Δ=Δ

( ) ( )[ ] ( ) ( )[ ]000000000 )()()()( BfAfDfCfreaction STBHbSTAHaSTDHdSTCHcG Δ−Δ+Δ−Δ−Δ−Δ+Δ−Δ=Δ

Change of free energy when a mol of a compound in standard state is formed from its antecedent elements in standard state

000fff STHG Δ−Δ=Δ

By definition, the free energy of formation of elements in standard state is zero



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Relation between Free Energy and equilibrium

VdPdG

dTSdTVdPdG

qTdSdqdq

PdVdqdE

SdTTdSVdPPdVdEdGTSPVETSHG

rev

rev

=

=

−=

=

=

−=

−−++=

−+=−=

Tconstantat0

For a reversible process

For a change of pressure at constant T



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PnRTGnGn

PRTPPRTGG

PdPRTdG

PdPRTGd

VdPdG

P

P

G

G

ln~~lnln~~

~

0

00

1

~

~ 00

=−

==−

=

=

=

∫∫=

For 1 mole of ideal gas

Integrating between a standard state and a non standard state

For n moles of ideal gas



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QRTGPPPPRTGG

PbRTPaRTPdRTPcRTBGbAGaDGdCGcG

BGbAGaDGdCGcG

PdDPcCPbBPaA

bB

aA

dD

cC

BADC

DCBA

ln~ln~~

]lnlnlnln[)](~)(~)(~)(~[~

)(~)(~)(~)(~~)()()()(

00

0000

+Δ=+Δ=Δ

−−++

+−−+=Δ

−−+=Δ

+→+

Let us assume a general reaction between ideal gases each with a partial pressure Pi

We apply the equation found previously

Important equation since allows to calculate the free energy change for any arbitrary set of partial pressures



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QRTGG ln~~ 0 +Δ=Δ

Let us assume that our general reaction is at equilibrium. Partial pressures correspond to the equilibrium ones. Since initial and final states are equal the free energy change will be 0 and

The functional relation between equilibrium partial pressures is called Equilibrium Constant

P

mequilibriu

KRTG

QRTG

ln~ln~0

0

0

−=Δ

+Δ=

RTH~

RS~

RTG~

P

000

eeeKΔ

−Δ

+Δ

−==

- No units - Only depends on T - Reaction entropy and reaction enthalpy



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Temperature dependence of Equilibrium Constant

If the reaction takes place between gases, we can assume that the difference in heat capacities between reagents and products will be very small, so entropy change can be disregarded

⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ−=

Δ−≈

Δ+

Δ−=

=Δ

−Δ

+

12

0

2

1

000

~~

11~

)()(ln

~~~ln

00

TTRH

TKTK

RTH

RTS

RTHK

eeK

P

P

P

RTH

RTS

P

This equation allows to calculate the equilibrium constant at any T if reaction enthalpy is known and if K is also known at a given temperature. These equations for KP can be extended to reactions in which reactants and products are in liquid phase. We will not demonstrate it. No units, same temperature dependence.

TOPIC 4: Thermodynamics

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