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Thermodynamics of Corrosion

Equilibrium constants

Basic thermodynamicrelationships

H, G, S, Cp

Gibbs free energy

Free energy of formation

Corrosion is the destructive attack of a metal by chemical

or electrochemical reaction with its environment.

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Chemical Thermodynamics

S = Entropy, the degree of disorder in a system

A chemical reaction is not favoured when

A chemical reaction is favoured when theentropy increases, i.e., S > 0

A chemical reaction is favoured when

A chemical reaction is not favoured when the

entropy decreases, i.e., S < 0

H = Enthalpy, the chemical heat in a reaction

G = H TS the Gibbs Free Energy, the chemical

energy of the system

G < 0, Spontaneous

G > 0, Not spontaneous

G = 0, Equilibrium

it requires heat, i.e., H > 0

it releases heat, i.e., H < 0

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Consider the generalized reaction:xM + yO2 MxO2y

Greac =

Gf(MxO2y) - (

xGf(M) +

yGf(O2))

= Gf(MxO2y)

Free Energy of Formation

Since the free energy is a state function, we can calculate

the change in free energy from the free energy of the initialand final states.

That is: 0 0

oxidation reaction

= GfStandard

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-(GT-H298)/T

Hf ,0 Hf,298 Gf,298 H298-H0 S298 Cp,298 500K 700K 1000K

kJ/mol kJ/mol kJ/mol kJ/mol J/mol/K J/mol/K J/mol/K J/mol/K J/mol/K

C(s, graphite) 0 0 0 1.05 5.74 8.53 6.887 9.063 12.636

O2 0 0 0 8.68 205.03 29.35 208.413 213.501 220.769

CO2 -393.14 -393.51 -394.36 9.363 213.64 37.11 218.187 225.287 235.806

Consider the reaction:

C(s, graphite) + O2(g) CO2(g)

We can look up the basic thermodynamic values for the

reactants and products,

EXAMPLE

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Element -(GT-H298)/T

Hf ,0 Hf,298 Gf,298 H298-H0 S298 Cp,298 500K 700K 1000K

kJ/mol kJ/mol kJ/mol kJ/mol J/mol/K J/mol/K J/mol/K J/mol/K J/mol/K

C(s, graphite) 0 0 0 1.05 5.74 8.53 6.887 9.063 12.636

O2 0 0 0 8.68 205.03 29.35 208.413 213.501 220.769

CO2 -393.14 -393.51 -394.36 9.363 213.64 37.11 218.187 225.287 235.806

From this table, we can extract the free energy of

reaction:

ELEMENT Gf / kJmol-1

C(s, graphite) 0.00

O2(g) 0.00

CO2(g) -394.36

C(s, graphite) + O2(g) CO2(g)

EXAMPLE

At room temperature, this reaction is spontaneous and

graphite is unstable in oxygen.

Greac = Gf(CO2) = -394.36 kJmol-1

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Temperature dependence?

The data are for the reaction at room temperature

The free energy values are determined for a

constant temperature and pressure

These are the normal conditions for mostchemical reactions studied

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Temperature dependence

Recall the definition of Gibbs free energy:

G = H - TS

Where H is the enthalpy, S is the entropy, and T is temperature.

dG = dH - SdT - TdS

Recall that: dH = CpdTand dS = (Cp/T) dT

dG = -SdT

Cp is the heat capacityat constant pressure

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Temperature dependence

G = H - TS

Starting from the definition of G,

Substituting S gives an expression to determine the

temperature dependence for the free energy

dG = -SdT

=dGdT T

G - H

Gibbs-Helmholtz Equation

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-(GT-H298)/T

Hf ,0 Hf,298 Gf,298 H298-H0 S298 Cp,298 500K 700K 1000K

kJ/mol kJ/mol kJ/mol kJ/mol J/mol/K J/mol/K J/mol/K J/mol/K J/mol/K

C(s, graphite) 0 0 0 1.05 5.74 8.53 6.887 9.063 12.636

O2 0 0 0 8.68 205.03 29.35 208.413 213.501 220.769

CO2 -393.14 -393.51 -394.36 9.363 213.64 37.11 218.187 225.287 235.806

Often, we want to know the thermodynamics of the reaction

at temperatures other than room temperature.

C(s, graphite) + O2(g) CO2(g)

To do this, we use the Gibbs-Helmholtz relationship,

and look up the basic thermodynamic values for the reactants

and products,

=

dGdT T

G - H= F The Free Energy Function-(GT-H298)/T

500K 700K 1000K

J/mol/K J/mol/K J/mol/K

6.887 9.063 12.636

208.413 213.501 220.769

218.187 225.287 235.806

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C(s, graphite) + O2(g) CO2(g)

ELEMENT -(Gf-H298)/T /

Jmol-1K-1

500 K

C(s, graphite) 6.887

O2(g) 208.413

CO2(g) 218.187

Freac = Ff(CO2) - (Ff(C) + Ff(O2))Once again, the value of the function (F) is given by:

So,

-(Gf - H298)/T = F

218.187 - (6.887 + 208.413)

2.887 Jmol-1K-1

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C(s, graphite) + O2(g) CO2(g)

-(Gr - H298)/T = 2.887 Jmol-1K-1 at 500K

Element -(GT-H298)/T

Hf ,0 Hf,298 Gf,298 H298-H0 S298 Cp,298 500K 700K 1000K

kJ/mol kJ/mol kJ/mol kJ/mol J/mol/K J/mol/K J/mol/K J/mol/K J/mol/K

C(s, graphite) 0 0 0 1.05 5.74 8.53 6.887 9.063 12.636

O2 0 0 0 8.68 205.03 29.35 208.413 213.501 220.769

CO2 -393.14 -393.51 -394.36 9.363 213.64 37.11 218.187 225.287 235.806

or:

Gr - H298 = -2.887 x 500 = -1443.5 Jmol-1

So:

= -1.4435Gr,500 + -393.51 = -394.95 kJmol-1

Hreac =

Hf(CO2) - (

Hf(C) +

Hf(O2))

-394.36 kJmol-1

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C(s, graphite) + O2(g) CO2(g) Summary

These calculations can be repeated to determinethe free energy at different temperatures.

=((GT-H298)/T)

C K -(GT-H298)/T Gr,T

J/mol/K J/mol kJ/mol

24.85 298 -394.36

226.85 500 2.887 1443.5 -394.95

426.85 700 2.723 1906.1 -395.42726.85 1000 2.401 2401.0 -395.91

Temperature

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=dGdT T

G - H

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Plot the free energy

when one mole ofgaseous oxygen at

1 atm pressure

combines with a pure

element to form oxide.

Ellingham diagrams

From A. Cottrell,

An Introduction to

Metallurgy, 1980.

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Thermodynamics of Corrosion

Equilibrium constants

Basic thermodynamicrelationships

H, G, S, Cp

Gibbs free energy

Free energy of formation

Corrosion is the destructive attack of a metal by chemical

or electrochemical reaction with its environment.

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Plot the free energy

when one mole ofgaseous oxygen at

1 atm pressure

combines with a pure

element to form oxide.

Ellingham diagrams

From A. Cottrell,

An Introduction to

Metallurgy, 1980.

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Ellingham diagrams

Free energy of

formation for the metaloxides have positive

slopes,

Free energy becomes

more positive withincreasing temperature

Smaller driving force

with increasing

temperature

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Ellingham diagrams

At a sufficiently high

temperature, the metal

oxide will release oxygenand exist as a pure

metal

M + xO2 MO2x

M + xO2 MO2x

Me

talStable

OxideSta

ble

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Ellingham diagrams

The lower the line, the

more stable the oxide

M + xO2 MO2x

Me

talStable

OxideSta

ble

=dGdT

-SKinks in thediagram indicatephase changes