Metallurgical Thermodynamics 2B - MTDMTB2
Dr BM Kanyane
Thermodynamics – Gibbs Free Energy
15 September 2022
TOPICS
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Matter: gases, liquids and solids
Heat capacity
Enthalpy & 1st law of thermodynamics
Carnot cycle & heat balances
Entropy & 2nd & 3rd law of thermo
Gibbs Free Energy & equilibrium constant
Feasibility of reactions
Equilibria & non-standard conditions
Ellingham diagram
Application to Metallurgical processes
GIBBS FREE ENERGY
GIBBS FREE ENERGY
Reactions that occur without the need for any input of energy are referred to as
spontaneous reactions. Spontaneous reactions can be exothermic, or endothermic.
Gibbs free energy is the maximum amount of free energy available to do useful work,
at constant temperature and pressure:
Gibbs free energy is equal to the enthalpy of the system minus the (temperature-entropy)
product, at constant temperature and pressure
G = H - TS
A = U – TS
Helmholtz energy @ constant volume and temperature
GIBBS FREE ENERGY
The change in Gibbs free energy (ΔG) predicts if a reaction will occur
spontaneously or be nonspontaneous.
ΔG = ΔH - TΔS
If ΔG > 0, the reaction is nonspontaneous in the direction written.
If ΔG = 0, the reaction is in a state of equilibrium.
If ΔG < 0, the reaction is spontaneous in the direction written.
ΔG° OF A REACTION
ΔG° is the standard Gibbs free energy change, if the reactants are present
in their standard state
ΔG° = ΔH° - TΔS°
ΔG°, for a reaction can be calculated from the standard free energies of formation, ΔG°f.
In the same way as the enthalpy change, the free energy change of a system
depends upon the initial and final states of the system,
The free energy change of a reaction can be added to, or subtracted from that of another reaction only
when both free energy data are given for the same conditions of temperature and pressure
In order to calculate the standard free energy change of a reaction, the standard free energies of the
reactants and products must be known.
However, as in the case of enthalpy, the absolute values of the free energies of substances are not
known, and only differences can be dealt with.
Therefore, free energies like enthalpies must have some reference point, with respect to which the
actual values for various substances can be calculated.
NOTE: The standard free energy of elements at 25°C and 1 atm pressure is zero
E.g. the standard free energy of formation of solid NiO at 25°C is -212480 J
It follows that the standard free energy change of the reaction is:
<Ni> + 0.5(02) = <NiO>
at 25° dG° = -212480 J
<NiO> = <Ni> + 0.5(02)
at 25° dG° = 212480 J
Notice the energy for the reverse reaction is positive
Calculate the standard free energy change of the reaction at 727°C (1,000 K) and 1 atm pressure from the
given data
dG° 1000K <MoO > = (-502080 J/mol).
3
dG° 1000K <H2O> = (-190372 J/mol).
<MoO3> + 3(H2) = <Mo> + 3(H20)
Since dG° of formation is equal to dG° of reaction, it follows that:
3
< 𝑀𝑜 > + (𝑂2 ) =< 𝑀𝑜𝑂3 >
dG°= −502080 J/mol
2
1
(𝐻2 ) + (𝑂2 ) =< 𝐻2 𝑂 >
dG°= −1902372 J/mol
2
If we reverse the Mo formation equation and multiply the H2O formation equation by 3 we have:
3
< 𝑀𝑜𝑂3 > = (𝑂2 ) +< 𝑀𝑜 >
2
3
3(𝐻2 ) + (𝑂2 ) = 3 < 𝐻2 𝑂 >
2
dG°= 502080 J/mol
dG°= 3 × [−1902372] J/mol
< 𝑴𝒐𝑶𝟑 > +𝟑(𝑯𝟐 ) = < 𝑴𝒐 > + 𝟑(𝑯𝟐 𝑶)
dG° = 502080 + (−5707116) = −5205036 J/mol
Given the following data, determine which metal has the greater probability of oxidation in steam
at 827°C (1100 K) and 1 atm pressure
<Ni0> + (H2) = <Ni> + (H20); dG° = (-2301 - 42.59 T J).
1
2
< 𝐶𝑟2 𝑂3 > + 𝐻2 = < 𝐶r > + 𝐻2 𝑂
3
3
dG° = (126566 − 30.67𝑇 𝐽
Solution:
Since we want the oxidation of metals the we have to reverse the given the given reactions and their
dG° signs
<Ni0> + (H2) = <Ni> + (H20)
<Ni> + (H20) = <Ni0> + (H2)
dG° = (+2301 + 42.59 T J).
1/3<Cr203> + (H2) = 2/3Cr> + (H20)
2<Cr> + 3(H20) = <Cr203> + 3(H2)
dG° = 3 × (−126566 + 30.67𝑇) 𝐽
NiO dG° = 49153.63 J/mol
Cr2O3 dG° = -278491.22 J/mol
It means NiO will not form because dG° is positive
Cr2O3 will form by the oxidation of Cr by steam since dG° is negative
Calculation of dG° at high temperature
Since the standard enthalpies of formation and standard entropies of compounds are usually
reported in the literature at 25°C, the standard enthalpy and entropy change, and hence standard
free energy change, of a reaction at 25°C can be easily calculated.
However, metallurgists are generally interested in calculating dG° and dG for reactions at high
temperatures. It is possible to calculate dG° of a reaction at a high temperature
∆𝐺 𝑜 = ∆𝐻 𝑜 − 𝑇∆𝑆 𝑜
𝑇
∆𝐺𝑇𝑜 =
𝑜
[∆𝐻298
+
𝑇
∆𝐶𝑝 𝑑𝑇]
298
𝑜
− [∆𝑆298
+
298
∆𝐶𝑝
𝑑𝑇
𝑇
This equation makes it possible to calculate dG° for a reaction at any temperature T from dH°298,
dS°298 and dCp values, for transformation the relevant enthalpies must be added
Calculate the standard free energy change of the reaction
at 327°C (600 K) from the following data:
1
< 𝑁𝑖 > + 𝑂2 = < 𝑁𝑖𝑂 >
2
𝑜
∆𝐻298
< 𝑁𝑖𝑂 > = (−240580 J/mol)
𝑜
∆𝑆298
< 𝑁𝑖 >= (29.79 J/K/mol)
𝑜
∆𝑆298
< 𝑂2 >= (205.09 J/K/mol)
𝑜
∆𝑆298
< 𝑁𝑖O>= 38.07 J/K/mol
𝐶𝑝 < 𝑁𝑖 >= (25.23 + 43.68 × 10−6 𝑇 2 −10.46 × 10−3 T J/K/mol)
𝐶𝑝 < 𝑂2 >= (29.96 + 4.18 × 10−3 𝑇−1.67 × 105 𝑇 −2 J/K/mol)
𝐶𝑝 < 𝑁𝑖𝑂 >= 54.01 J/K/mol)
1. Calculate dHo600 from dHo298 and dCp values
2. Calculate dSo600 from dSo298 and dCp values
3. Calculate dGo600 from 1 and 2 above
Answer:
[-185399.797] J/mol
The temperature plays an important role in determining the Gibbs free energy
and spontaneity of a reaction
Not all reactions occur at exactly 298 K
It can be helpful to determine the temperature when ΔG° = 0
∆𝐺 𝑜 = ∆𝐻 𝑜 − 𝑇∆𝑆 𝑜
when ΔG° = 0, …
∆𝐻 𝑜 = 𝑇∆𝑆 𝑜
∆𝐻 𝑜
𝑇=
∆𝑆 𝑜
Equilibrium temperature can be calculated form the above equation
Spontaneity and signs of dH & dS