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6.7 Heterogeneous Reaction
We will consider an equilibrium system with a gas phase and one or more condensed phases.
The mole fractions in the equilibrium relations refer to the mole fraction in a given phase, not
the total mole fraction. When the heterogeneous system is in equilibrium, there will be
chemical equilibrium in the gas phase and also phase equilibrium between species in the gas
phase and the pure condensed phases. When a pure condensed phase is present, the partial
pressure of that component in the gas phase will equal the vapor or saturation pressure of the
pure condensed phase. Hence a pure condensed phase cannot exist if the partial pressure of
that component is less than the vapor pressure.
The activity of a pure condensed phase is given by
a=
f
fo
In this definition, f is the fugacity of the pure solid in the equilibrium system and fo is the
fugacity of the pure condensed phase at 1 bar. For a pure substance at given T and P, we have
dμ = vdP − sdT
(6.7-1)
At constant temperature
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dμ = vdP = RTd(lnf)
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(6.7-2)
Integrating equation (6.7-2) from the standard pressure 1 bar to the pressure of the system
yields
f
RT ∫ d (lnf) =
fo
P
∫ vdP
1
⎛ f ⎞
RTln ⎜⎜ o ⎟⎟ = RTln a = v(P − 1)
⎝f ⎠
(6.7-3)
We have assumed incompressible for the volume of the condensed phase. The activity is
given by
⎡ v ( P − 1) ⎤
a = exp ⎢
⎣ RT ⎥⎦
(6.7-3)
Except for very high pressures the activity is close to unity. The state of the pure condensed
phase at equilibrium is not significantly different from the standard state.
Example 6.7-1 7 ---------------------------------------------------------------------------------Determine the activity of liquid water at 30oC and at the saturation pressure, 10 bar, and 100
bar.
Solution -----------------------------------------------------------------------------------------At 30oC, Psat = 0.0424 bar. We will assume the volume of liquid water remains constant at
the saturation temperature of 30oC.
v = 0.001004 m3/kg = (0.001004)(18.02) = 0.01809 m3/kmol
The activity of liquid water at 30oC and 0.0424 bar is
⎧ 0.01809(0.0424 − 1) × 105 ⎫
⎡ v ( P − 1) ⎤
a = exp ⎢
=
exp
⎨
⎬ = 0.9993
(8314.3)(303.15)
⎣ RT ⎥⎦
⎩
⎭
The activity of liquid water at 30oC and 10 bar is then
⎧ 0.01809(10 − 0.0424) × 105 ⎫
a = exp ⎨
⎬ = 1.0065
(8314.3)(303.15)
⎩
⎭
The activity of liquid water at 30oC and 100 bar is
7
Kyle, B.G., Chemical and Process Thermodynamics, Prentice Hall, 1999, pg. 257
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⎧ 0.01809(100 − 0.0424) × 105 ⎫
a = exp ⎨
⎬ = 1.0736
(8314.3)(303.15)
⎩
⎭
Hence the fugacity of a pure liquid varies little over a moderate pressure range.
Example 6.7-2 8 ---------------------------------------------------------------------------------Determine the equilibrium conversion for the isomerization reaction of methylcyclopentane
(CH3C5H9) to cyclohexane (C6H12) at 298oK. Gibbs energies of formation are given at 298oK
as: Δgof,CH3C5H9 = 31.72 [kJ/mol] and Δgof,C6H12 = 26.89 [kJ/mol]
Solution -----------------------------------------------------------------------------------------Determine the equilibrium constant at room temperature
∑ν Δg
Δg o rxn , 298 =
i
o
f , 298
Δg o rxn , 298 = (Δgof)C6H12 − (Δgof)CH3C5H9 = 26.89 − 31.72 = − 4.83 kJ/mol
The equilibrium constant is then
⎛ Δg o rxn, 298 ⎞
⎛
⎞
− 4,830
⎟⎟ = exp ⎜⎜ −
K = exp ⎜⎜ −
⎟⎟ = 7.0182
RT ⎠
⎝ (8.314)( 298.15) ⎠
⎝
The equilibrium constant is also given by the expression
ν
νi
i
⎛ fˆi ⎞
⎛xγ f ⎞
K = Π ⎜⎜ o ⎟⎟ = Π ⎜⎜ i io i ⎟⎟
⎝ fi ⎠
⎝ fi ⎠
Except for very high pressure the ratio
(E-1)
fi
⎡ v ( P − 1) ⎤
= exp ⎢
is close to unity. Equation (E-1)
o
fi
⎣ RT ⎥⎦
becomes
K = Π ( xi γ i ) i
ν
(E-2)
For ideal solution: K = Π( xi ) i =
ν
ζ=
8
xC6 H12
xCH 3C5H 9
=
ζ
1−ζ
= 7.0182
7.0182
= 0.8753 ⇒ 87.5% of the liquid exists as cyclohexane.
8.0182
Koretsky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 456
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Chapter 6
Example 6.7-3 9 ---------------------------------------------------------------------------------Calcium carbonate can dissociate according to the following reaction:
CaCO3(s) ⇔ CaO(s) + CO2(g)
Determine the equilibrium pressure for a closed system with pure CaCO3 in vacuum at
1000oK. Assume that the two solid phases are completely immiscible.
CaCO3(s)
− 951.25
Δgof,1000 (kJ/mol)
CaO(s)
− 531.09
CO2(g)
− 395.81
Solution -----------------------------------------------------------------------------------------The equilibrium constant K can be expressed in terms of the activities:
K=
⎛ fˆCaO ⎞⎛ fˆCO2
⎜ o ⎟⎜ o
⎜ f ⎟⎜ f
⎝ CaO ⎠⎝ CO2
=
⎛ fˆCaCO ⎞
3 ⎟
⎜
o
⎜ f CaCO
⎟
3 ⎠
⎝
aCaO aCO2
aCaCO3
⎞
⎟
⎟
⎠
(E-1)
Since we have pure solid phases:
aCaO =
f
fˆCaO
⎡ v ( P − 1) ⎤
= CaO
= exp ⎢
≈ 1 at low pressure
o
o
f CaO
f CaO
⎣ RT ⎥⎦
fˆCaCO3
aCaCO3 =
f
o
CaCO3
=
f CaCO3
f
o
CaCO3
⎡ v ( P − 1) ⎤
= exp ⎢
≈ 1 at low pressure
⎣ RT ⎥⎦
For the gas phase
a CO2 =
fˆCO2
f
o
CO2
=
yCO2 P
1 bar
= PCO2
⎛ Δg o rxn ,1000 ⎞
⎟⎟
Therefore K = PCO2 = = exp ⎜⎜ −
RT
⎝
⎠
The Gibbs energy of reaction is given by
9
Koretsky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 457
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Δg o rxn ,1000 =
∑ν Δg
i
o
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f ,1000
Δg o rxn ,1000 = (Δgof)CaO + (Δgof)CO2 − (Δgof)CaCO3
Δg o rxn ,1000 = (− 531.09)CaO + (− 395.81)CO2 − (− 951.25)CaCO3
Δg o rxn ,1000 = 24.35 kJ/mol
⎛ Δg o rxn ,1000 ⎞
⎛
⎞
24,350
⎟⎟ = exp ⎜⎜ −
K = exp ⎜⎜ −
⎟⎟ = 0.0535
RT
⎝ (8.314)( 298.15) ⎠
⎝
⎠
Therefore, calcium carbonate will dissociate at 1000oK until the pressure reaches 0.0535 bar
or until all the available CaCO3(s) reacted.