Physical Chemistry for Energy Engineering (20th: 2018/11/26)
Transcript of Physical Chemistry for Energy Engineering (20th: 2018/11/26)
Physical Chemistry for Energy Engineering
(20th: 2018/11/26)
Takuji Oda
Associate Professor, Department of Nuclear EngineeringSeoul National University
*The class follows the text book: D.A. McQuarrie, J.D. Simon, โPhysical Chemistry: A Molecular Approach", University Science Books (1997).
Course schedule (as of Nov. 5)
29-Oct No lecture31-Oct
14 5-Nov 2. Phase equilibrium-115 7-Nov 2. Phase equilibrium-216 12-Nov 3. Chemical equilibrium-117 14-Nov Answers of homework-218 19-Nov Exam-02 (2 hour)19 21-Nov 3. Chemical equilibrium-220 26-Nov 3. Chemical equilibrium-321 28-Nov 3. Chemical equilibrium-422 3-Dec 3. Chemical kinetics-123 5-Dec 3. Chemical kinetics-224 10-Dec 3. Chemical kinetics-325 12-Dec Answers of homework-326 17-Dec Exam-03 (2 hour)
3.1. Chemical equilibrium for gases- $26-6: The sign of โ๐๐๐บ๐บ and not that of โ๐๐๐บ๐บ๐บ ๐๐ determines the direction of reaction
spontaneity -
(Example-6) Consider โ H2 (g) + ยฝ O2 (g) โ H2O (l) โ, for which โ๐๐๐บ๐บ๐บ = โ237 kJ ๐๐๐๐๐๐โ1 at 298.15 K.
In this case, โ๐๐๐บ๐บ๐บ has a large negative value, thus basically H2O (l) is much more stable than the reactants at 298.15 K. However, a mixture of H2 (g) and O2 (g) remains unchanged.
If a spark or a catalyst is introduced, then the reaction occurs explosively. The โnoโ is emphatic in thermodynamics: If thermodynamics insists that a
certain process will not occur spontaneously, then it will not occur. On the other hand, the โyesโ is actually โmaybeโ. The fact that a process will
occur spontaneously does not imply that it will occur at a detectable rate. Diamond remains its form, although a graphite is more favorable
energetically, is another example.
The speed of reaction can be analyzed in the framework of rate theory. (next topic)
3.1. Chemical equilibrium for gases- $26-6: The sign of โ๐๐๐บ๐บ and not that of โ๐๐๐บ๐บ๐บ ๐๐ determines the direction of reaction
spontaneity -
Theory of equilibrium is powerful, but it cannot indicate how long it take to reach the equilibrium state.Kinetics and rate theory can analyze and give information on the speed of a reaction.
Index
time
Spontaneous process
Equilibrium
Kinetics and rate theory are often based on thermodynamics, but sometimes extended to microscopic systems (via statistics mechanics/thermodynamics).
Sufficiently long time
Insufficient
Kinetics / rate theory
3.1. Chemical equilibrium for gases- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
We utilize the Gibbs-Helmholtz equation:
Substitute โ๐๐๐บ๐บ๐บ = โ๐ ๐ ๐๐ ln๐พ๐พ๐๐ (note this is a definition of โ๐๐๐บ๐บ๐บ, not a condition achieved at equilibrium state) for this equation:
โ๐๐โ๐บ๐บ๐บ ๐๐๐๐๐๐ ๐๐
= โโ๐ป๐ป๐บ๐๐2
๐๐ ln๐พ๐พ๐๐(๐๐)๐๐๐๐ ๐๐
=๐๐ ln๐พ๐พ๐๐(๐๐)
๐๐๐๐=โ๐๐๐ป๐ป๐บ๐ ๐ ๐๐2
This means that If โ๐๐๐ป๐ป๐บ > 0 (endothermic reaction), ๐พ๐พ๐๐(๐๐) increases with temperature. If โ๐๐๐ป๐ป๐บ < 0 (exothermic reaction), ๐พ๐พ๐๐(๐๐) decreases with temperature.
Integrate the equation: ln๐พ๐พ๐๐(๐๐2)๐พ๐พ๐๐(๐๐1)
= ๏ฟฝ๐๐1
๐๐2 โ๐๐๐ป๐ป๐บ(๐๐)๐ ๐ ๐๐2
๐๐๐๐
If the temperature range is small enough to consider โ๐๐๐ป๐ป๐บ constant:
ln๐พ๐พ๐๐(๐๐2)๐พ๐พ๐๐(๐๐1)
= โโ๐๐๐ป๐ป๐บ๐ ๐
1๐๐2โ
1๐๐1
3.1. Chemical equilibrium for gases- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
(Example-7) Consider โ PCl3 (g) + Cl2 (g) โ PCl2 (g) โ.Given that โ๐๐๐ป๐ป๐บ has an average value of -69.8 kJ mol-1 over 500-700 K and ๐พ๐พ๐๐ is 0.0408 at 500K, evaluate ๐พ๐พ๐๐ at 700K.
As โ๐๐๐ป๐ป๐บ can be assumed as a constant over the concerned temperatures,
ln๐พ๐พ๐๐(๐๐2)๐พ๐พ๐๐(๐๐1)
= โโ๐๐๐ป๐ป๐บ๐ ๐
1๐๐2โ
1๐๐1
Substituting provided values gives:
ln๐พ๐พ๐๐(700 ๐พ๐พ)๐พ๐พ๐๐(500 ๐พ๐พ)
= ln๐พ๐พ๐๐(700 ๐พ๐พ)
0.0408= โ
โ69.8 ร 103
๐ ๐ 1
700โ
1500
๐พ๐พ๐๐ 700 ๐พ๐พ = 3.36 ร 10โ4
Note that since the reaction is exothermic, ๐พ๐พ๐๐ 700 ๐พ๐พ is less than ๐พ๐พ๐๐ 500 ๐พ๐พ ; namely less product (PCl2) at higher temperatures as ๐พ๐พ๐๐ = ๏ฟฝ๐๐๐๐๐ถ๐ถ๐ถ๐ถ2
๐๐๐๐๐ถ๐ถ๐ถ๐ถ3๐๐๐ถ๐ถ๐ถ๐ถ2.
3.1. Chemical equilibrium for gases- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
ln๐พ๐พ๐๐(๐๐2)๐พ๐พ๐๐(๐๐1)
= ๏ฟฝ๐๐1
๐๐2 โ๐๐๐ป๐ป๐บ(๐๐)๐ ๐ ๐๐2
๐๐๐๐
Finally, we consider the temperature dependence of โ๐๐๐ป๐ป๐บ constant in:
This equation can write down as:
ln๐พ๐พ๐๐(๐๐) = ln๐พ๐พ๐๐(๐๐1) + ๏ฟฝ๐๐1
๐๐ โ๐๐๐ป๐ป๐บ(๐๐๐)๐ ๐ ๐๐๐2
๐๐๐๐๐
For example, as we learned, the temperature dependence of โ๐๐๐ป๐ป๐บ may be written as:
โ๐๐๐ป๐ป๐บ ๐๐2 = โ๐๐๐ป๐ป๐บ ๐๐1 + ๏ฟฝ๐๐1
๐๐2โ๐ถ๐ถ๐บ๐๐ ๐๐ ๐๐๐๐
or expanded in respect to temperature as:โ๐๐๐ป๐ป๐บ ๐๐ = ๐ผ๐ผ + ๐ฝ๐ฝ๐๐ + ๐พ๐พ๐๐2 + ๐ฟ๐ฟ๐๐3 + โฏ
In this latter case, ln๐พ๐พ๐๐(๐๐) becomes:
ln๐พ๐พ๐๐(๐๐) = โ๐ผ๐ผ๐ ๐ ๐๐
+๐ฝ๐ฝ๐ ๐
ln๐๐ +๐พ๐พ๐ ๐ ๐๐ +
๐ฟ๐ฟ2๐ ๐
๐๐2 + (๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐_๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐) โฆ
3.1. Chemical equilibrium for gases- $26-7: The variation of an equilibrium constant with temperature is given by the
Vanโt Hoff Equation -
ln๐พ๐พ๐๐(๐๐) = ln๐พ๐พ๐๐(๐๐1) + ๏ฟฝ๐๐1
๐๐ โ๐๐๐ป๐ป๐บ(๐๐๐)๐ ๐ ๐๐๐2
๐๐๐๐๐
โ๐๐๐ป๐ป๐บ ๐๐2 = โ๐๐๐ป๐ป๐บ ๐๐1 + ๏ฟฝ๐๐1
๐๐2โ๐ถ๐ถ๐บ๐๐ ๐๐ ๐๐๐๐
โ๐ถ๐ถ๐๐๐บ = ๐ถ๐ถ๐๐๐บ ๐๐๐ป๐ป3 ๐๐ โ ๏ฟฝ12๐ถ๐ถ๐๐๐บ ๐๐2 ๐๐ โ ๏ฟฝ3
2๐ถ๐ถ๐๐๐บ ๐ป๐ป2 ๐๐
Thus
ln๐พ๐พ๐๐(๐๐) = ln๐พ๐พ๐๐(725 ๐พ๐พ) + ๏ฟฝ725
๐๐ โ๐๐๐ป๐ป๐บ(๐๐๐)๐ ๐ ๐๐โฒ2
๐๐๐๐โฒ
= 12.06 + โ4583 ๐๐ โ 3.749 ln๐๐ + 1.857 ร 10โ3๐๐ โ 0.118 ร 10โ6๐๐2
(Example-8) Consider โ ยฝ N2 (g) + 3/2 H2 (g) โ NH3 (g) โ.The molar heat capacities of these species are give over 300-1500K as:
โ๐ถ๐ถ๐๐๐บ ๐๐2 ๐๐ ๐ฝ๐ฝ ๐พ๐พโ1 ๐๐๐๐๐๐โ1 = 24.98 + 5.912 ร 10โ3๐๐ โ 0.03376 ร 10โ6๐๐2โ๐ถ๐ถ๐๐๐บ ๐ป๐ป2 ๐๐ ๐ฝ๐ฝ ๐พ๐พโ1 ๐๐๐๐๐๐โ1 = 29.07 โ 0.8368 ร 10โ3๐๐ + 2.012 ร 10โ6๐๐2โ๐ถ๐ถ๐๐๐บ ๐๐๐ป๐ป3 ๐๐ ๐ฝ๐ฝ ๐พ๐พโ1 ๐๐๐๐๐๐โ1 = 25.93 + 32.58 ร 10โ3๐๐ โ 3.046 ร 10โ6๐๐2
Given that โ๐๐๐ป๐ป๐บ ๐๐๐ป๐ป3(๐๐) = โ46.11 ๐๐๐ฝ๐ฝ ๐๐๐๐๐๐โ1๐๐๐๐ 300 ๐พ๐พ ๐๐๐๐๐๐ ๐พ๐พ๐๐ = 6.55 ร 10โ3 at 725 K, derive the expression for the variation of ๐พ๐พ๐๐ ๐๐ with temperature.
โ๐๐๐ป๐ป๐บ ๐๐ = โ46.11 ร 10โ3 + ๏ฟฝ300
๐๐โ๐ถ๐ถ๐บ๐๐ ๐๐ ๐๐๐๐
Review of Chapter 3:Comparison between phase equilibrium and chemical equilibriumWe consider a system consisting of two phases of a pure substance (1-component) in equilibrium each other. (liquid and gas, for example here) under const.-T and const.-P condition..
The Gibbs energy of this system is given by๐บ๐บ = ๐บ๐บ๐๐ + ๐บ๐บ๐๐
where ๐บ๐บ๐๐ and ๐บ๐บ๐๐ are the Gibbs energies of the liquid and the gas phase.
Now, if the liquid phase increases with ๐๐๐๐๐๐ mole (thus, the solid phase decreases with ๐๐๐๐๐๐ (๐๐๐๐๐๐ = โ๐๐๐๐๐๐)), the Gibbs energy change is:
๐๐๐บ๐บ = ๐๐๐บ๐บ๐๐ + ๐๐๐บ๐บ๐๐ = ๐๐๐บ๐บ๐๐
๐๐๐๐๐๐ ๐๐,๐๐๐๐๐๐๐๐ + ๐๐๐บ๐บ๐ถ๐ถ
๐๐๐๐๐ถ๐ถ ๐๐,๐๐๐๐๐๐๐๐ = ๐๐๐บ๐บ๐๐
๐๐๐๐๐๐ ๐๐,๐๐โ ๐๐๐บ๐บ๐ถ๐ถ
๐๐๐๐๐ถ๐ถ ๐๐,๐๐๐๐๐๐๐๐
Here, we define chemical potentials, ๐๐๐๐ = ๐๐๐บ๐บ๐๐
๐๐๐๐๐๐ ๐๐,๐๐and ๐๐๐๐ = ๐๐๐บ๐บ๐ถ๐ถ
๐๐๐๐๐ถ๐ถ ๐๐,๐๐, then
๐๐๐บ๐บ = ๐๐๐๐ โ ๐๐๐๐ ๐๐๐๐๐๐ (constant T and P)As ๐๐๐บ๐บ = 0 is the condition for an equilibrium state, ๐๐๐๐ = ๐๐๐๐ is the condition.
As the condition is cont.-T const.-P, the equilibrium system has the minimum Gibbs energy. This is the ground rule. Here, we consider conditions with which a two-phase system (liquid & gas here) as the minimum Gibbs energy.
Review of Chapter 3:Comparison between phase equilibrium and chemical equilibrium
http://cft.fis.uc.pt/eef/FisicaI01/fluids/thermo20.htm
(state-1) 300 K, 1 atm
(1)
Phase diagram of water
The phase diagram indicates that โat 300 K and 1 atm, the equilibrium phase of water is liquid water.โ
It means โif water is at 300 K and 1 atm (and with a fixed amount), liquid phase has a smaller Gibbs energy than solid and gas phases.โ
[Quiz] If there is a liquid water in this class room (assuming 300 K), do you think there are some ice (liquid) or vapor water (gas) coexisting? Why do you think so and what is the mechanism of it?
Review of Chapter 3:Comparison between phase equilibrium and chemical equilibrium
๐๐๐บ๐บ๐๐๐๐ ๐๐
= โ๐๐๐๐๐บ๐บ๐๐๐๐ ๐๐
= ๐๐[J]/[K]=[J/K] [J]/[Pa]=[m3]
Substance ๏ฟฝ๐๐ (m3/mol) S๐บ (J/mol/K)H2O l 0.018E-3 250.5
H2O g 22.4E-3 188.8
Typical pressure changes in chemistry are โ1 atm โ 0.5 atmโ, โ1atm โ 10 atmโ, for example, (โ1E5 Pa โ 0.5E5 Paโ, โ1E5 Pa โ 1E6 Paโ).
Typical temperature changes are such as โ300 K โ 600 Kโ, etc. Thus, the difference between typical pressure and typical temperature (the
their changes) are 3-4 orders of magnituede. For gas, the difference between ๏ฟฝ๐๐ and S๐บ are 4 orders of magnitude. Thus,
pressure change affect the Gibbs energy of gas in a comparable scale with temperature change.
For liquid (solid), on the other hand, the difference between ๏ฟฝ๐๐ and S๐บ are 7orders of magnitude. Thus, the Gibbs energy is hardly affected by pressure.
Review of Chapter 3:Comparison between phase equilibrium and chemical equilibrium
H2O(l) H2O(l)
H2O (g)
300 K, V0 m3
(a Pa)
๐ด๐ด0 ๐ด๐ด1
๐๐๐๐ 300 ๐พ๐พ,๐๐= ๐๐๐๐ (300 ๐พ๐พ, ๐๐)
300 K, V0 m3
(0 Pa)
H2O(s)
300 K, V0 m3
(0 Pa)
๐ด๐ด2
๐๐๐๐ 300๐พ๐พ, 0๐๐๐๐< ๐๐๐ ๐ (300๐พ๐พ, 0๐๐๐๐)
๐จ๐จ๐๐ > ๐จ๐จ๐๐๐จ๐จ๐๐ > ๐จ๐จ๐๐
Const.-T const.-V condition
H2O(l)
Initial state(a non-equilibrium state)
State-1(a non-equilibrium state)
State-2(the equilibrium state)
Review of Chapter 3:Comparison between phase equilibrium and chemical equilibrium
๐บ๐บ0๐ฎ๐ฎ๐๐ > ๐ฎ๐ฎ๐๐
Const.-T const.-P condition
H2O(l)
300 K, 3E3 PaH2O(g)
300 K, 1E5 Pa
H2O(s)
๐บ๐บ1
๐บ๐บ2
๐ฎ๐ฎ๐๐ > ๐ฎ๐ฎ๐๐
H2O(g)
H2O(l)
H2O(g)
H2O(l)
๐บ๐บ3
๐บ๐บ4
๐ฎ๐ฎ๐๐ > ๐ฎ๐ฎ๐๐
๐ฎ๐ฎ๐๐ > ๐ฎ๐ฎ๐๐
Indeed, ๐ฎ๐ฎ๐๐ = ๐ฎ๐ฎ๐๐ if the states where 2 coexisting phases exist are equilibrium states at this const.-T const.-P.
In this case, ๐๐๐๐ 300 ๐พ๐พ, 3๐ธ๐ธ3 ๐๐๐๐ = ๐๐๐๐ (300 ๐พ๐พ, 3๐ธ๐ธ3 ๐๐๐๐)
H2O(l)
Review of Chapter 3:Comparison between phase equilibrium and chemical equilibrium
The saturation pressure is the pressure where chemical potential of liquid and gas phases become identical at a given temperature.
๐๐๐๐ ๐๐ ๐พ๐พ,๐๐๐ ๐ ๐ ๐ ๐ ๐ ๐๐๐๐ = ๐๐๐๐ (๐๐ ๐พ๐พ,๐๐๐ ๐ ๐ ๐ ๐ ๐ ๐๐๐๐) For practice, it may also be defined as ๐๐๐๐ ๐๐ ๐พ๐พ,๐๐๐ ๐ ๐ ๐ ๐ ๐ ๐๐๐๐ = ๐๐๐๐ ๐๐ ๐พ๐พ, 1 ๐๐๐๐๐๐ ~๐๐๐๐ ๐๐ ๐พ๐พ,๐๐๐ ๐ ๐ ๐ ๐ ๐ ๐๐๐๐๐๐ This equation is more useful when we want to
know the saturation pressure of water vapor in a class room, for example.
Note that (molar) Gibbs energy of water vapor depends on partial pressure of water vapor, not on the total pressure of system. (assuming ideal gas)
<Saturation pressure of water vapor>
So, even when the total pressure of the system is fixed, we may change the Gibbs energy (of the system) by adjusting partial pressures. For example, if we initially have 1 mol of H2O(g), a state where H2O(g),
H2(g) and O2(g) coexist may have the minimum Gibbs energy under const.-P const.-T condition.
0
20000
40000
60000
80000
100000
260 280 300 320
Satu
ratio
n pr
essu
re (
Pa)
Temperature (K)
3.0. Pressure dependence of chemical potential--๐๐ ๐ป๐ป,๐ท๐ท = ๐๐๐บ ๐ป๐ป + ๐น๐น๐ป๐ป ๐๐๐๐ โ๐ท๐ท๐๐ ๐ท๐ท๐บ for pure substance of an ideal gas -
Assuming const.-T condition, and dividing both sides with dP
For reversible processes (thus for thermodynamical equilibrium states)๐๐๐บ๐บ = ๐๐๐๐๐๐ โ ๐๐๐๐๐๐
๐๐ ๐๐,๐๐ = ๐๐๐บ ๐๐ + ๐ ๐ ๐๐ ln โ๐๐ ๐๐๐บ
We consider the pressure dependence of gas of pure substance (single-component).
๐๐๐บ๐บ๐๐๐๐ ๐๐
= ๐๐
Then, keeping T and P constant, taking derivative regarding n
๐๐๐๐๐๐
๐๐๐บ๐บ๐๐๐๐ ๐๐ ๐๐,๐๐
=๐๐๐๐๐๐๐๐ ๐๐,๐๐
๐๐๐๐๐๐
๐๐๐บ๐บ๐๐๐๐ ๐๐ ๐๐,๐๐
=๐๐๐๐๐๐
๐๐๐บ๐บ๐๐๐๐ ๐๐,๐๐ ๐๐
=๐๐๐๐๐๐๐๐ ๐๐
Note that up to here ideal gas is not assumed.
Here, we assume ideal gas. Then, using ๐๐๐๐๐๐๐๐ ๐๐,๐๐
= ๐ ๐ ๐๐๐๐
๐๐ ๐๐,๐๐ = ๐๐๐บ ๐๐ + ๐ ๐ ๐๐ ln โ๐๐ ๐๐๐บ
We consider the pressure dependence of gas of pure substance (single-component).
๐๐๐๐๐๐๐๐ ๐๐
=๐๐๐๐๐๐๐๐ ๐๐,๐๐
๐๐๐๐ = ๐ ๐ ๐๐๐๐๐๐๐๐ (const.-T)
Taking the integral from the standard pressure (๐๐๐บ = 1 ๐๐๐๐๐๐) to an arbitrary pressure ๐๐,
๐๐ โ ๐๐๐บ = ๐ ๐ ๐๐ ln๐๐๐๐๐บ
Be careful that chemical potential ๐๐ is molar Gibbs energy for pure substance, but it is not the case when multiple substances are mixed.
3.0. Pressure dependence of chemical potential--๐๐ ๐ป๐ป,๐ท๐ท = ๐๐๐บ ๐ป๐ป + ๐น๐น๐ป๐ป ๐๐๐๐ โ๐ท๐ท๐๐ ๐ท๐ท๐บ for pure substance of an ideal gas -
3.1. Chemical equilibrium for gases-Difference between ๐๐๐ฎ๐ฎ
๐๐๐๐๐จ๐จ ๐ป๐ป,๐ท๐ท,๐๐๐ฉ๐ฉ,๐๐๐๐,๐๐๐๐and ๐๐๐ฎ๐ฎ
๐๐๐๐ ๐ป๐ป,๐ท๐ท-
These two partial derivative are both definitions of chemical potential, but for
different conditions: ๐๐๐บ๐บ๐๐๐๐๐๐ ๐๐,๐๐,๐๐๐๐,๐๐๐ด๐ด,๐๐๐ต๐ต
is for a mixture (A,B,Y,Z) and ๐๐๐บ๐บ๐๐๐๐ ๐๐,๐๐
for
pure substance. For example, ๐๐๐บ๐บ
๐๐๐๐ ๐๐,๐๐is the change in Gibbs energy for the pure gas Y
when an infinitesimal amount of gas Y (๐๐๐๐) is added keeping other
parameters constant. In this case, ๐๐๐บ๐บ๐๐๐๐๐๐ ๐๐,๐๐
= ๐บ๐บ๐๐๐๐
, because this process
only change the amount of Y with fixing other thermodynamical conditions. Thus, chemical potential is equal to molar Gibbs energy (const.-T and const.-P) for pure substance.
๐๐๐บ๐บ๐๐๐๐๐๐ ๐๐,๐๐,๐๐๐๐,๐๐๐ด๐ด,๐๐๐ต๐ต
is the change in Gibbs energy for the gas mixture when
an infinitesimal amount of gas Y (๐๐๐๐๐๐) is added keeping other parameters constant. The chemical potential is equal to partial molar Gibbs energy(const.-T and const.-P [total P]) for a mixture.
These two partial derivative are both definitions of chemical potential, but for
different conditions: ๐๐๐บ๐บ๐๐๐๐๐๐ ๐๐,๐๐,๐๐๐๐,๐๐๐ด๐ด,๐๐๐ต๐ต
is for a mixture (A,B,Y,Z) and ๐๐๐บ๐บ๐๐๐๐ ๐๐,๐๐
for
pure substance. (1) For example, if [T= 300 K and P= 1 atm, n=1 mol], โ ๐๐๐บ๐บ
๐๐๐๐ ๐๐,๐๐๐๐๐๐โ
corresponds to the Gibbs energy change by increasing ๐๐๐๐ mol of the substance.
(2) On the other hand, if [T= 300 K, P= 1 atm, n=1 mol],
โ ๐๐๐บ๐บ๐๐๐๐๐๐ ๐๐,๐๐,๐๐๐๐,๐๐๐ด๐ด,๐๐๐ต๐ต
๐๐๐๐๐๐โ corresponds to the Gibbs energy change by
increasing ๐๐๐๐๐๐ mol of Y. In this case, the partial pressure of the gas is not 1 atm, but ๐๐๐๐
๐๐๐๐+๐๐๐๐+๐๐๐ด๐ด+๐๐๐ต๐ตatm. As the Gibbs energy of gas largely depends on
the partial pressure, the Gibbs energy change should be different from that of (1).
How can we define Gibbs energy of gas mixture?
3.1. Chemical equilibrium for gases-Difference between ๐๐๐ฎ๐ฎ
๐๐๐๐๐จ๐จ ๐ป๐ป,๐ท๐ท,๐๐๐ฉ๐ฉ,๐๐๐๐,๐๐๐๐and ๐๐๐ฎ๐ฎ
๐๐๐๐ ๐ป๐ป,๐ท๐ท-
3.1. Chemical equilibrium for gases-Gibbs energy in gas mixture (ideal gas)-
T0 K, V0 m3, P0 PaA(g): na molB(g): nb mol๐๐๐ด๐ด = ๐๐0
๐๐๐ด๐ด๐๐๐ด๐ด+๐๐๐ต๐ต
๐๐๐ต๐ต = ๐๐0๐๐๐ต๐ต
๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, V0 m3, PB Pa
B(g): nb mol
๐๐๐ต๐ต = ๐๐0๐๐๐ต๐ต
๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, V0 m3, PA Pa
A(g): na mol
P๐๐๐ด๐ด = ๐๐0๐๐๐ด๐ด
๐๐๐ด๐ด+๐๐๐ต๐ต
The 1st law: ๐๐๐๐ = ๐๐๐๐๐๐ โ ๐๐๐๐๐๐As const.-T const.-V condition, โ๐๐ = โ๐๐โ๐๐If the gases are ideal gases, because const.-T is assumed, โ๐๐ = โ๐๐ ๐๐ = 0, hence, โ๐๐ = 0.In addition, for ideal gas, โ๐ป๐ป = โ๐ป๐ป ๐๐ = 0. Hence, โ๐บ๐บ = โ ๐ป๐ป โ ๐๐๐๐ = 0.This result means that the Gibbs energy of mixed gas can be defined as the summation of Gibbs energies of constituent pure gasses holding corresponding partial pressure (for ideal gas).
๐๐๐ด๐ด =๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐,๐๐๐ต๐ต
=๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐๐ด๐ด
= ๐๐๐บ๐ด๐ด ๐๐ + ๐ ๐ ๐๐ ln โ๐๐๐ด๐ด ๐๐๐บ
Mixture Pure substance
T0 K, V0 m3, P0 Pa
A(g): na molB(g): nb mol
๐๐๐ด๐ด = ๐๐0๐๐๐ด๐ด
๐๐๐ด๐ด+๐๐๐ต๐ต๐๐๐ต๐ต = ๐๐0
๐๐๐ต๐ต๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, V0 m3, PB Pa
A(g): na molB(g): nb mol
๐๐๐ต๐ต = ๐๐0๐๐๐ต๐ต
๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, V0 m3, PA Pa
A(g): na mol
P๐๐๐ด๐ด = ๐๐0๐๐๐ด๐ด
๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, V0 m3, P0 Pa
A(g): na molB(g): nb mol
๐๐๐ด๐ด = ๐๐0๐๐๐ด๐ด
๐๐๐ด๐ด+๐๐๐ต๐ต๐๐๐ต๐ต = ๐๐0
๐๐๐ต๐ต๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, VBm3, P0 Pa
B(g): nbmol
๐๐๐ต๐ต =๐๐0
๐๐๐ต๐ต๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, VAm3, P0 Pa
A(g): namol
P๐๐๐ด๐ด =๐๐0
๐๐๐ด๐ด๐๐๐ด๐ด+๐๐๐ต๐ต
In this case, we need to consider the entropy of mixing.
Indeed, the entropy of mixing is related to entropy change due to volume change.
โ๐บ๐บ โ 0
โ๐๐ โ 0
โ๐บ๐บ = 0
โ๐๐ = 0
3.1. Chemical equilibrium for gases-Gibbs energy in gas mixture (ideal gas)-
3.1. Chemical equilibrium for gases-review-
Regarding the pressure effect, it is largely different between gas phase and condensed phases (solid, liquid)
Solid Liquid GasEffect of pressure on Gibbs energy
Negligible (< GPa level)
Negligible(< GPa level)
Important (comparable with temperature effect)
Depending on partial pressureIf the total pressure is1 barโฆ
Any solid phase feels1 bar.
Any liquid phase feels 1 bar
Total gas pressure is 1 bar, butpartial pressures are changeable.
*Note there is no โpartial pressureโ for solid and liquid. For pure substance, the chemical potential, which is molar Gibbs energy for pure
substance, is as follows, where โ๐บโ denotes the standard state
๐๐ ๐๐,๐๐ =๐บ๐บ๐๐
= ๐๐๐บ ๐๐ + ๐ ๐ ๐๐ ln โ๐๐ ๐๐๐บ For example, โ๐ ๐ ๐๐ ln โ๐๐ ๐๐๐บ โ term at 300 K becomes -1.72 kJ/mol for
0.5x105 Pa, -28.7 kJ/mol for 1 Pa, -57.4 kJ/mol for 1x10-5 Pa. However, the contribution to Gibbs energy of the system (๐๐๐๐ , not ๐๐) is not
so large because ๐๐ is very small for a low pressure.
3.1. Chemical equilibrium for gases-review-
The following equation is for pure substance:
๐๐ ๐๐,๐๐ =๐๐๐บ๐บ๐๐๐๐ ๐๐,๐๐
=๐บ๐บ๐๐
= ๐๐๐บ ๐๐ + ๐ ๐ ๐๐ ln โ๐๐ ๐๐๐บ
Now we want to know the related equation for mixture.
๐๐๐ด๐ด =๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐,๐๐๐ต๐ต,๐๐๐๐,๐๐๐๐
<(system-1) Mixture>T0 K, P0 Pa
A(g): nA mol, B(g): nB molY(g): nY mol, Z(g): nZ mol
<(system-2) Pure A substance>T0 K, P0 PaA(g): nA mol
For each system, if we add โ๐๐๐ด๐ด of A with keeping other quantities fixed (T0, and P0,and nB, nY and nZ for system-1), Gibbs energy of the system should be also changed, โ๐บ๐บ. However, because the composition of the system different, โ๐บ๐บ๐ ๐ ๐ ๐ ๐ ๐ โ1 โ โ๐บ๐บ๐ ๐ ๐ ๐ ๐ ๐ โ2. And
limโ๐๐๐ด๐ดโ0
โ๐บ๐บ๐ ๐ ๐ ๐ ๐ ๐ โ1โ๐๐๐ด๐ด
=๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐,๐๐๐ต๐ต,๐๐๐๐,๐๐๐๐
= ๐๐๐ด๐ด โ ๐บ๐บ๐๐๐ด๐ด
limโ๐๐๐ด๐ดโ0
โ๐บ๐บ๐ ๐ ๐ ๐ ๐ ๐ โ2โ๐๐๐ด๐ด
=๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐
= ๐๐๐ด๐ด =๐บ๐บ๐๐๐ด๐ด
*Note both are definitions of chemical potential, but for different systems.
3.1. Chemical equilibrium for gases-review-
Mixture Pure substance
Indeed, if we assume ideal gases, the chemical potential for a gas mixture can be replaced with the chemical potential for a pure gas substance with the corresponding partial pressure.
T0 K, V0 m3, P0 PaA(g): nA molB(g): nB mol๐๐๐ด๐ด = ๐๐0
๐๐๐ด๐ด๐๐๐ด๐ด+๐๐๐ต๐ต
๐๐๐ต๐ต = ๐๐0๐๐๐ต๐ต
๐๐๐ด๐ด+๐๐๐ต๐ต
where ๐บ๐บ๐ ๐ ๐ ๐ ๐ ๐ โ๐๐๐๐๐๐ = ๐บ๐บ๐๐๐๐๐๐๐๐โ๐ด๐ด + ๐บ๐บ๐๐๐๐๐๐๐๐โ๐ต๐ต [for ideal gas].
๐๐๐ด๐ด =๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐,๐๐๐ต๐ต
=๐๐๐บ๐บ๐๐๐๐๐ด๐ด ๐๐,๐๐๐ด๐ด
= ๐๐๐บ๐ด๐ด ๐๐ + ๐ ๐ ๐๐ ln โ๐๐๐ด๐ด ๐๐๐บ
T0 K, V0 m3, PB Pa
B(g): nB mol
๐๐๐ต๐ต = ๐๐0๐๐๐ต๐ต
๐๐๐ด๐ด+๐๐๐ต๐ต
T0 K, V0 m3, PA Pa
A(g): nA mol
P๐๐๐ด๐ด = ๐๐0๐๐๐ด๐ด
๐๐๐ด๐ด+๐๐๐ต๐ต
<System of mixture> <System of pure A > <System of pure B >
This idea can be extended to more than binary systems, then we can evaluate the pressure dependence of chemical potential of mixture as:
3.1. Chemical equilibrium for gases-review-
Using the obtained pressure dependence of Gibbs energy of gas phase in a mixture, we can further write down โdG = โ๐๐ ๐๐๐๐๐๐๐๐๐๐โ to find dG = 0 condition, which is the condition for equilibrium state.
Then, as a result, following chemical equilibrium equations are obtained. Again, please recall that these equations give us the condition where the
Gibbs energy of the system is minimized. So, we can get the same result if we directly evaluate the Gibbs energy of the system and find the condition with which it is minimized.
โ๐๐๐บ๐บ๐บ = โ๐ ๐ ๐๐ ln๐พ๐พ๐๐(๐๐)
๐พ๐พ๐๐ ๐๐ = ๐๐๐๐๐๐ =โ๐๐๐๐ ๐๐๐บ ๐๐๐๐ โ๐๐๐๐ ๐๐๐บ ๐๐๐๐
โ๐๐๐ด๐ด ๐๐๐บ ๐๐๐ด๐ด โ๐๐๐ต๐ต ๐๐๐บ ๐๐๐ต๐ต๐๐๐๐
=๐๐๐๐๐๐๐๐๐๐๐๐
๐๐๐๐
๐๐๐ด๐ด๐๐๐ด๐ด๐๐๐ต๐ต
๐๐๐ต๐ต ร ๐๐๐บ ๐๐๐ด๐ด+๐๐๐ต๐ตโ๐๐๐ต๐ตโ๐๐๐ต๐ต๐๐๐๐
*the subscript eq emphasizes that the partial pressures are in an equilibrium.
๐๐๐บ = 1 ๐๐๐๐๐๐
โ๐๐๐บ๐บ๐บ = ๐๐๐๐๐๐๐บ๐๐ ๐๐ + ๐๐๐๐๐๐๐บ๐๐ ๐๐ โ ๐๐๐ด๐ด๐๐๐บ๐ด๐ด(๐๐) โ ๐๐๐ต๐ต๐๐๐บ๐ต๐ต(๐๐)
๐พ๐พ๐๐ ๐๐ = ๐๐๐๐๐๐ โโ๐๐๐บ๐บ๐บ๐ ๐ ๐๐
For โ๐๐๐ด๐ดA(g) + ๐๐๐ต๐ตB(g) โ ๐๐๐๐Y(g) + ๐๐๐๐Z(g)โ
Quiz
(Quiz-1) Please determine the equilibrium constant for the following reaction at 1000 K: H2(g) + CO2(g) โ CO(g) + H2O(g)
(Quiz-2) Suppose that we have a mixture of the gases H2(g), CO2(g), CO(g) and H2O(g) at 1000 K at 1 bar, with ๐๐๐ป๐ป2 = 0.200 ๐๐๐๐๐๐, ๐๐๐ถ๐ถ๐ถ๐ถ2 = 0.200 ๐๐๐๐๐๐, ๐๐๐ถ๐ถ๐ถ๐ถ = 0.300 ๐๐๐๐๐๐, and ๐๐๐ป๐ป2๐ถ๐ถ = 0.300 ๐๐๐๐๐๐. Is the reaction described by the equation "H2(g) + CO2(g) โ CO(g) + H2O(g)" at equilibrium? If not, in what direction (left or right) will the reaction proceed to attain equilibrium for const.-T cons.t-P condition?
(Quiz-3) Please determine the equilibrium partial pressures at 1000 K at 1 bar.
โ๐๐๐ฎ๐ฎ๐บ (kJ/mol) @ 1000 K
H2(g) 0CO2(g) -394.9CO(g) -155.4H2O(g) -192.6
Quiz-answer
(Quiz-1) Please determine the equilibrium constant for the following reaction at 1000 K: H2(g) + CO2(g) โ CO(g) + H2O(g)
โ๐๐๐บ๐บ๐บ = ๐๐๐๐๐๐๐บ๐๐ ๐๐ + ๐๐๐๐๐๐๐บ๐๐ ๐๐ โ ๐๐๐ด๐ด๐๐๐บ๐ด๐ด ๐๐ โ ๐๐๐ต๐ต๐๐๐บ๐ต๐ต ๐๐= ๐๐๐บ๐ถ๐ถ๐ถ๐ถ ๐๐ + ๐๐๐บ๐ป๐ป2๐ถ๐ถ ๐๐ โ ๐๐๐บ๐ป๐ป2 ๐๐ โ ๐๐๐บ๐ถ๐ถ๐ถ๐ถ2 ๐๐= โ200.3 + โ192.6 โ 0 โ 395.9 = 3 kJ/mol
๐พ๐พ๐๐ ๐๐ = ๐๐๐๐๐๐ โโ๐๐๐บ๐บ๐บ๐ ๐ ๐๐
= ๐๐๐๐๐๐ โ3 ร 103
8.31 ร 1000= 0.697
โ๐๐๐ฎ๐ฎ๐บ (kJ/mol) @ 1000 K
H2(g) 0CO2(g) -395.9CO(g) -200.3H2O(g) -192.6
Quiz answer(Quiz-2) Suppose that we have a mixture of the gases H2(g), CO2(g), CO(g) and H2O(g) at 1000 K at 1 bar, with ๐๐๐ป๐ป2 = 0.200 ๐๐๐๐๐๐, ๐๐๐ถ๐ถ๐ถ๐ถ2 = 0.200 ๐๐๐๐๐๐, ๐๐๐ถ๐ถ๐ถ๐ถ = 0.300 ๐๐๐๐๐๐, and ๐๐๐ป๐ป2๐ถ๐ถ = 0.300 ๐๐๐๐๐๐. Is the reaction described by the equation "H2(g) + CO2(g) โCO(g) + H2O(g)" at equilibrium? If not, in what direction (left or right) will the reaction proceed to attain equilibrium for const.-T cons.t-P condition?
๐พ๐พ๐๐ ๐๐ = ๐๐๐๐๐๐ =โ๐๐๐๐ ๐๐๐บ ๐๐๐๐ โ๐๐๐๐ ๐๐๐บ ๐๐๐๐
โ๐๐๐ด๐ด ๐๐๐บ ๐๐๐ด๐ด โ๐๐๐ต๐ต ๐๐๐บ ๐๐๐ต๐ต๐๐๐๐
=๐๐๐๐๐๐๐๐๐๐๐๐
๐๐๐๐
๐๐๐ด๐ด๐๐๐ด๐ด๐๐๐ต๐ต
๐๐๐ต๐ต ร ๐๐๐บ ๐๐๐ด๐ด+๐๐๐ต๐ตโ๐๐๐ต๐ตโ๐๐๐ต๐ต๐๐๐๐
We need to use an equation like:
First, we evaluate โreaction quotientโ using the following equation:
๐๐ =โ๐๐๐๐ ๐๐๐บ ๐๐๐๐ โ๐๐๐๐ ๐๐๐บ ๐๐๐๐
โ๐๐๐ด๐ด ๐๐๐บ ๐๐๐ด๐ด โ๐๐๐ต๐ต ๐๐๐บ ๐๐๐ต๐ต=
0.300 ร 0.3000.200 ร 0.200
= 2.25
Because this value is larger than ๐พ๐พ๐๐(= 0.697), the system should evolve to decrease ๐๐ (non-equilibrium state now). Thus, partial pressure of reactants will increase, which means the reaction will proceed from the right to the left.
*Be careful that ๐๐ is not identical to ๐๐๐๐๐๐. ๐๐ = ๐๐๐๐๐๐ = ๐พ๐พ๐๐ ๐๐ is achieved when the system (thus partial pressures) are of equilibrium state.
Quiz answer
(Quiz-3) Please determine the equilibrium partial pressures at 1000 K at 1 bar.
๐พ๐พ๐๐ ๐๐ = ๐๐๐๐๐๐ =โ๐๐๐๐ ๐๐๐บ ๐๐๐๐ โ๐๐๐๐ ๐๐๐บ ๐๐๐๐
โ๐๐๐ด๐ด ๐๐๐บ ๐๐๐ด๐ด โ๐๐๐ต๐ต ๐๐๐บ ๐๐๐ต๐ต๐๐๐๐
=(0.300 + ๐๐๐๐๐๐)(0.300 + ๐๐๐๐๐๐)(0.200 โ ๐๐๐๐๐๐)(0.200 โ ๐๐๐๐๐๐)
๐๐๐๐= 0.697
If we define the extent of reaction with ๐๐, the partial pressures at equilibrium state is given as follows (๐๐ becomes ๐๐๐๐๐๐):
๐๐๐ป๐ป2 = (0.200 โ ๐๐๐๐๐๐) ๐๐๐๐๐๐ ๐๐๐ถ๐ถ๐ถ๐ถ2 = (0.200 โ ๐๐๐๐๐๐) ๐๐๐๐๐๐
๐๐๐ถ๐ถ๐ถ๐ถ = (0.300 + ๐๐๐๐๐๐) ๐๐๐๐๐๐ ๐๐๐ป๐ป2๐ถ๐ถ = (0.300 + ๐๐๐๐๐๐) ๐๐๐๐๐๐
Then,
0.300 + ๐๐๐๐๐๐ = 0.697 0.200 โ ๐๐๐๐๐๐๐๐๐๐๐๐ = โ0.0725
๐๐๐ป๐ป2 = ๐๐๐ถ๐ถ๐ถ๐ถ2 = 0.273 ๐๐๐๐๐๐ ๐๐๐ถ๐ถ๐ถ๐ถ = ๐๐๐ป๐ป2๐ถ๐ถ = 0.227 ๐๐๐๐๐๐
Applying this value to the partial pressure expressions, we can obtain the following equilibrium partial pressures.
3.1. Chemical equilibrium for gases-an example-
(Problem) Initially, we have 1 mol of CH4(g) and 1 mol of H2O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH4(g), H2O(g), CO(g), H2(g), C2H2(g). Please determine the equilibrium amounts of these species.
3.1. Chemical equilibrium for gases-an example-
(Problem) Initially, we have 1 mol of CH4(g) and 1 mol of H2O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH4(g), H2O(g), CO(g), H2(g), C2H2(g). Please determine the equilibrium amounts of these species.
CH4(g) + H2O(g) โnCH4 CH4(g) + nH2O H2O(g) + nCO CO(g) + nH2 H2(g) + nC2H2 C2H2(g)
For mass conservation, we have 3 conditions. [H] 4+2=4 nCH4 + 2 nH2O +2 nH2 +2 nC2H2 [C] 1= nCH4 + nCO + 2nC2H2 [O] 1= nH2O + nCO
We need 2 additional equations to determine 5 parameters. At equilibrium states, any balance reactions should be of equilibrium states. Then, we consider 2 of them. [eq.1] CH4(g) + H2O(g) โ 3H2(g) + CO(g) [eq.2] CH4(g) + CO(g) โ C2H2(g) + H2O(g)
Now we have 5 equations for 5 unknown variables, then we can determine the 5 variables.
3.1. Chemical equilibrium for gases-an example -
(Problem) Initially, we have 1 mol of CH4(g) and 1 mol of H2O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH4(g), H2O(g), CO(g), H2(g), C2H2(g). Please determine the equilibrium amounts of these species.
[eq.1] CH4(g) + H2O(g) โ 3H2(g) + CO(g)
[eq.2] CH4(g) + CO(g) โ C2H2(g) + H2O(g)
๐พ๐พ๐๐,1 ๐๐ = ๐๐๐๐๐๐ โโ๐๐๐บ๐บ๐บ1๐ ๐ ๐๐
=๐๐๐ป๐ป23 ๐๐๐ถ๐ถ๐ถ๐ถ๐๐๐ถ๐ถ๐ป๐ป4๐๐๐ป๐ป2๐ถ๐ถ
๐พ๐พ๐๐,2 ๐๐ = ๐๐๐๐๐๐ โโ๐๐๐บ๐บ๐บ2๐ ๐ ๐๐
=๐๐๐ถ๐ถ2๐ป๐ป2๐๐๐ป๐ป2๐ถ๐ถ๐๐๐ถ๐ถ๐ป๐ป4๐๐๐ถ๐ถ๐ถ๐ถ
For example, if the balance reaction given in [eq.1] is not of equilibrium state, some reaction (to right or to left) should happen to decrease the Gibbs energy of the system.
Hence, any kind of balance reaction between constituent gases should achieve equilibrium conditions.
where partial pressures are proportional to molar fractions.
Quiz
(Quiz-1) Please determine the equilibrium constant for the following reaction at 1000 K: H2(g) + CO2(g) โ CO(g) + H2O(g)
(Quiz-2) Suppose that we have a mixture of the gases H2(g), CO2(g), CO(g) and H2O(g) at 1000 K at 1 bar, with ๐๐๐ป๐ป2 = 0.200 ๐๐๐๐๐๐, ๐๐๐ถ๐ถ๐ถ๐ถ2 = 0.200 ๐๐๐๐๐๐, ๐๐๐ถ๐ถ๐ถ๐ถ =0.300 ๐๐๐๐๐๐, and ๐๐๐ป๐ป2๐ถ๐ถ = 0.300 ๐๐๐๐๐๐. Is the reaction described by the equation "H2(g) + CO2(g) โ CO(g) + H2O(g)" at equilibrium? If not, in what direction (left or right) will the reaction proceed to attain equilibrium for const.-T cons.t-P condition?
(Quiz-3) Please determine the equilibrium partial pressures at 1000 K at 1 bar.
โ๐๐๐ฎ๐ฎ๐บ (kJ/mol) @ 1000 K
H2(g) 0CO2(g) -394.9CO(g) -155.4H2O(g) -192.6
3.1. Chemical equilibrium for gases-an example-
(Problem) Initially, we have 1 mol of CH4(g) and 1 mol of H2O(g) in the system and keep const.-T (1000 K) const.-P (1 bar) condition. When the system arrived at the equilibrium state, we found CH4(g), H2O(g), CO(g), H2(g), C2H2(g). Please determine the equilibrium amounts of these species.