3
Property relations and Thermochemistry
3.1
Intensive and extensive properties
Extensive properties depend on the amount of mass or number of moles in the
system. The extensive properties are usually in capital letters: U, H, S etc.
Intensive properties are expressed per unit mass or mole (kmole). The intensive
properties are in lower-case letters. The mass intensive property is written
u, h, s... and the molar intensive property as ū, h̄, s̄....
3.2
Ideal-gas mixtures
The system contains a mixture of ideal gases. Each component or species Xi
has a molar mass Mi . The number of moles of component i is Ni . The total
mass of species i is:
mi = Ni Mi
The sum of partial pressures pi is the total pressure p:
�
p=
pi
(38)
(39)
i
The mole fraction, mass fraction and mixture molar weight are defined in
the following way:
Mole fraction χi :
Mass fraction Yi :
Ni
χi = �
.
i Ni
Mixture molar weight:
mi
Yi = �
.
i mi
M=
�
χ i Mi .
(40)
(41)
(42)
i
The partial pressure pi may be expressed using molar fraction (40):
pi = χ i p
(43)
Ideal gas mixtures properties could be calculated as sums of the individual
mass or mole fraction i.e.:
enthalpy/mass:
�
h=
Y i hi
(44)
i
enthalpy/mole:
h̄ =
�
χi h̄i
i
The same for internal energy u and s. Though entropy is:
18
(45)
s(T, p) =
�
Yi si (T, pi )
(46)
χi s̄i (T, pi )
(47)
i
s̄(T, p) =
�
i
(note partial pressure pi )
We have:
� p �
i
pref
� p �
i
s̄i (T, pi ) = s̄i (T, pref ) − R ln
pref
si (T, pi ) = si (T, pref ) − R ln
(48)
(49)
If the molar enthalpy h̄ is given for a system the enthalpy/mass, h is readily
given provided that the molar fraction of the different species is known:
h=
H
H
=
Σmi
ΣMi Ni
h̄ =
H
ΣNi
ΣNi
ΣMi Ni
1
h = h̄
Ni
ΣMi ΣN
i
⇒ h = h̄
and finally
h = h̄
3.3
1
ΣMi χi
(50)
Molar absolute (or standardized) enthalpy and molar
enthalpy of formation
The molar absolute enthalpy of a species i at temperature T is h̄◦i (T ). The
superscript ◦ denotes that the pressure is at standard state pressure (p◦ ), usually
1 atm. The molar absolute enthalpy at T of species i is related to the molar
�
absolute enthalpy at an other temperature T through:
�
h̄◦i (T ) = h̄◦i (T ) +
�
T
T
�
dh̄◦i
(51)
Now we define the enthalpy difference (which is tabulated) ∆h̄◦i (T ) as
∆h̄◦i (T )
Hence the integral
�
T
T
�
�T
dh̄◦i =
T�
�
≡
�
T
Tref
dh̄◦i
(52)
dh̄◦i in eq.(51) may be expressed:
T
Tref
dh̄◦i +
�
Tref
T
�
�
dh̄◦i = ∆h̄◦i (T ) − ∆h̄◦i (T )
19
and used in eq.(51):
�
�
h̄◦i (T ) = h̄◦i (T ) + ∆h̄◦i (T ) − ∆h̄◦i (T ).
(53)
�
The term h̄◦i (T ) comes in part from the forming of the species and is best
explained in the context of an example. Let us look at the simple situation
where water is formed from the elements hydrogen and oxygen
H2 + 12 O2 → H2 O
and write the absolute enthalpy for water (eq.(53))
�
�
h̄◦H2 O (T ) = h̄◦H2 O (T ) + ∆h̄◦H2 O (T ) − ∆h̄◦H2 O (T ).
(54)
�
The absolute enthalpy h̄◦H2 O (T ) is related to the enthalpy of the elements of
water in the following way
�
�
�
�
1
h̄◦H2 O (T ) = h̄◦f,H2 O (T ) + h̄◦H2 (T ) + h̄◦O2 (T ).
2
�
(55)
�
h̄◦f,H O (T ) is called the enthalpy of formation of water at temperature T and
2
is the difference between the enthalpy of the product (H2 O) and the enthalpy
�
of the elements (H2 , O2 ) forming the product at T and pressure p◦ . The
term arises from the shift in the enthalpy scales between product and reactants
and is due to the difference in the chemical energy stored in the constituents.
There is no natural zero level for the enthalpy scale (as it is for entropy (perfect
crystals of a single pure isotope of a single element at T = 0K)) and an absolute
scale of the enthalpy has to be decided. The convention is that the absolute
enthalpy is zero for an element in its most stable state at Tref (= 298.15K) and
at pref (= 1atm) (for hydrogen and oxygen it is H2 and O2 in gas phase). For
such an element i the absolute enthalpy is:
h̄◦i (T ) = ∆h̄◦i (T )
(56)
and hence zero for T = Tref . Now we can write the absolute enthalpy for water
at temperature T . Use eq.(56) in eq.(55) and put the result into equation (54)
and we have:
�
�
�
�
1
h̄◦H2 O (T ) = h̄◦f,H2 O (T )+∆h̄◦H2 (T )+ ∆h̄◦O2 (T )+∆h̄◦H2 O (T )−∆h̄◦H2 O (T ) (57)
2
�
If we put T = Tref (∆h̄◦i (Tref ) = 0) eq.(57) becomes:
h̄◦H2 O (T ) = h̄◦f,H2 O (Tref ) + ∆h̄◦H2 O (T )
(58)
which gives us the common form of the absolute or standardized enthalpy:
h̄◦i (T ) = h̄◦f,i (Tref ) + ∆h̄◦i (T )
(59)
As the enthalpy is a state function the absolute enthalpy of a species is only
dependent on its temperature, pressure and phase and route to the final temperature is of no importance. Figure(7) shows a simple example: hydrogen
�
dissociating into atoms. Independent of at what temperature T the enthalpy
20
of formation of the monatomic hydrogen is taken we get the same absolute
��
h̄◦H (T ) as long as we take into account also the change of the sensible enthalpy
�
of both H and H2 when the temperature T varies. Thus eq.(57) rewritten for
this case:
��
�
�
��
�
1
h̄◦H (T ) = h̄◦f,H (T ) + ∆h̄◦H2 (T ) + ∆h̄◦H (T ) − ∆h̄◦H (T )
(60)
2
∆hoH(T')
0
T'
Tref =298
T''
T
Figure 7: Figure showing the absolute enthalpy change due to dissociation of a
hydrogen molecule into atoms
�
��
Exercise: Evaluate the reaction at a few different T and T .
When we have the absolute enthalpy of the reactants and products of a
given reaction at a given temperature and pressure we may determine the enthalpy change the mixture will encounter due to the reaction. The enthalpy of
the products minus the enthalpy of the reactants is called the heath of reaction (or enthalpy�
of reaction, enthalpy of combustion) ∆HR (or per
mass ∆hR = ∆HR /( prod νi Mi )) at constant pressure and temperature for
the reaction in question. (Unfortunately there is also a heath of combustion
(= −∆hR ))
An example: Determine ∆HR for the reaction:
CH4 + 2 O2 → CO2 + 2 H2 O
at T =298K and T =4000K. Thus
∆H(T ) = h̄◦CO2 (T ) + 2h̄◦H2 O (T ) − 1h̄◦CH4 (T ) − 2h̄◦O2 (T )
With equations (59) and (56) we get:
∆H(T ) = h̄◦f,CO2 (Tref ) + ∆h̄◦CO2 (T )
+2(h̄◦f,H2 O (Tref ) + ∆h̄◦H2 O (T ))
−(h̄◦f,CH4 (Tref ) + ∆h̄◦CH4 (T ) + 2∆h̄◦O2 (T ))
21
At T = 298K and T = 4000K we get ∆H�= −191.749 kcal and ∆H = −201.641
kcal respectively. The mass involved is prod νi Mi = 80g
3.4
Adiabatic flame temperature
The adiabatic flame temperature is the temperature the products in the reaction
raise to if we technically have a complete reaction to the right. From the first
law of thermodynamics we have for an adiabatic system at constant pressure p:
Hreact (T, p) = Hprod (Tad , p)
(61)
The temperature Tad thus obtained is the adiabatic flame temperature at
constant pressure. Continuing with the previous example of burning methane
in oxygen we start with reactants at T = Tref and look for the Tad which satisfies
the relation ∆H = 0:
∆H = 0 = h̄◦f,CO2 (Tref ) + ∆h̄◦CO2 (Tad )
+2(h̄◦f,H2 O (Tref ) + ∆h̄◦H2 O (Tad ))
−(h̄◦f,CH4 (Tref ) + ∆h̄◦CH4 (Tref ) + 2∆h̄◦O2 (Tref ))
The matching to get a Tad that sets ∆H to 0 is left as an exercise.
The enthalpy tables can also be used to determine the adiabatic flame temperature at constant volume. Again from the first law of thermodynamics we
have for an adiabatic system at constant volume:
Ureact (T, V ) = Uprod (Tad , V ).
(62)
From the definition of enthalpy we have:
U = H − pV.
We use the relation in eq.(62) and set T = Tref :
Hreact (Tref ) − (pV )reac = Hprod (Tad ) − (pV )prod
or
Hprod (Tad ) − Hreact (Tref ) = (pV )prod − (pV )reac
The ideal gas law: pV = N RT and the restriction V = constant simplifies the
equation:
Hprod (Tad ) − Hreact (Tref ) = (Nprod Tad − Nreac Tref )R
(63)
where Nprod and Nreac are number of moles of products and reactants respectively. If we use the relation in the previous example we see that the number
of moles does not change during the reaction Nprod = Nreac = N and the
adiabatic flame temperature at constant volume can be determined from the
following relation:
Hprod (Tad ) − Hreact (Tref ) = N (Tad − Tref )R.
Compared with the adiabatic temperature at constant pressure the adiabatic
temperature at constant volume will be higher.
22