Download Thermochemistry

Combustion Chemistry
Hai Wang
Stanford University
2015 Princeton-CEFRC Summer School On Combustion
Course Length: 3 hrs
June 22 – 26, 2015
Copyright ©2015 by Hai Wang
This material is not to be sold, reproduced or distributed without prior written
permission of the owner, Hai Wang.
Lecture 1
1. THERMOCHEMISTRY
Thermodynamics is the foundation of a large range of physical science problems. It
provides us with a basic understanding about the driving force of a physical process and the
limits of such a process. The development of thermodynamic theory was intimately related
to combustion. In particular, the second law of thermodynamics was conceived largely to
prove that it is impossible to operate a perpetual motion machine.
1.1 Thermodynamic First Law
The first law of thermodynamics states that the energy is conserved when this energy is
transformed from one form to another. In the context of combustion analysis, we state that
for a control mass (or the working fluid),
Q −W = ΔU
(1.1)
where Q (kJ or kcal) is the heat transferred from the working fluid to the surrounding, W (kJ
or kcal) is the work done by the working fluid to the surrounding, and U (kJ or kcal) is the
internal energy of the working fluid.
It is important to mention that energy transformation always involves a process that has a
initial state (1) and a final state (2). ΔU = U2 – U1 is therefore the change of internal energy
of the working fluid from the initial state to the final state. Thermodynamic analysis follow
the convention that if the working fluid gives off heat to the surrounding, Q < 0 , and if the
working fluid receives heat from the surrounding, Q is positive. For example, in a simple
cooling process, a fluid loses its internal energy to the surrounding (i.e., lowering its
temperature) and thus ΔU < 0. Assuming that no work is done (W = 0), then Q < 0.
Likewise, if the working fluid does net work to the surrounding, W is positive, and if the
surrounding does net work to the working fluid, W < 0 (e.g., for adiabatic compression (Q =
0), the work done by the surrounding to the working fluid serves to raise the internal energy
of the working fluid, since –W = ΔU > 0.
The symbol U designates the internal energy of a given mass of a substance. Internal energy
is a measure of the total energy of the control mass. For example, excluding nuclear energy
the internal energy of air is a sum of the kinetic energy of each atom. Therefore, the internal
energy can be made a material property if it is defined as the internal energy per mass, an
intensive property, denoted here as u (kJ/kg) or u (kJ/kmol). In this course, we shall follow
the notation that intensive properties are expressed in lower cases.
Assuming that we are running a thermodynamic process that the only work done during the
process is that associated with boundary work under a constant pressure P1 = P2 = P (e.g., a
piston work), the work done may be calculated from
1-1
Stanford University
Version 1.2
©Hai Wang
2
W = ∫ PdV = P (V2 − V1 ) .
1
(1.2)
Putting Eq. (1.2) into Eq. (1.1), we have
Q = (U 2 + P2V2 ) − (U 1 + P1V1 ) ,
(1.3)
where V is the volume of the working fluid. In this case, the heat transferred during the
process corresponds to a net change of the controlled mass in the quantity U + PV between
the initial and final states. We find it convenient to define a new thermodynamic property,
the enthalpy
H = U + PV ,
h = u + Pv ,
h = u + Pv .
(1.4a)
(1.4b)
(1.4c)
Here v and v are the specific volumes, having the units (m3/kg) and (m3/kmol) respectively.
Clearly, these specific volumes are related to the mass density ρ and molar density (or
concentration) c, respectively, i.e., v = 1/ρ and v = 1/c.
In general, the internal energy u and enthalpy h depend on only two independent properties
that specifying the thermodynamic state, e.g., (T, P), (T, v), or (P, v). For a low-density gas
like air or combustion gases, T, P, and v are related by the ideal gas law or the equation of
state,
Pv = R 'T ,
Pv = Ru T ,
(1.5a)
(1.5b)
where Ru is the universal gas constant (8.314 kJ/kmol-K), R’ is the specific gas constant and
equal to Ru/MW, and MW is the molecular weight of the substance.
For a low-density gas, the internal energy is primarily a function of T, i.e., u ≅ u (T ) . This
relationship may be expressed by defining a constant-volume specific heat cv (kJ/kmol-K)
⎛ ∂u ⎞ .
cv = ⎜
⎟
⎝ ∂T ⎠v
(1.6)
For an ideal gas we have du = cv dT . Likewise, the relationship between enthalpy and
temperature may be established by defining a constant-pressure specific heat c p (kJ/kmol-K)
" ∂h %
cp = $ ' ,
# ∂T & p
1-2
(1.7)
Stanford University
Version 1.2
©Hai Wang
and dh = c p dT . In other words, the two specific heats defined above characterize the heat
required to raise the temperature of a substance by 1 K. Since for an ideal gas
dh = du + d ( pv ) = du + Ru dT and u ≅ u (T ) , we see that h and c p are also function of
temperature only. The relation between dh and du also yields c p = cv + Ru .
Here it is important to note that the enthalpy discussed thus far involves only the heating or
cooling a substance. This type of enthalpy is known as the sensible enthalpy or sensible heat.
Later, we will introduce two other types of enthalpy, one of which is critical to combustion
problems.
The first law of thermodynamics is quite insufficient to describe energy conversion.
Equation (1.1) states that it is possible to cool a substance of a given mass spontaneously (i.e.,
lowering its internal energy U), and transfer this energy to the surrounding. In other words,
within the first law of thermodynamics, it is possible to transform heat from a lowtemperature body to a high temperature body. We know that this cannot be true. The
second law of thermodynamics, to be discussed below, will address this problem.
1.2 Thermodynamic Second Law and Entropy
In contrast to the first law of thermodynamics, the second law is more difficult to
understand. The Kelvin-Planck statement of this law is It is impossible to construct a device that
will operate in a cycle and produce no effect other than the raising of a weight and the exchange of heat with a
single reservoir. In other words, it is impossible to construct a heat engine that (a) receives heat
continuously from a heat reservoir, (b) turns the heat transferred entirely to work, (c)
without having to leave any marks on the surrounding. Without diverging into a lengthy
discussion of the second law of thermodynamics, let us define entropy S (kJ/K) as
⎛δ Q ⎞
.
S =⎜
⎟
⎝ T ⎠ int rev
(1.8)
where (δQ)int rev is the heat a control mass received during an infinitesimal, internally
reversible process. Based on an analysis of thermodynamic cycles, it may be shown that for
a spontaneous process to occur, the entropy of the control mass must be equal to or greater
than zero,
ΔS = S2 – S1 ≥ 0.
(1.9)
Neither the Kelvin-Planck statement nor Eq. (1.8) really tells us what entropy is. An
understanding of entropy will have to come sometime later when we introduce statistical
thermodynamics. Here let us place some discussion about entropy in a non-rigorous fashion.
Entropy is a measure of molecular randomness. This randomness may be measured by the
predictability of the positions of atoms in a substance. A crystal material would have a small
entropy because atoms are more or less “locked” into the crystal lattice. In fact, the third
law of thermodynamics states that the entropy of a pure crystalline substance at absolute
1-3
Stanford University
Version 1.2
©Hai Wang
zero temperature is zero. In other words, the atoms in a pure crystal are “frozen” (no
oscillation) at 0 K. Therefore, their spatial position is completely predictive. In contrast, a
gas would have a large entropy because molecules that make up the gas constantly move
about in the space, resulting in small predictability regarding their positions. Moreover, an
increase in temperature of the gas leads an increase in the speed of molecular motion and
smaller predictability of the molecular positions. In other words, entropy increases with an
increase in temperature. In contrast, an increase in pressure leads to closer spacing among
molecules. As a result, the molecules become more confined spatially and the entropy is
smaller at higher pressures. The dissociation of a chemical substance into gaseous fragments
always leads to an increase in entropy since it is harder to predict the spatial positions of the
fragments than the molecules of their parent substance.
The inequality expressed by Eq. (1.9) basically says that for a spontaneous process to occur,
the entropy of the control volume must increase, i.e., natural processes favor more
randomness than orderness. Conceptually this makes sense since our experience tells us that
a building can spontaneously collapse into a pile of rubble, but a pile of rubble would not
spontaneously transform into a building (not without our intervention). Two different gases,
say, N2 and O2, would always mix and they never spontaneously separate spatially, leading to
better predictability of their positions.
The concept of entropy is also deeply rooted in our life. Take the life of a workaholic as an
example, the first law of thermodynamics states that it is possible for him/her to receive heat
Q in the form of food and hopefully without gaining weight (ΔU = 0), to transform this heat
entirely to work W. The second law says that he/she really cannot do this. That is, my
office always gets messier over time and I will need to clean it (i.e., not all the heat goes to
useful work) as time goes by.
Because entropy is a measure of randomness, which in turn, is determined by T and P, it is
also a material property. It follows that we can define and denote the entropy of a substance
by s (kJ/kg-K) or s (kJ/kmol-K). Although we do not know for the time being how to
directly measure entropy, we may develop some relationships that can help us to determine
the entropy value. Here we apply the first law to a constant T and P, internally reversible
process (e.g., compress a volume immersed in a temperature bath by a piston very slowly),
δ Qint rev − δ Wint rev = dU ,
(1.10)
but since δ Qint rev = TdS and δ Wint rev = PdV , we have
TdS = dU + PdV
(1.11)
or
ds =
du Pdv
dT Pdv
.
+
= cv
+
T
T
T
T
1-4
(1.12)
Stanford University
Version 1.2
©Hai Wang
Replacing u by h − Ts and rearranging, we obtain
ds =
dh vdP
dT vdP
.
−
= cp
−
T
T
T
T
(1.13)
Applying the ideal gas law, we may rewrite equations (12) and (13) as
dT
dv
,
+ Ru
T
v
dT
dP
.
ds = c p
− Ru
T
P
ds = cv
(1.14)
(1.15)
One may integrate the above equations to show that
v
dT
+ Ru ln 2 ,
T
v1
2
P
dT
Δ s = s 2 − s1 = ∫ c p
− Ru ln 2 .
1
T
P1
2
Δ s = s2 − s1 = ∫ cv
1
(1.16)
(1.17)
Equation (1.17) states that if c p is a constant, an increase of temperature by ΔT from T
causes the entropy to increase by c p ln (1 + ΔT T ) and an increase of pressure by ΔP from P
leads to the entropy to decrease by Ru ln (1 + ΔP P ) . Given the third law of
thermodynamics, which establish the absolute zero for entropy, the entropy of an ideal gas at
a given thermodynamic state (i.e., known T and P) can be easily determined if c p is known.
Equation (1.17) also states that unlike enthalpy and internal energy, the entropy of an ideal
gas is a function of both temperature and pressure. In application, we define the standard
entropy s o as
2
Δ s o = s2 o − s1o = ∫ c p
1
dT ⎡ T2 dT
⎤ ⎡ T1 dT
⎤
= ⎢∫ c p
− Ru ln ( P o )⎥ − ⎢ ∫ c p
− Ru ln ( P o )⎥ , (1.18)
0
0
T ⎣
T
T
⎦ ⎣
⎦
or
T
s o = ∫ cp
0
dT
− R u ln ( P o ) ,
T
(1.19a)
o
where P is the standard pressure of 1 atm. Hence,
T
Δs o = ∫ c p
0
1-5
dT
,
T
(1.19b)
Stanford University
Version 1.2
©Hai Wang
By tabulating this standard entropy, we may easily determine the entropy change of an ideal
gas under an arbitrary condition by
s (T , P ) = s o (T ) − Ru ln
P
.
Po
(1.20)
1.3 Chemical Reactions
Before we apply the above thermodynamic principles to combustion analysis, we need to
take a moment to review a few aspects of chemical reactions. From a process point of view,
a chemical reaction may be viewed as the conversion of reactants that enter into a reactor
(the initial state) to products that leaves the reactor (the final state). For example, methane
(CH4) flows into a reactor with air (21%O2 and 79% N2). Suppose the molar ratio of oxygen
to methane is 2-to-1. We may write that to burn 1 mole of methane,
CH4 + 2 O2 + (2×79/21) N2 →
Reactor
→ CO2 + 2H2O + (2×79/21) N2 .
Here the products include 1 mole CO2, 2 moles of H2O and (2×79/21) moles of N2. Of
course, in writing the above process reaction, we may neglect the box and simply write
CH4 + 2 O2 + (2×79/21) N2 → CO2 + 2H2O + (2×79/21) N2 .
(1.21)
The above reaction is known as the complete combustion reaction as all the carbon in the fuel is
oxidized to CO2 and all the hydrogen is converted to H2O. These compounds are called the
complete combustion products. If there is no excess oxygen (i.e., all oxygen is consumed in
the oxidation process), the characteristic fuel-to-oxygen molar ratio is known as the
stoichiometric ratio (equal to ½ for methane). The stoichiometric ratio for an arbitrary fuel
CmHn may be readily determined by writing out the complete, stoichiometric reaction,
CmHn + (m+ n 4 ) O2 + (m+ n 4 )×(79/21) N2 → m CO2 + n
2
H2O + (m+ n 4 )×(79/21) N2 .
(1.22)
which gives the stoichiometric ratio equal to 1/(m+ n 4 ) .
In a practical combustion process, however, the fuel-to-oxygen molar ratio needs not to be
the stoichiometric ratio. For example, a gasoline engine often runs slightly above the
stoichiometric ratio at the cold start, for reasons to be discussed later. To characterize fuelto-oxygen ratio in a practical combustion process, we introduce the equivalence ratio,
defined as the molar ratio of fuel-to-oxygen for an actual combustion process by that of
stoichiometric combustion:
φ=
( moles of fuel moles of oxygen )act.
( moles of fuel moles of oxygen )stoi.
1-6
.
(1.23)
Stanford University
Version 1.2
©Hai Wang
Of course, it may be shown that the equivalence ratio may be calculated using the molar
ratio of fuel-to-air or the mass ratio of fuel-to-oxygen or fuel-to-air.
By examining the equivalence ratio, we can quickly tell the nature of the fuel/air mixture.
That is, if φ = 1, we have stoichiometric reaction; if φ < 1 we have excess oxygen that is not
completely used in a reaction process and the combustion is called fuel-lean combustion; if
φ > 1 we have excess fuel and the combustion is called fuel-rich combustion.
⎧< 1 fuel lean
⎪
φ = ⎨1
stoichiometric
⎪> 1 fuel rich
⎩
Under the fuel rich combustion, the combustion reaction inherently yields incomplete
combustion products, like CO, H2 etc.
1.4 Enthalpy of Formation, Enthalpy of Combustion
As we discussed in section 1.1, there are 3 types of enthalpy. The first type is associated with
heating or cooling of a substance. The second type is latent enthalpy (or heat). This is the
enthalpy associated with the phase change of a substance. For example, the latent heat of
evaporation of H2O, hlg , is
hlg = h g − hl ,
(1.24)
where h g and hl are the enthalpy of water in its vapor and liquid states, respectively. What
is perhaps more important to combustion analysis is the reaction enthalpy. For example,
reaction (21) releases an amount of heat due to chemical bond rearrangements. Combining
Eqs (1.3) and (1.4a), we have
Q = H 2 − H1 .
Since state 1 corresponds to the reactants, and state 2 corresponds the products, the above
equation states that (a) in a non-adiabatic reactor, the heat released from the reactor is equal
to the total enthalpy of the combustion products subtracted by the total enthalpy of the
reactant, and (b) since for a combustion process Q < 0, H2 < H1, i.e., the total enthalpy of
the products is lower than that of the reactants.
The nature of reaction enthalpy is very different from the sensible enthalpy, as the former is
due to re-arranging chemical bonds and the latter is simply due to heat and cooling without
changing the chemical nature of the substance. To calculate the exact amount of reaction
enthalpy and therefore the amount of heat release, we need to first understand the concept
and application of enthalpy of formation.
1-7
Stanford University
Version 1.2
©Hai Wang
The enthalpy of formation h f at a given temperature is defined as the heat released from producing 1 mole of
a substance from its elements at that temperature. By this definition, the enthalpy of formation is
zero for the reference elements. These elements are, for example, graphite [denoted by C(S)
hereafter), molecular hydrogen H2, molecular oxygen (O2), molecular nitrogen (N2), and
molecular chlorine (Cl2).
The enthalpy of formation of CO2, say at 298 K, may be conceptually measured by reacting
1 mole of graphite and 1 mole of O2 at 298 K, producing 1 mole of CO2 at the same
temperature:
C(S) + O2 → CO2 + Q,
where Q is the heat released from the above process (Q = –393.522 kJ). Using Eq. (1.24),
we have
Q = -393.522 (kJ)=H 2 − H1 = 1× h f ,298 K ( CO2 ) − 1× ⎡⎣h f ,298 K ( C(S)) + h f ,298 K (O2 )⎤⎦
= h f ,298 K (CO2 )
.
The enthalpy of formation of CO2 is therefore h fo =–393.522 kJ/mol at 298 K. Likewise the
enthalpy of formation of CO is determined by measuring the heat release from
C(S) + ½ O2 → CO + Q (–110.53 kJ at 298 K),
(1.25)
and h fo (CO) = –110.53 kJ/mol at 298 K.
The conceptual definition uses the same temperature for the reactor, reactants, and products,
and this condition is known as the standard condition. For this reason, we use a superscript “o”,
i.e., h fo , to designate this standard condition and the corresponding enthalpy of formation is
termed as the standard enthalpy of formation.
Obviously if reaction (25) is carried out at a different temperature under the standard
condition, we do not necessarily get the same heat release. In other words, the enthalpy of
formation is dependent on temperature, yet this temperature dependence is related to
sensible heat. To illustrate the relationship of enthalpy of formation of a substance at two
different temperatures, we sketch the following diagram:
1-8
Stanford University
Version 1.2
©Hai Wang
1
∑
elements
ΔH =
∑
elements
’
h fo,T
’
2
⎡⎣ H (T ) − H ( 298 )⎤⎦
= ⎡⎣h (T ) − h ( 298 )⎤⎦ C(S) + ⎡⎣h (T ) − h ( 298 )⎤⎦ O
T
ΔH CO2 = ⎣⎡h (T ) − h ( 298 ) ⎦⎤ CO
2
2
1
298 K
2
h fo,298 K
Elements
(1 mole C(s) + 1 mole O2)
Substance
(1 mole CO2)
Recognizing that enthalpy is a state function, i.e., for an ideal gas the enthalpy of a substance
is fully defined if the temperature is known, and the change of enthalpy for a process is
independent of the path, we may write
H 2' − H 1 = H 2' − H 2 + ( H 2 − H 1 ) = "#h (T ) − h (298)$%
CO2
+ h f ,298 (CO2 )
= H 2' − H 1' + ( H 1' − H 1 ) = h f ,T (CO2 ) + "#h (T ) − h (298)$%
C(S)
+ "#h (T ) − h (298)$%
.
O2
It follows that
h f ,T (CO2 ) = h f ,298 (CO2 ) + "#h (T ) − h (298)$%
CO2
{
− "#h (T ) − h (298)$% + "#h (T ) − h (298)$%
C(S)
O2
}
,
which may be generalized to
h fo,T = h fo,298 + ⎡⎣h (T ) − h(298)⎦⎤ substance −
∑ν
elements
i
⎡⎣h (T ) − h (298)⎤⎦ i ,
(1.25a)
where ν i is the stoichiometric coefficients of the ith elements in the reaction that forms the
substance. Therefore if the specific heat or sensible enthalpy of a substance is known, we
only need to know the value of enthalpy of formation at a particular temperature, and in
general this temperature is equal to 298 K.
In combustion analysis, we often group the first and second term of Eq. (1.25a) by defining
a total enthalpy as
hT ≡ h f ,298 + #$h (T ) − h(298)%&
,
substance
(1.25b)
and
h f ,T = hT −
∑
elements
ν i "#h (T ) − h(298)$% .
i
(1.25c)
Table 1.2 lists the enthalpy of formation of some important species for combustion analysis.
1-9
Stanford University
Version 1.2
©Hai Wang
Table 1.2 Standard enthalpy of formation of key combustion species in the vapor state
h fo,298 (kJ/mol)
Species
Name
H 2O
CO
CO2
CH4
C 2H 6
C 3H 8
C4H10
C8H18
C 2H 4
C 2H 2
CH3OH
C 6H 6
H•
O•
OH•
Water
Carbon monoxide
Carbon dioxide
Methane
Ethane
Propane
Butane
Octane
Ethylene
Acetylene
Methanol
Benzene
Hydrogen atom
Oxygen atom
Hydroxyl radical
-241.8
-110.5
-393.5
-74.9
-84.8
-104.7
-125.6
-208.4
52.5
226.7
-201.0
82.9
218.0
248.2
39.0
The standard enthalpy of combustion hc (kJ/mol-fuel) is defined as the heat released from the
complete combustion of 1 mole of fuel at 298 K. Consider the complete combustion of
methane (Eq. 1.21). We will again utilize the path independent property to illustrate that hc
may be determined by the sum of enthalpy of formation of the products (multiplied by the
molar ratio of the product-to-fuel) subtracted by the sum of enthalpy of formation of the
reactants:
CH4 + 2O2 + 2×79/21 N2
hc
ΔH 1
CO2 + 2H2O + 2×79/21 N2
ΔH 2
C(S) + 2H2 + 2O2 + 2×79/21 N2
hc (kJ/kmol-fuel) = ΔH1 + ΔH 2 = −1 × h fo,298 ( CH 4 ) + 1 × h fo,298 ( CO2 ) + 2 × h fo,298 ( H 2 O ) .
For an arbitrary fuel (CmHn) undergoing combustion (1.22), we determine its enthalpy of
combustion as
1-10
Stanford University
Version 1.2
©Hai Wang
n
hc (kJ/kmol-fuel) = m × h fo,298 ( CO2 ) + × h fo,298 ( H 2 O ) − h fo,298 ( Cm Hn ) .
2
In addition, for an arbitrary reaction given by
∑ν A ⎯⎯→ ∑ν
i
i
react.
prod.
i'
Ai'' ,
(1.26)
where Ai and Ai' are the ith reactants and products, respectively, and ν i are termed as the
stoichiometric coefficients, we determine the enthalpy of reaction at an arbitrary temperature
T by
ΔH ro,T = ∑ ν i ' hT ( Ai ' ) − ∑ ν i hT ( Ai )
prod.
react.
⎪⎧
⎪⎫
= ∑ ν i ' h fo,298 ( Ai ' ) − ∑ ν i h fo,298 ( Ai ) + ⎨ ∑ ν i ' ⎡⎣h (T ) − h (298)⎤⎦ i ' − ∑ ν i ⎡⎣h (T ) − h (298)⎤⎦ i ⎬
prod.
react.
react.
⎩⎪ prod.
⎭⎪
⎧⎪
⎫⎪
= ΔH ro,298 + ⎨ ∑ ν i ' ⎡⎣h (T ) − h (298)⎤⎦ i ' − ∑ ν i ⎡⎣h (T ) − h (298)⎤⎦ i ⎬
⎪⎩ prod.
react.
⎭⎪
Since the total numbers of the elements in the reactants and products are identical, the
sensible enthalpy terms for the elements in Eq. (1.25c) are canceled out. If ΔHro,T is positive,
the reaction absorbs heat. This type of reactions is known to be endorthermic.
If ΔHro,T < 0, the reaction releases heat as it proceeds to completion. This type of reactions
is known to be exothermic. Conversely, If ΔHro,T > 0, the reaction requires heat to achieve
completion. This type of reactions is known to be endothermic.
1.5 Chemical Equilibrium
The complete combustion reactions given by Eqs. (1.21) and (1.22) essentially correspond to
maximum heat release. That is, if products other than CO2 and H2O are formed, the
enthalpy of reaction will be decidedly lower. In practical combustion processes, a
combustion reaction can never reach completion. Rather the products of combustion will
acquire the state of chemical equilibrium. Although often than not the products will be
dominated by the complete combustion products, incomplete combustion products (CO, H2,
soot, NO etc) are inherent to a combustion process.
Our experience tells us that a process or reaction would be spontaneous if it releases heat.
For example, the combustion of methane spontaneously produces CO2 and H2O ( ΔHro,298
<0), but a mixture of CO2 and H2O would not spontaneously react and produce methane
and O2. On the other hand, the entropy of 1 mole of CO2 is decidedly smaller than the
entropy for a mixture made of 1 mole of CO and 0.5 mole of O2. Likewise the entropy of 1
1-11
Stanford University
Version 1.2
©Hai Wang
mole of H2O is smaller than the entropy for a mixture made of 1 mole of H2 and 0.5 mole of
O2. Therefore it may be said that reaction (1.21)
CH4 + 2 O2 + (2×79/21) N2 → CO2 + 2H2O + (2×79/21) N2 .
(1.21)
produces the largest heat but with a smaller entropy change, whereas reaction (1.27)
produces less heat but with a larger change of entropy upon reaction:
CH4 + 2 O2 + (2×79/21) N2 → CO + 2H2 + O2 + (2×79/21) N2 .
(1.27)
We learned from the second law of thermodynamics that without heat release or absorption,
a spontaneous process will proceed in the direction to increase entropy. Therefore a
compromise between enthalpy and entropy
releases
mustofbe
Figure 1.1.
Variation
themade.
enthalpy, entropy, and Gibbs
an exothermic
This compromise is responsible for function
chemicalforequilibrium.
It reaction.
may be quantified by the Gibbs
G(T,P) = H(T) - TS(T,P)
function or Gibbs free energy,
G(T,P)
-TS(T,P)
H(T)
0
100% Reactants
Equilibrium
Reaction Progress, ε
1
100% Products
G ≡ H − TS ,
(1.28a)
g ≡ h − Ts ,
(1.28b)
g ≡ h − Ts ,
(1.28c)
Figure 1.1 shows a possible scenario for variation of the enthalpy, entropy times temperature,
Gibbs function for an exothermic reaction as it progresses to completion at a given T and P.
We define here a reaction progress variable ε, such that ε = 0 for pure reactants and ε = 1
for pure products. If the reaction is exothermic, the total enthalpy of the reacting gases
(reactants and products) decreases as ε increases. Here the spontaneous release of chemical
energy is driving the reaction towards completion.
If the reaction is accompanied by a decrease in the entropy (e.g., H2 + 1/2O2 → H2O), the –
TS(T,P) function would monotonically increase as reaction progresses. This entropy
reduction gives resistance towards the completion of reaction. Overall the Gibbs function
must decrease initially, but because of the rise of the –TS term, it eventually will have to go
1-12
Stanford University
Version 1.2
©Hai Wang
up as ε increases. In other words, the Gibbs function must reach a minimum at some point.
The definition of chemical equilibrium is therefore
dG (T , P )
=0,
dε
(1.29a)
dG (T , P ) = 0 .
(1.29b)
or simply
Again, the above equilibrium criterion represents a compromise of H and –ST, since both of
them prefer to minimize themselves. Therefore, the driving force of chemical reaction lies
in the minimization of the Gibbs function.
Now let us consider an arbitrary reaction given by Eq. (1.26). The Gibbs function of the
reacting gas may be written as
G (T , P ) = ∑ ni g i (T , P ) + ∑ ni ' g i ' (T , P ) ,
react.
(1.30)
prod.
where ni is the molar number of the ith species. Putting Eq. (1.30) into (1.29a), we obtain, for
constant T and P,
dG (T , P )
dn
dn
= ∑ g i ( T , P ) i + ∑ g i ' (T , P ) i ' = 0 ,
dε
d ε prod.
dε
react.
(1.31)
Conservation of mass requires that
−
1 dn1
1 dn 2
1 dn N
1 dn1'
1 dn 2 '
1 dn N '
=−
... = −
=
=
=
= γ (ε ) ,
ν1 dε
ν 2 dε
ν n d ε ν 1' d ε ν 2 ' d ε ν n ' d ε
(1.32)
where N and N’ are the total numbers of reactants and products, respectively, and γ (ε ) is a
function that depends only on ε. Combining equations (1.31) and (1.32), we obtain
⎡
⎤
γ ( ε ) ⎢ − ∑ ν i g i ( T , P ) + ∑ ν i ' g i ' (T , P ) ⎥ = 0 .
prod.
⎣ react.
⎦
(1.33)
Since γ (ε ) ≠ 0 , we see that equilibrium state is given by
− ∑ν i g i (T , P ) + ∑ν i ' g i ' (T , P ) = 0 .
react.
prod.
The function g i is the Gibbs function of species i, which may be expressed by
1-13
(1.34)
Stanford University
Version 1.2
©Hai Wang
⎡
⎛ P ⎞⎤
g i (T , P ) = h fo (T ) − Tsi (T ) = h fo (T ) − T ⎢ si o (T ) − Ru ln ⎜ o ⎟ ⎥ .
⎝ P ⎠⎦
⎣
(
(1.35)
)
We now define a standard Gibbs function g o T , P o = 1 atm as
g o (T ) = h fo (T ) − Tsi o (T ) ,
(1.36)
⎛P ⎞
g i (T , P ) = g o (T ) + Ru T ln ⎜ o ⎟ .
⎝P ⎠
(1.37)
and re-write Eq. (1.35) as
Putting Eq. (1.37) into (1.34) and rearranging, we have
∑ν
prod.
⎡
⎛P ⎞
⎛ P ⎞⎤
g (T ) − ∑ν i g io (T ) = −Ru T ⎢ ∑ν i ' ln ⎜ io' ⎟ − ∑ν i ln ⎜ io ⎟ ⎥ ,
⎝ P ⎠ react.
⎝ P ⎠⎦
react.
⎣ prod.
o
i' i'
(1.38)
o
where Pi is the partial pressure of species i, and of course, P = 1 atm. The left-hand side of
the above equation may be defined as the standard Gibbs function change of reaction,
ΔGro (T ) ≡ ∑ν i ' g io' (T ) − ∑ν i g io (T )
prod.
(1.39)
react.
The right-hand side of Eq. (1.38) may be re-arranged to yield
⎛ ∏ Piν' i '
⎜ prod.
ΔGro (T ) = −Ru T ln ⎜
Piν i
⎜∏
⎝ react.
⎞
⎟
⎟.
⎟
⎠
(1.40)
or
K p (T ) ≡
∏ Pν
i'
i'
prod.
∏ Pν
i
react.
i
⎡ ΔGro (T ) ⎤
= exp ⎢ −
⎥.
Ru T ⎦
⎣
(1.41)
o
where Kp(T) is the equilibrium constant of the reaction. Note that by neglecting P in Eqs.
(1.40) and (1.41), we have forced Pi to take the unit of atm.
The equilibrium constant may also be defined by the concentrations of the reactants and
products,
1-14
Stanford University
Version 1.2
K c (T ) ≡
©Hai Wang
∏ cν
i'
i'
prod.
∏c
react.
νi
i
∏ Pν
i'
=
i'
prod.
∏P
νi
( Ru T )
−Δν
= K p (T ) ( R u T )
−Δν
,
(1.42)
i
react.
where
Δν = ∑ v i ' − ∑ v i .
prod.
react.
There are several important facts about the equilibrium constant.
(a) While Kp is defined as the pressure ratio of the products and reactants (Eq. 1.41), this
equilibrium constant is a function of temperature only.
(b) Consider the reaction
H 2O = H 2 + ½ O 2.
(1.43f)
The equilibrium constant for the forward direction of the reaction is
PH2 PO122
K p , f (T ) =
PH2O
.
We may also write the reaction in the back direction,
H 2 + ½ O 2 = H 2O ,
(1.43b)
and its equilibrium constant
K p ,b (T ) =
PH2O
PH2 PO122
.
Obviously,
K p , f (T ) =
1
K p ,b (T )
.
(c) Reaction (43f) may be written alternatively as
2H2O = 2H2 + O2,
with its equilibrium constant
K
'
p, f
(T ) =
PH22 PO2
1-15
PH22O
.
(1.43f’)
Stanford University
Version 1.2
©Hai Wang
Comparing the equilibrium constants for the two forward reactions, we see that
2
K 'p , f (T ) = ⎡⎣K p , f (T )⎤⎦ .
(d) Consider the following two reactions
H2 = 2 H•.
H2O = H• + OH•
(1.44f)
(1.45f)
(where the • denotes that the species is a free radical). We have
K p ,44 f (T ) =
P P
PH2•
and K p ,45 f (T ) = H• OH• .
PH2
PH2O
A linear combination of reactions (43f-45f) yield
H2O = ½ H2 + OH• .
(1.46f)
A little algebra tells us that
(e) While Kp is not a function of pressure, Kc generally is dependent on pressure so long as
Δν ≠ 0 . On the other hand, if Δν = 0 , Kc(T) = Kp(T).
(f) The equilibrium constant of a given reaction may be determined if the enthalpy of
formation and the entropy of reactants and products are known through Eqs. (1.36),
(1.39) and (1.41).
(g) The definition of Kp tells us that the reaction would be more complete if Kp is larger. A
larger Kp may be accomplished with a larger, negative ΔGro (T ) . Combining Eqs. (1.36)
and (39), we see that
⎡
⎤
⎡
⎤
ΔGro (T ) ≡ ⎢ ∑ ν i ' h fo, i ' (T ) − ∑ ν i h fo,i (T )⎥ − T ⎢ ∑ ν i ' si o' (T ) − ∑ ν i si o (T )⎥
react.
react.
⎣ prod.
⎦
⎣ prod.
⎦ , (1.47)
o
o
= ΔH r (T ) − T ΔSr (T )
where ΔSro (T ) is termed as the entropy of reaction. Therefore a large, negative
ΔHro (T ) (reaction being highly exothermic) favors a large Kp , whereas a large, positive
ΔSro (T ) (reaction creating a large amount of entropy) also favors a large Kp or promotes
the completion of the reaction.
1-16
Stanford University
Version 1.2
©Hai Wang
1.6 Adiabatic Flame Temperature
With the concepts of chemical equilibrium understood, we may now try to calculate the
equilibrium composition of a combustion reaction. In doing so, we wish to define the
adiabatic flame temperature. Consider an adiabatic combustion process whereby the
reactants enters into a combustor at temperature T0, and products exit the combustor at the
adiabatic flame temperature Tad. Since the process is adiabatic (Q = 0 ), we have
H prod. (Tad ) − H react. (T0 ) = 0 .
(1.48)
We now expand Eq. (1.48) using the total enthalpy equation for each species (Eq. 1.25b),
∑ν
prod.
h
i ' Tad , i '
⎡
⎤
− ∑ν i hT0 , i = ⎢ ∑ν i ' h fo,298, i ' − ∑ν i h fo,298, i ⎥ + ∑ν i ' ⎡⎣h (Tad ) − h ( 298 )⎤⎦ i '
react.
react.
⎣ prod.
⎦ prod.
.(1.49)
− ∑ν i ⎡⎣h (T0 ) − h ( 298 )⎤⎦ i = 0
react.
Obviously the first term on the right-hand side of Eq. (1.49) is the standard enthalpy of
reaction ΔHro,298 . The second term determines the sensible heat needed to heat the products
from 298 K to the adiabatic flame temperature Tad. To simplify our analysis, we shall assume
that the reactants enter into the reactor at T0 = 298 K so the third term becomes 0.
Rearranging Eq. (1.49), we see that
−ΔHro,298 = ∑ν i ' ⎡⎣h (Tad ) − h ( 298 )⎤⎦ i ' .
(1.50)
prod.
In other words, the adiabatic flame temperature is obtained when all the heat released from a
combustion reaction is used to raise the product temperature from 298 to Tad.
The existence of chemical equilibrium makes the calculation of this adiabatic flame
temperature a bit more involved. Specifically, while the values of ν i are always well defined,
ν i' is not since it is dependent on the equilibrium composition of the products.
Consider the combustion of 1 mole of carbon (graphite) in 1 mole of oxygen at a pressure of
1 atm.
1 mole C(S) + 1 mole O2 → x CO2 + yCO + zO2.
(1.51)
The reactant temperature is 298 K. The principle of chemical equilibrium states that the
products cannot be entirely CO2. Rather, a small amount of CO (y moles) must be produced
along with z moles of O2 unused. These products are in equilibrium at the adiabatic flame
temperature among themselves through
1-17
Stanford University
Version 1.2
©Hai Wang
CO2 = CO + ½ O2 ,
(1.52)
with its equilibrium constant given by
K p (Tad ) =
PCOPO122
PCO2
⎞
yz 1 2 ⎛
P
=
⎜
x ⎝ x + y + z ⎟⎠
12
(
)
= exp −ΔGr0 R u Tad .
(1.53)
(We need to recognize that the products of a combustion process cannot be in equilibrium
with the reactants of the process. Rather it is the products that are in equilibrium among
themselves.)
Since there are four unknowns in Eq. (1.53) (i.e., x, y, z and Tad), we need to provide three
more equations to solve this problem. Two of these equations come from mass
conservation:
Carbon: x + y = 1 mol ,
Oxygen: 2x + y + 2z = 2 mol .
(1.54)
(1.55)
The last equation is given by Eq. (1.50), which may be expanded to give
− ⎡⎣ xh fo,298 (CO2 ) + yh fo,298 (CO )⎤⎦ = x ⎡⎣h (Tad ) − h ( 298 )⎤⎦ CO
2
+ y ⎡⎣h (Tad ) − h ( 298 )⎤⎦ CO + z ⎡⎣h (Tad ) − h ( 298 )⎤⎦ O
.(1.56)
2
Clearly the coupled Eqs (1.53-1.56) cannot be solved analytically. We shall defer to section
1.8 and use Excel to solve the equation. Any more realistic combustion equilibrium
problems will have to be solved numerically by a computer—a topic we will discuss also in
section 1.8.
Figure 1.2 shows the variation of the adiabatic flame temperature as a function of
equivalence ratio for the combustion of methane, propane, ethylene and benzene in air at 1
atm pressure. As expected, the flame temperature peaks at an equivalence ratio around unity,
slightly on the fuel rich side. The decrease of Tad towards smaller φ is caused by dilution of
unused oxygen as well as the greater amount of nitrogen brought into the combustor with
oxygen. The decrease of the flame temperature towards larger φ is because of oxygen
deficiency, which does not allow the fuel to be fully oxidized.
Figure 1.3 shows the effect of pressure on the adiabatic flame temperature. It is seen that as
pressure increases, (a) the adiabatic flame temperature becomes higher and (b) the peak
shifts towards φ = 1. Here it may be noted that the pressure serves to reduce the extent of
dissociation of CO2 and H2O, and in doing so, force the reaction towards better completion.
To explain the variation of Tad as a function of pressure, we plot in Figure 1.4 the
equilibrium composition at φ = 1. It is seen that as pressure decreases, the extent of CO2
and H2O dissociation into H2, O2, CO, and even highly unstable free radical species,
including H•, O•, and OH• becomes more and more significant.
1-18
Stanford University
Version 1.2
©Hai Wang
Adiabatic Flame Temperature, Tad (K)
2500
Figure 1.3.
Adiabatic
flame
temperature
ofof
Figure
1.4. 1.2.Mole
fractions
Figure
Adiabatic
propane
combustion
inflame
air
equilibrium
productsofcomputed
for
temperature
methane,
at several pressures.
propane
propane,
combustion
ethyleneinand
air (benzene
φ=1).
at 1 atm pressure.
2000
ethylene
1500
benzene
propane
methane
1000
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Adiabatic Flame Temperature, Tad (K)
Equivalence Ratio, φ
100 atm
10 atm
2300
1 atm
2200
0.1 atm
2100
0.85
0.90
0.95
1.00
1.05
1.10
1.15
Equivalence Ratio, φ
H2O
Mole Fraction
10-1
CO2
CO
10-2
O2
OH
H2
10-3
H
O
10-4
10-5
0.1
1
10
100
Pressure, P (atm)
1-19
1.20
Stanford University
Version 1.2
©Hai Wang
1.7 Tabulation and Mathematical Parameterization of Thermochemical Properties
Key thermodynamic or thermochemical properties discussed so far include the constantpressure specific heat, sensible enthalpy, enthalpy of formation, and entropy under the
standard condition. It is important to recognize that under the ideal gas condition, c p is not
a function of pressure, but it is a function of temperature.
A common method is to tabulate, among others, c p (T ) , s o (T ) , h(T)–h(298) and h fo (T ) as
a function of temperature. Such table is known as the JANAF table.1 Appendix 1.A gives a
truncated version of these tables for species listed in Table 1.2. In research publications,
these tables are usually condensed to something that looks like Table 1.3.
Table 1.3. Thermochemical properties of selected species in the vapor state.
Species
h fo ( 298K )
s o ( 298K )
(kJ/mol)
(J/mol-K)
c p (T ) (J/mol-K)
298
500
1000
1500
2000
2500
C(S)
0
5.730
8.523
14.596
21.624
23.857
25.167
25.976
H2
0
130.663
28.834
29.297
30.163
32.358
34.194
35.737
O2
0
205.127
29.377
31.082
34.881
36.505
37.855
38.999
H 2O
-241.8
188.810
33.587
35.214
41.294
47.333
51.678
54.731
CO
-110.5
197.640
29.140
29.811
33.163
35.132
36.288
36.917
CO2
-393.5
213.766
37.128
44.628
54.322
58.222
60.462
61.640
CH4
-74.6
186.351
35.685
46.494
73.616
90.413
100.435
106.864
C 2H 6
-83.9
229.051
52.376
77.837
122.540
144.761
158.280
165.774
C 3H 8
-103.8
270.141
73.530
112.409
174.614
204.334
222.359
232.305
C4H10
-125.8
309.686
98.571
148.552
227.379
264.424
286.823
299.106
C8H18
-208.7
466.772
187.486
286.282
431.399
494.910
534.404
557.447
52.3
219.156
42.783
62.321
93.860
109.190
118.563
123.799
C 2H 4
C 2H 2
227.4
200.892
43.989
54.715
67.908
75.906
81.045
84.262
-200.9
239.785
44.030
59.526
89.656
105.425
113.891
118.560
82.8
269.020
82.077
138.240
210.948
240.242
257.667
267.023
H•
218.0
114.706
20.786
20.786
20.786
20.786
20.786
20.786
O•
249.2
161.047
21.912
21.247
20.924
20.846
20.826
20.853
39.3
183.722
29.887
29.483
30.694
32.948
34.755
36.077
CH3OH
C 6H 6
OH•
The thermochemical data of a large number of species has been compiled by the National
Institute of Standards and Technology.
These data may be found at
http://webbook.nist.gov/chemistry/. Another web-based source of data is the Active
Tables (ATcT): https://cmcs.ca.sandia.gov/cmcs/portal/user/anon/js_pane.
1
Chase, M. W., Jr., J. Phys. Chem. Ref. Data, 4th Edition, Mono. 9, Suppl. 1 (1998a).
1-20
Stanford University
Version 1.2
©Hai Wang
For combustion calculations, a very good source of thermochemical data is: Alexander
Burcat and Branko Ruscic “Third Millennium Thermodynamic Database for Combustion
and
Air-Pollution
Use,”
2005
(http://www.technion.ac.il/~aer0201/
or
http://garfield.chem.elte.hu/Burcat/burcat.html).
The database is the result of extensive work by Professor Alexander Burcat of Technion
University, Israel over the last 20 years. In this database, the thermochemical data are
expressed in the form of polynomial function and are thus more compact than the JANAF
table. A typical record of thermochemical data may look like the following:
Species name
TLow, Thigh, Tmid
Composition
Reference source
Phase
CO2
L 7/88C
1O
2
0
0G
200.000 6000.000 1000.
0.46365111E+01 0.27414569E-02-0.99589759E-06 0.16038666E-09-0.91619857E-14
-0.49024904E+05-0.19348955E+01 0.23568130E+01 0.89841299E-02-0.71220632E-05
0.24573008E-08-0.14288548E-12-0.48371971E+05 0.99009035E+01-0.47328105E+05
ai (i = 1,7) for Tmid < T < Thigh
1
2
3
4
ai (i = 1,7) for Tlow < T < Tmid
The polynomial fits are made for two separate temperature ranges (Tlow ≤ T ≤ Tmid and Tmid
≤ T ≤ Thigh). There are 7 polynomial coefficients, ai (i,=1,..7), for each temperature range.
The thermochemical data are calculated from these fits as
c p Ru = a1 + a 2T + a 3T 2 + a 4 T 3 + a 5T 4 ,
(1.57)
s Ru = a1 ln T + a2T + a3 T 2 + a4 T 3 + a5 T 4 + a7 .
hT Ru = a1T + a2 T 2 2 + a3 T 3 3 + a4 T 4 4 + a5 T 5 5 + a6
(1.58)
o
2
3
4
(1.59)
where hT is the total enthalpy (Eq. 1.25b), s o is the standard entropy (1 atm). Note that
these equations are in strict agreement with known relationships among the thermochemical
properties, i.e.,
hT (T ) = ∫
T
c p dT + h fo,298.15 ,
(1.60)
c p d ln T + s o ( 298.15 ) .
(1.61)
298.15
s o (T ) = ∫
T
298.15
To calculate and tabulate the thermochemical data properties in the JANAF type form, c p
and s o are directly calculated with Eqs. (1.57) and (1.58). The sensible enthalpy is
determined as
h (T ) − h ( 298.15 ) = hT (T ) − hT ( 298.15 ) ,
(1.62)
where hT is calculated using Eq. (1.59). For enthalpy of formation, we use
h fo (T ) = hT (T ) −
∑ ν h (T )
i T ,i
elements
1-21
(1.63)
Stanford University
Version 1.2
©Hai Wang
where ν i represents the molecular composition of the substance. For example, a CxHyOz
species has ν C = x , ν H2 = y 2 and ν O2 = z 2 .
An EXCEL spreadsheet has been prepared for the JANAF like tabulation. The file may be
downloaded from the course web site
http://melchior.usc.edu/public/AME579/Week_2/NASA_poly to JANAF.xls.
Burcat’s database may also be downloaded in text form
http://melchior.usc.edu/public/AME579/Week_2/Burcat Thermo.txt,
and in Excel form
http://melchior.usc.edu/public/AME579/Week_2/Burcat Thermo.xls.
1.8 Solution of an Equilibrium and Adiabatic Flame Temperature Problem
We shall now return to the problem of carbon (graphite) oxidation in section 1.6. We wish
to calculate the adiabatic flame temperature for combustion of 1 mole of carbon (graphite)
in 1 mole of oxygen at a pressure of 1 atm. The initial temperature is 298 K. The four
equations are
⎧x + y = 1
⎪2 x + y + 2z = 2
⎪
12
⎪ yz 1 2 ⎛
⎞
P
⎪
= exp ( −ΔGr0 Ru Td )
.
⎨ x ⎜⎝ x + y + z ⎟⎠
⎪
⎪− ⎡⎣ xh fo,298 ( CO2 ) + yh fo,298 ( CO )⎤⎦ = x ⎡⎣h (Tad ) − h ( 298 )⎤⎦
CO2
⎪
⎪+ y ⎡h (Tad ) − h ( 298 )⎤ + z ⎡h (Tad ) − h ( 298 )⎤
⎦ CO
⎣
⎦ O2
⎩ ⎣
Solution of the above problem is provided in an Excel sheet downloadable from
http://melchior.usc.edu/public/AME579/Week_2/carbon oxidation.xls. Note that to run
the Excel solver requires the user to download the thermochemical property tables from
http://melchior.usc.edu/public/AME579/Week_2/NASA_poly to JANAF.xls. This file
should be placed in the same directory as the carbon oxidation.xls file.
The solution of this set of nonlinear algebraic equations gives
⎧x = 0.233 mol
⎪ y = 0.767 mol
⎪
⎨
⎪z = 0.384 mol
⎪
⎩Tad = 3537 K
1-22
Stanford University
Version 1.2
©Hai Wang
Now suppose that the carbon is burned in air, instead of pure oxygen, the set of nonlinear
algebraic equations may be revised by including the molar number of N2 (=1 mole
O2×79/21) and the sensible enthalpy required to heat up the nitrogen,
⎧x + y = 1
⎪2 x + y + 2z = 2
⎪
12
⎪ yz 1 2 ⎛
⎞
P
⎪
= exp ( −ΔGr0 Ru Td )
.
⎨ x ⎜⎝ x + y + z + 79 21 ⎟⎠
⎪
⎪− ⎡⎣ xh fo,298 ( CO2 ) + yh fo,298 ( CO )⎤⎦ = x ⎡⎣h (Tad ) − h ( 298 )⎤⎦
CO2
⎪
⎪+ y ⎡h (Tad ) − h ( 298 )⎤ + z ⎡h (Tad ) − h ( 298 )⎤ + 79 21 ⎡h (Tad ) − h ( 298 )⎤
⎦ CO
⎣
⎦ O2
⎣
⎦ N2
⎩ ⎣
The solution is
⎧x = 0.893 mol
⎪ y = 0.107 mol
⎪
⎨
⎪z = 0.053 mol
⎪
⎩Tad = 2312 K
Comparing the two sets of solution, we find that (a) the adiabatic flame temperature is
notably lower when air is used, and (b) the reaction is less complete when pure oxygen is
used because the higher adiabatic flame temperature forces a greater extent of CO2
dissociation into CO and O2.
Two commonly used equilibrium solvers are Stanjan (or the equilibrium solver – EQUIL of the ChemKin suite of package) and the NASA Equilibrium code (cec86). We will use the
equilibrium solver of the ChemKin suite of package for the current class. Instructions about
the computer code can be found on p. 30.
1-23
Stanford University
Version 1.2
©Hai Wang
Appendix A1. Truncated JANAF tables
Graphite (C(S))
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
8.523
8.592
11.832
14.596
16.836
18.559
19.835
20.794
21.624
22.169
22.660
23.102
23.499
23.857
24.177
24.465
24.724
24.957
25.167
25.358
25.531
25.691
25.838
25.976
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
(kJ/mol)
0.000
0.017
1.042
2.368
3.944
5.717
7.640
9.674
11.795
13.985
16.227
18.515
20.846
23.214
25.616
28.048
30.508
32.992
35.498
38.025
40.569
43.131
45.707
48.298
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
5.730
5.787
8.713
11.659
14.526
17.257
19.823
22.217
24.451
26.538
28.489
30.320
32.047
33.681
35.231
36.705
38.111
39.454
40.740
41.972
43.156
44.295
45.391
46.449
1-24
h fo (T )
g of (T )
(kJ/mol)
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
Stanford University
Version 1.2
©Hai Wang
Hydrogen (H2)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
28.834
28.850
29.277
29.297
29.254
29.344
29.615
29.970
30.163
30.634
31.089
31.527
31.950
32.358
32.752
33.132
33.499
33.853
34.194
34.525
34.843
35.151
35.449
35.737
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
(kJ/mol)
130.663
130.856
139.229
145.768
151.105
155.619
159.553
163.062
166.232
169.130
171.815
174.320
176.672
178.891
180.992
182.989
184.893
186.714
188.459
190.135
191.749
193.305
194.807
196.260
0.000
0.058
2.969
5.900
8.827
11.755
14.702
17.681
20.690
23.730
26.817
29.948
33.122
36.337
39.593
42.887
46.219
49.586
52.989
56.425
59.893
63.393
66.923
70.483
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1-25
h fo (T )
g of (T )
(kJ/mol)
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
Stanford University
Version 1.2
©Hai Wang
Oxygen (O2)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
29.377
29.387
30.120
31.082
32.080
32.990
33.747
34.355
34.881
35.232
35.569
35.893
36.205
36.505
36.795
37.074
37.343
37.604
37.855
38.098
38.334
38.562
38.784
38.999
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
(kJ/mol)
205.127
205.323
213.870
220.692
226.448
231.463
235.919
239.930
243.578
246.919
249.999
252.859
255.530
258.038
260.404
262.643
264.769
266.795
268.731
270.584
272.361
274.070
275.716
277.304
0.000
0.059
3.031
6.090
9.249
12.503
15.842
19.248
22.710
26.216
29.756
33.329
36.934
40.569
44.234
47.928
51.649
55.396
59.169
62.967
66.789
70.634
74.501
78.390
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1-26
h fo (T )
g of (T )
(kJ/mol)
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
Stanford University
Version 1.2
©Hai Wang
Water vapor (H2O(v))
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
33.587
33.596
34.268
35.214
36.320
37.508
38.733
39.986
41.294
42.659
43.941
45.145
46.275
47.333
48.324
49.251
50.117
50.925
51.678
52.380
53.034
53.641
54.206
54.731
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
188.810
189.034
198.783
206.527
213.044
218.731
223.819
228.453
232.733
236.734
240.501
244.066
247.454
250.683
253.770
256.727
259.567
262.299
264.930
267.469
269.921
272.292
274.587
276.811
0.000
0.067
3.458
6.930
10.506
14.197
18.008
21.944
26.007
30.206
34.536
38.991
43.563
48.244
53.027
57.907
62.876
67.928
73.059
78.262
83.533
88.867
94.260
99.707
1-27
h fo (T )
(kJ/mol)
-241.821
-241.841
-242.849
-243.836
-244.767
-245.632
-246.435
-247.182
-247.859
-248.454
-248.979
-249.442
-249.847
-250.199
-250.504
-250.766
-250.989
-251.178
-251.336
-251.468
-251.576
-251.664
-251.735
-251.792
g of (T )
(kJ/mol)
-228.585
-228.496
-223.896
-219.043
-213.996
-208.798
-203.481
-198.066
-192.571
-187.013
-181.404
-175.753
-170.070
-164.359
-158.626
-152.875
-147.110
-141.334
-135.548
-129.756
-123.957
-118.154
-112.348
-106.539
Stanford University
Version 1.2
©Hai Wang
Carbon monoxide (CO)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
29.140
29.143
29.375
29.811
30.415
31.133
31.894
32.606
33.163
33.637
34.069
34.460
34.813
35.132
35.418
35.674
35.903
36.107
36.288
36.447
36.589
36.713
36.822
36.917
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
197.640
197.835
206.246
212.844
218.330
223.071
227.278
231.077
234.543
237.726
240.672
243.414
245.981
248.394
250.671
252.826
254.872
256.818
258.675
260.449
262.148
263.778
265.342
266.848
0.000
0.058
2.982
5.940
8.950
12.027
15.178
18.404
21.694
25.035
28.420
31.847
35.311
38.808
42.336
45.891
49.470
53.071
56.691
60.327
63.979
67.645
71.321
75.009
1-28
h fo (T )
(kJ/mol)
-110.529
-110.518
-110.105
-110.002
-110.147
-110.472
-110.912
-111.423
-111.985
-112.588
-113.214
-113.862
-114.531
-115.220
-115.926
-116.651
-117.392
-118.149
-118.922
-119.710
-120.514
-121.332
-122.166
-123.014
g of (T )
(kJ/mol)
-137.155
-137.334
-146.344
-155.421
-164.495
-173.529
-182.509
-191.428
-200.288
-209.089
-217.835
-226.527
-235.168
-243.761
-252.308
-260.810
-269.268
-277.685
-286.062
-294.400
-302.699
-310.962
-319.189
-327.381
Stanford University
Version 1.2
©Hai Wang
Carbon dioxide (CO2)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
37.128
37.218
41.286
44.628
47.359
49.589
51.425
52.970
54.322
55.268
56.124
56.899
57.596
58.222
58.782
59.281
59.724
60.116
60.462
60.765
61.031
61.263
61.464
61.640
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
213.766
214.015
225.297
234.881
243.268
250.741
257.487
263.635
269.288
274.510
279.356
283.880
288.122
292.118
295.894
299.473
302.874
306.114
309.206
312.164
314.997
317.715
320.326
322.839
0.000
0.074
4.006
8.307
12.911
17.762
22.816
28.038
33.404
38.884
44.454
50.106
55.831
61.623
67.473
73.377
79.328
85.320
91.349
97.411
103.501
109.616
115.753
121.908
1-29
h fo (T )
(kJ/mol)
-393.505
-393.506
-393.572
-393.655
-393.786
-393.963
-394.171
-394.389
-394.606
-394.821
-395.033
-395.243
-395.453
-395.665
-395.882
-396.104
-396.334
-396.573
-396.823
-397.086
-397.362
-397.653
-397.960
-398.285
g of (T )
(kJ/mol)
-394.372
-394.377
-394.658
-394.920
-395.162
-395.378
-395.566
-395.727
-395.865
-395.980
-396.076
-396.154
-396.217
-396.264
-396.297
-396.316
-396.322
-396.314
-396.294
-396.262
-396.216
-396.157
-396.086
-396.001
Stanford University
Version 1.2
©Hai Wang
Methane (CH4)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
35.685
35.760
40.530
46.494
52.730
58.650
63.998
68.850
73.616
77.713
81.404
84.728
87.720
90.413
92.839
95.028
97.007
98.802
100.435
101.929
103.302
104.573
105.756
106.864
186.351
186.590
197.488
207.161
216.190
224.769
232.957
240.778
248.277
255.489
262.412
269.061
275.452
281.597
287.511
293.206
298.695
303.989
309.099
314.036
318.809
323.430
327.906
332.246
c p (T )
0.000
0.071
3.871
8.217
13.179
18.752
24.889
31.535
38.656
46.226
54.185
62.495
71.120
80.029
89.194
98.589
108.192
117.984
127.947
138.066
148.329
158.724
169.241
179.872
1-30
h fo (T )
(kJ/mol)
-74.594
-74.655
-77.704
-80.544
-83.013
-85.070
-86.749
-88.096
-89.114
-89.814
-90.269
-90.510
-90.564
-90.454
-90.202
-89.828
-89.348
-88.775
-88.124
-87.403
-86.622
-85.788
-84.908
-83.986
g of (T )
(kJ/mol)
-50.544
-50.383
-41.830
-32.527
-22.685
-12.462
-1.970
8.711
19.525
30.425
41.378
52.360
63.353
74.344
85.323
96.282
107.217
118.122
128.994
139.833
150.635
161.401
172.130
182.821
Stanford University
Version 1.2
©Hai Wang
Ethane (C2H6)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
52.376
52.642
65.627
77.837
89.040
99.101
107.983
115.743
122.540
127.823
132.662
137.081
141.106
144.761
148.069
151.053
153.735
156.138
158.280
160.183
161.866
163.347
164.644
165.774
229.051
229.402
246.340
262.313
277.511
292.007
305.833
319.010
331.564
343.495
354.827
365.623
375.932
385.794
395.244
404.311
413.023
421.400
429.465
437.234
444.725
451.954
458.934
465.679
c p (T )
0.000
0.105
6.024
13.205
21.558
30.974
41.338
52.533
64.455
76.977
90.005
103.495
117.408
131.704
146.348
161.307
176.549
192.045
207.768
223.693
239.797
256.059
272.460
288.982
1-31
h fo (T )
(kJ/mol)
-83.854
-83.956
-88.822
-93.084
-96.664
-99.580
-101.902
-103.711
-105.061
-106.039
-106.753
-107.232
-107.503
-107.589
-107.516
-107.305
-106.978
-106.553
-106.050
-105.486
-104.877
-104.236
-103.579
-102.916
g of (T )
(kJ/mol)
-31.884
-31.534
-13.313
6.072
26.250
46.975
68.077
89.438
110.975
132.629
154.359
176.139
197.949
219.771
241.593
263.407
285.204
306.981
328.733
350.459
372.156
393.825
415.466
437.079
Stanford University
Version 1.2
©Hai Wang
Propane (C3H8)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
73.530
73.949
94.097
112.409
128.644
142.749
154.863
165.312
174.614
181.688
188.162
194.072
199.452
204.334
208.751
212.732
216.308
219.508
222.359
224.889
227.124
229.090
230.809
232.305
270.141
270.634
294.707
317.709
339.671
360.587
380.460
399.318
417.224
434.203
450.294
465.592
480.174
494.104
507.435
520.212
532.475
544.257
555.590
566.502
577.016
587.157
596.944
606.397
c p (T )
0.000
0.147
8.564
18.906
30.976
44.563
59.459
75.480
92.483
110.303
128.801
147.917
167.598
187.791
208.449
229.527
250.982
272.776
294.872
317.237
339.840
362.653
385.649
408.807
1-32
h fo (T )
(kJ/mol)
-103.842
-103.977
-110.281
-115.638
-120.004
-123.452
-126.111
-128.108
-129.506
-130.416
-130.989
-131.262
-131.268
-131.042
-130.612
-130.009
-129.259
-128.389
-127.422
-126.380
-125.285
-124.155
-123.008
-121.861
g of (T )
(kJ/mol)
-23.472
-22.932
5.058
34.533
64.993
96.111
127.667
159.515
191.554
223.707
255.927
288.184
320.450
352.709
384.947
417.151
449.316
481.436
513.507
545.529
577.499
609.418
641.288
673.110
Stanford University
Version 1.2
©Hai Wang
Normal butane (C4H10)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
98.571
99.112
125.064
148.552
169.281
187.205
202.523
215.684
227.379
236.202
244.275
251.642
258.344
264.424
269.920
274.871
279.314
283.287
286.823
289.958
292.722
295.148
297.267
299.106
309.686
310.347
342.467
372.947
401.906
429.383
455.407
480.039
503.378
525.470
546.373
566.221
585.119
603.154
620.398
636.913
652.752
667.962
682.585
696.656
710.210
723.277
735.884
748.058
c p (T )
0.000
0.198
11.425
25.128
41.043
58.890
78.397
99.323
121.485
144.670
168.700
193.502
219.006
245.150
271.872
299.116
326.829
354.963
383.472
412.314
441.451
470.847
500.470
530.291
1-33
h fo (T )
(kJ/mol)
-125.767
-125.926
-133.355
-139.607
-144.631
-148.521
-151.440
-153.544
-154.914
-155.689
-156.057
-156.064
-155.751
-155.158
-154.322
-153.280
-152.063
-150.705
-149.234
-147.677
-146.061
-144.409
-142.742
-141.081
g of (T )
(kJ/mol)
-16.535
-15.802
22.056
61.658
102.403
143.898
185.882
228.181
270.675
313.275
355.927
398.595
441.252
483.876
526.452
568.969
611.420
653.800
696.104
738.333
780.486
822.565
864.572
906.509
Stanford University
Version 1.2
©Hai Wang
Normal octane (C8H18)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
187.486
188.566
240.137
286.282
326.349
360.219
388.302
411.539
431.399
446.430
460.225
472.856
484.394
494.910
504.469
513.136
520.974
528.044
534.404
540.110
545.217
549.775
553.836
557.447
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
466.772
468.029
529.457
588.108
643.939
696.868
746.862
793.980
838.389
880.221
919.665
957.009
992.480
1026.265
1058.515
1089.363
1118.919
1147.279
1174.529
1200.743
1225.989
1250.328
1273.813
1296.497
0.000
0.376
21.852
48.222
78.905
113.285
150.756
190.783
232.950
276.852
322.195
368.858
416.730
465.703
515.680
566.567
618.279
670.736
723.865
777.596
831.867
886.621
941.805
997.373
1-34
h fo (T )
(kJ/mol)
-208.731
-209.011
-221.937
-232.547
-240.815
-246.981
-251.414
-254.468
-256.354
-257.332
-257.701
-257.523
-256.861
-255.773
-254.313
-252.534
-250.484
-248.210
-245.755
-243.159
-240.461
-237.695
-234.894
-232.087
g of (T )
(kJ/mol)
16.269
17.780
95.386
175.994
258.516
342.252
426.749
511.715
596.959
682.343
767.789
853.243
938.664
1024.023
1109.297
1194.470
1279.530
1364.470
1449.285
1533.973
1618.536
1702.974
1787.291
1871.491
Stanford University
Version 1.2
©Hai Wang
Ethylene (C2H4)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
42.783
42.992
53.098
62.321
70.529
77.685
83.843
89.153
93.860
97.500
100.836
103.885
106.664
109.190
111.479
113.546
115.406
117.074
118.563
119.889
121.062
122.097
123.005
123.799
219.156
219.443
233.213
246.070
258.174
269.597
280.383
290.572
300.213
309.332
317.961
326.154
333.956
341.403
348.524
355.346
361.889
368.174
374.218
380.036
385.640
391.045
396.261
401.299
c p (T )
0.000
0.086
4.897
10.676
17.327
24.746
32.831
41.487
50.641
60.212
70.131
80.369
90.899
101.694
112.729
123.982
135.431
147.057
158.840
170.764
182.813
194.972
207.228
219.569
1-35
h fo (T )
(kJ/mol)
52.298
52.234
49.173
46.440
44.085
42.100
40.445
39.075
37.968
37.079
36.342
35.742
35.262
34.890
34.610
34.410
34.276
34.198
34.164
34.163
34.185
34.222
34.265
34.306
g of (T )
(kJ/mol)
68.280
68.387
74.241
80.832
87.938
95.408
103.141
111.063
119.123
127.283
135.517
143.807
152.138
160.500
168.884
177.282
185.691
194.105
202.522
210.940
219.358
227.774
236.188
244.601
Stanford University
Version 1.2
©Hai Wang
Acetylene (C2H2)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
43.989
44.134
50.268
54.715
58.110
60.916
63.429
65.775
67.908
69.787
71.517
73.108
74.568
75.906
77.130
78.248
79.269
80.199
81.045
81.816
82.516
83.153
83.733
84.262
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
200.892
201.187
214.772
226.492
236.780
245.954
254.254
261.862
268.905
275.467
281.614
287.402
292.874
298.065
303.004
307.714
312.216
316.527
320.662
324.636
328.458
332.140
335.692
339.121
0.000
0.088
4.825
10.086
15.734
21.688
27.907
34.369
41.055
47.941
55.008
62.240
69.625
77.150
84.802
92.572
100.449
108.423
116.486
124.629
132.847
141.131
149.475
157.875
1-36
h fo (T )
(kJ/mol)
227.397
227.393
227.170
226.848
226.417
225.896
225.322
224.737
224.172
223.638
223.135
222.659
222.209
221.782
221.375
220.986
220.612
220.249
219.897
219.552
219.211
218.873
218.535
218.194
g of (T )
(kJ/mol)
209.884
209.766
203.922
198.146
192.444
186.822
181.279
175.808
170.402
165.051
159.748
154.485
149.258
144.062
138.894
133.751
128.631
123.530
118.449
113.385
108.338
103.306
98.288
93.285
Stanford University
Version 1.2
©Hai Wang
Methanol vapor (CH3OH(v))
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
44.030
44.155
51.400
59.526
67.334
74.146
79.807
84.681
89.656
93.611
97.125
100.239
102.994
105.425
107.570
109.459
111.124
112.593
113.891
115.042
116.067
116.984
117.810
118.560
239.785
240.080
253.735
266.074
277.628
288.534
298.817
308.502
317.675
326.410
334.709
342.608
350.140
357.330
364.204
370.783
377.088
383.136
388.945
394.530
399.906
405.086
410.083
414.908
c p (T )
0.000
0.088
4.852
10.397
16.746
23.830
31.536
39.764
48.475
57.642
67.183
77.054
87.219
97.642
108.294
119.148
130.178
141.366
152.691
164.139
175.696
187.349
199.089
210.908
1-37
h fo (T )
(kJ/mol)
-200.932
-201.005
-204.576
-207.747
-210.407
-212.581
-214.360
-215.827
-216.987
-217.843
-218.487
-218.953
-219.269
-219.463
-219.556
-219.571
-219.523
-219.429
-219.301
-219.151
-218.987
-218.817
-218.647
-218.482
g of (T )
(kJ/mol)
-162.241
-161.981
-148.427
-134.013
-119.008
-103.597
-87.902
-72.003
-55.958
-39.811
-23.597
-7.336
8.955
25.264
41.583
57.905
74.226
90.543
106.854
123.158
139.455
155.744
172.025
188.300
Stanford University
Version 1.2
©Hai Wang
Benzene (C6H6)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
s o (T )
h(T)-h(298)
(J/mol-K)
(J/mol-K)
(kJ/mol)
82.077
82.718
112.650
138.240
159.404
176.423
189.948
200.995
210.948
217.961
224.361
230.183
235.466
240.242
244.545
248.408
251.863
254.939
257.667
260.074
262.187
264.033
265.637
267.023
269.020
269.572
297.567
325.536
352.677
378.577
403.052
426.081
447.775
468.215
487.458
505.651
522.905
539.317
554.961
569.905
584.203
597.905
611.052
623.683
635.832
647.528
658.800
669.673
c p (T )
0.000
0.165
9.968
22.550
37.468
54.292
72.636
92.198
112.798
134.249
156.370
179.101
202.388
226.178
250.421
275.072
300.089
325.432
351.065
376.955
403.070
429.383
455.869
482.503
1-38
h fo (T )
(kJ/mol)
82.818
82.707
77.627
73.463
70.145
67.540
65.507
63.930
62.775
61.966
61.377
60.985
60.767
60.701
60.766
60.940
61.203
61.538
61.925
62.349
62.791
63.237
63.673
64.085
g of (T )
(kJ/mol)
129.707
130.023
146.585
164.325
182.823
201.817
221.144
240.698
260.405
280.209
300.077
319.986
339.918
359.860
379.803
399.738
419.660
439.566
459.452
479.318
499.163
518.988
538.794
558.582
Stanford University
Version 1.2
©Hai Wang
Hydrogen Atom (H•)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
20.786
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
114.706
114.845
120.825
125.463
129.253
132.457
135.232
137.681
139.871
141.852
143.660
145.324
146.865
148.299
149.640
150.900
152.088
153.212
154.278
155.293
156.260
157.184
158.068
158.917
0.000
0.042
2.120
4.199
6.277
8.356
10.435
12.513
14.592
16.670
18.749
20.828
22.906
24.985
27.063
29.142
31.221
33.299
35.378
37.456
39.535
41.614
43.692
45.771
1-39
h fo (T )
(kJ/mol)
217.995
218.007
218.630
219.243
219.859
220.473
221.078
221.667
222.241
222.801
223.340
223.858
224.355
224.830
225.284
225.718
226.131
226.525
226.900
227.257
227.596
227.920
228.227
228.520
g of (T )
(kJ/mol)
203.281
203.183
198.146
192.955
187.639
182.220
176.714
171.133
165.488
159.785
154.032
148.235
142.399
136.529
130.627
124.698
118.743
112.766
106.769
100.754
94.722
88.675
82.614
76.541
Stanford University
Version 1.2
©Hai Wang
Oxygen atom (O•)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
21.912
21.900
21.462
21.247
21.138
21.062
20.987
20.923
20.924
20.903
20.885
20.870
20.857
20.846
20.838
20.832
20.828
20.826
20.826
20.827
20.831
20.837
20.844
20.853
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
161.047
161.194
167.427
172.190
176.053
179.306
182.113
184.581
186.785
188.778
190.596
192.267
193.813
195.252
196.597
197.860
199.051
200.177
201.245
202.261
203.230
204.156
205.043
205.894
0.000
0.044
2.209
4.344
6.462
8.572
10.675
12.770
14.861
16.953
19.042
21.130
23.216
25.301
27.385
29.469
31.552
33.635
35.717
37.800
39.883
41.966
44.050
46.135
1-40
h fo (T )
(kJ/mol)
249.171
249.185
249.864
250.469
251.008
251.491
251.924
252.316
252.677
253.014
253.329
253.626
253.905
254.169
254.418
254.653
254.875
255.085
255.283
255.471
255.647
255.813
255.968
256.113
g of (T )
(kJ/mol)
231.743
231.626
225.669
219.548
213.312
206.991
200.604
194.165
187.684
181.168
174.623
168.052
161.458
154.846
148.216
141.571
134.913
128.242
121.561
114.871
108.171
101.464
94.750
88.029
Stanford University
Version 1.2
©Hai Wang
Hydroxyl radical (OH•)
T
(K)
298
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2100
2200
2300
2400
2500
c p (T )
(J/mol-K)
29.887
29.878
29.573
29.483
29.534
29.684
29.916
30.240
30.694
31.174
31.641
32.093
32.529
32.948
33.349
33.730
34.091
34.433
34.755
35.056
35.339
35.602
35.848
36.077
s o (T )
h(T)-h(298)
(J/mol-K)
(kJ/mol)
183.722
183.922
192.469
199.056
204.434
208.997
212.975
216.516
219.724
222.672
225.405
227.955
230.350
232.608
234.748
236.781
238.719
240.572
242.346
244.049
245.686
247.263
248.784
250.252
0.000
0.060
3.030
5.982
8.931
11.892
14.871
17.878
20.923
24.017
27.157
30.344
33.575
36.850
40.165
43.519
46.910
50.336
53.796
57.287
60.806
64.354
67.926
71.523
1-41
h fo (T )
(kJ/mol)
39.346
39.348
39.376
39.333
39.240
39.108
38.945
38.759
38.569
38.389
38.217
38.052
37.895
37.744
37.599
37.459
37.324
37.193
37.065
36.939
36.813
36.687
36.560
36.432
g of (T )
(kJ/mol)
34.631
34.599
33.010
31.422
29.848
28.292
26.758
25.246
23.754
22.282
20.825
19.382
17.952
16.533
15.124
13.723
12.331
10.946
9.568
8.196
6.830
5.470
4.116
2.767