CHEMICAL THERMODYNAMICS
A. ENTHALPY OF FORMATION AND COMBUSTION
(In search of the Criterion of Feasibility)
1. First Law of Thermodynamics (Joule 1843 - 48)
dU = đQ – đW
or
đQ = dU + đW
The energy of a system is equal to the sum of the heat and the work done on
the system. Note dU = dE in some texts.
Remember Eq. of state and exact differentials
1. Define Enthalpy (H)
dH = dU + d(PV)
dH = đQ - đW + PdV + VdP
Copyright © 2014 R. R. Dickerson
1
For example, consider the burning of graphitic carbon in
oxygen, the change in enthalpy is:
Cgraph + O2 → CO2
Ho
-94.0 kcal/mole
In the combustion of 1 mole (12 g) of pure carbon as graphite (a
reasonable approximation of coal) 94 kcal are released. This is
enough to raise the temperature of 1.0 L of water 94 °C.
The superscript ° stands for standard conditions (25 oC and
1.00 atm). Because we started with elements in their standard
state H° = Hfo the standard heat of formation for CO2.
There is a table of Hfo in Pitts & Pitts, Appendix I, p. 1031.
Copyright © 2009 R. R.
Dickerson & Z.Q. Li
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At constant pressure and if the only work is done
against the atmosphere, i.e. PdV work, then
đW = PdV
dHp = đQp
and đQ is now an exact differential - that is
independent of path. Enthalpy is an especially
useful expression of heat.
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The burning of graphitic carbon might proceed
through formation of CO:
Ho or Hfo
Cgraph + O2 → CO2
-94.0 kcal/mole
Cgraph + 1/2 O2 → CO
-26.4
CO + 1/2 O2 → CO2
-67.6
---------------------------------------------------NET Cgraph + O2 → CO2
-94.0 kcal/mole
This is Hess' law – the enthalpy of a reaction is independent
of the mechanism (path). The units of kcal are commonly
used because Hfo is usually measured with Dewars and
change in water temperature.
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Dickerson & Z.Q. Li
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Heat capacity:
The amount of heat required to produce a one degree
change in temp in a given substance.
C = đQ/dT
Cp = (∂Q/∂T)p = (∂H/∂T)p
Cv = (∂Q/∂T)v = (∂U/∂T)v
Because đQp = dH and đQv = dU
For an ideal gas PV = nRT
Cp = Cv + R
Where R = 2.0 cal mole-1 K-1 =
287(J kg-1 K-1)* 0.239 (cal/J) * 29.0E-3 (kg/mole)
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Dickerson & Z.Q. Li
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The heat capacity depends on degrees of freedom in the
molecule enjoys.
Translation = 1/2 R each
(every gas has 3 translational degrees of freedom)
Rotation = 1/2 R
Vibration = R
For a gas with N atoms you see 3N total degrees of
freedom and 3N - 3 internal (rot + vib) degrees of
freedom.
Equipartition principle: As a gas on warming takes up
energy in all its available degrees of freedom.
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Dickerson & Z.Q. Li
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Measured Heat Capacities (cal mole-1 K-1)
Cv
Cp
He
3.0
5.0
Ar
3.0
5.0
O2
5.0
7.0
N2
4.95
6.9
CO
5.0
6.9
CO2
6.9
9.0
SO2
7.3
9.3
H2O
6.0
8.0
Cv = R/2 x (T.D.F.) + R/2 x (R.D.F.) + R x (V.D.F.)
Cp = Cv + R
Translational degrees of freedom: always 3.
Internal degrees of freedom: 3N - 3
Where N is the number of atoms in the molecule.
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Dickerson & Z.Q. Li
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Test Calculation:
Cv(He): 3R/2 = 3.0 cal/(mole K) good!
Cv(O2): 3R/2 + 2(R/2) + 1(R) = (7/2)R ≈ 7.0 cal/(mole K)?
The table shows 5.0; what's wrong?
Not all energy levels are populated at 300 K.
Not all the degrees of freedom are active (vibration).
O2 vibration occurs only with high energy; vacuum uv radiation.
At 2000K Cv (O2) approx 7.0 cal/mole K
Students: show that on the primordial Earth the dry adiabatic lapse rate
was about 12.6 K/km.
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Dickerson & Z.Q. Li
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For one mole of an ideal gas,
P/T = R/V
At constant volume,
dU = Cv dT
Thus
dФ = {Cv dT}/T + {RdV}/V
Integrating
Ф = Cv ln(T) + R ln(V) + Фo
Where Фo is the residual entropy. This equation
lets you calculate entropy for an ideal gas at a
known T and V.
Copyright © 2011 R. R. Dickerson & Z.Q. Li
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GIBBS FREE ENERGY
The Second Law states that for a reversible reaction:
dФ = đQ/T
For an irreversible reaction,
dФ > đQ/T
At constant temperature and pressure for reversible and irreversible
reactions:
dU = đQ - PdV - VdP
dU ≤ đQ - PdV - VdP
dU - TdФ + PdV ≤ 0
Because dP and dT are zero we can add VdP and ФdT to the equation.
dU - TdФ - Ф dT + PdV + VdP ≤ 0
d(U + PV - TФ) ≤ 0
We define G as (U + PV − TФ) or (H − TФ)
Copyright © 2011 R. R. Dickerson & Z.Q. Li
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dG = dH − TdФ
G = H − TФ
G is the Gibbs free energy that stands as the criterion of feasibility. G
tends toward the lowest values, and if G for a reaction is positive, the
reaction cannot proceed!
The Gibbs free energy of a reaction is the sum of the Gibbs free energy
of formation of the products minus the sum of the Gibbs free energy of
formation of the reactants.
Grxn° =
 Gf°(products) -  Gf°(reactants)
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Dickerson & Z.Q. Li
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2. ENTHALPY OF REACTIONS
The heat of a reaction is the sum of the heats of formation of the
products minus the sum of the heats of formation of the reactants.
Hrxn =  Hfo(products) -  Hfo(reactants)
The change of enthalpy of a reaction is fairly independent of
temperature.
EXAMPLE: ENTHALPY CALCULATION
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Dickerson & Z.Q. Li
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3. BOND ENERGIES
See Appendix III of Pitts for a table of bond energies. The
quantity is actually heat not energy.
Definitions:
Bond Dissociation Energy - The amount of energy required to break a
specific bond in a specific molecule.
Bond Energy - The average value for the amount of energy required to
break a certain type of bond in a number of species.
EXAMPLE: O-H in water
We want
H2O → 2H + O
+221 kcal/mole
We add together the two steps:
H2O → OH + H +120
OH → O + H
+101
--------------------------------NET
+221
Bond energy (enthalpy) for the O-H bond is 110.5 kcal/mole, but this is not
Copyright © 2009 R. R.
13
the b.d.e. for either O-H bond.
Dickerson & Z.Q. Li
Another example: What is the C-H bond enthalpy in methane?
We want Ho for the reaction: CH4 → Cgas + 4H
Any path will do (equation of state.)
Ho (kcal/mole)
CH4 + 2 O2 → CO2 + 2H2O
-193
CO2 → Cgraph + O2
+94
2H2O → 2 H2 + O2
+116
2H2 → 4H
+208
Cgraph → Cgas
+171
------------ ----------------------------------NET CH4 → Cgas + 4H
+ 396 kcal/mole
The bond energy for C-H in methane is +396/4 = +99 kcal/mole.
Bond energies are useful for "new" compounds and substances for which b.d.e. can't be
directly measured such as radical.
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Dickerson & Z.Q. Li
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FREE ENERGY
We have a problem, neither internal energy
(E or U) nor enthalpy (H) is the "criterion of
feasibility". Chemical systems generally tend
toward the minimum in E and H, but not
always.
Everyday experience tells us that water
evaporates at room temperature, but this is
uphill in terms of the total energy.
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Dickerson & Z.Q. Li
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Example 1
H2O(l)  H2O(g)
P = 10 torr, T = 25oC
U = + 9.9 kcal/mole
The enthalpy, H, is also positive, about 10
kcal/mole, and PdV is too small to have an
impact.
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Dickerson & Z.Q. Li
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Example 2.
The formation of nitric oxide from nitrogen and
oxygen occurs at combustion temperature.
We know that H >>0 at room temperature, also at
combustion temperatures.
We can calculate H as a function of temperature
with heat capacities, Cp, found in tables; but this is
more detail than is necessary. Remember that
R = 1.99 cal/moleK and dH = CpdT
N2 + O2  2NO
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Dickerson & Z.Q. Li
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N2 + O2  2NO
At room temperature:
H298 = + 43.14 kcal/mole
The reaction is not favored, but combustion
and lightning heat the air, and Cp ≡ (∂H/∂T)p.
Copyright © 2013 R. R.
Dickerson & Z.Q. Li
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Gibbs Free Energy, G, and Equilibrium Constants, Keq
Consider the isothermal expansion of an ideal gas.
dG = VdP
From the ideal gas law,
dG = (nRT/P)dP
Integrating both sides,
2
ò dG
1
=
2
ò
1
æ P2 ö
nRT
dP = nRT ln ç ÷
P
è P1 ø
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Dickerson & Z.Q. Li
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Consider the following reaction,
aA + bB ↔ cC + dD
Where small case letters represent coefficients. For each gas
individually:
G = nRT ln(P2/P1)
æ PAa PBb ö
æ PCc PDd ö
DG = nRT ln ç c d ÷ = -nRT ln ç a b ÷
è PA PB ø
è PC PD ø
Remember what an equilibrium constant is:
æ PCc PDd ö
Keq = ç a b ÷
è PA PB ø
thus
G = -nRT ln (Keq)
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Dickerson & Z.Q. Li
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Gibbs Free Energy and Equilibrium
aA + bB ↔ cC + dD
G° = -nRT ln (Keq)
This holds only for reactants that start (state 1) at standard conditions and
products that finish (state 2) at standard conditions (1.00 atm). Standard
conditions are 25 °C and 1.00 atm pressure. Watch out for units – Gibbs
Free Energy of formation is tabulated for these standard conditions.
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Dickerson & Z.Q. Li
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Example: Lightning
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Dickerson & Z.Q. Li
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Example: Lightning
In the absence of industrial processes, lightning is a major source of
odd nitrogen (NOx) and thus nitrate (NO3-) in the atmosphere. Even today,
lightning is a major source of NOx in the upper free troposphere.
N2 + O2  2NO
How can this happen? Let’s calculate the Gibbs Free Energy for the reaction for
298 K and again for 2000 K.
G° = – nRT ln (Keq)
Gf° = Hf° 0.0 for N2 and O2 Gf° (NO) = 20.7 kcal mole-1
Hf° (NO) = 21.6 kcal mole-1
R = 1.98 cal mole-1 K-1
G° = 2*(20.7) = 41.4 kcal mole-1
Keq = exp (– 41.4E3/1.98*298)
= 3.4 E-31
Copyright © 2011 R. R. Dickerson
æ P2 ö
Keq = ç NO ÷
è PN 2 PO2ø
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G° = 2*(20.7) = 41.4 kcal mole-1
æ P2 ö
Keq = ç NO ÷
è PN 2 PO2ø
= exp (-G°/RT) = 3.4 E-31
Assume PN2 = 0.8 atm; PO2 = 0.2 atm
PNO = - Keq × 0.8× 0.2
= 2.3E-16 atm
(pretty small)
Let’s try again at a higher temperature (2000 K).
Remember H and φ are independent of temperature.
GT ≈ H° – Tφ°
41.4 = 43.2 – 298 φ°
φ° = +6.04E-3 kcal mole-1 K-1
GT ≈ 43.2 – 2000* 6.04E-3
= 31.12 kcal mole-1
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Dickerson & Z.Q. Li
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G(2000) ≈ 31.12 kcal mole-1
Keq = = PNO2/PN2*PO2 = exp (-GT/RT)
= exp (-31.12E3/1.98*2000)
= 3.87E-4
If the total pressure is 1.00 atm
PNO = 7.9E-3 atm
= [0.79% by volume]
You can show that the mole fraction of NO at equilibrium is nearly
independent of pressure. Try repeating this calculation for 2500 K; you
should obtain Keq = 3.4E-3 and [NO] = 2.3%.
In high temperature combustion, such as a car engine or power plant,
NO arises from similar conditions.
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Dickerson & Z.Q. Li
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References
Allen, D. J., and K. E. Pickering, Evaluation of lightning flash rate
parmaterization For use in global chemical transport models, J. Geophys.
Res., 107(23), Art. No. 4711, 2002.
Chameides W. L., et al., NOx production in lightning, J. Atmos. Sci., 34, 143149, 1977.
Rakov V., and M. A. Uman, Lightning: Physics and Effects, University Press,
Cambridge, 2003.
Copyright © 2013 R. R.
Dickerson & Z.Q. Li
26
What is ∆H at 1500K?
If dH = CpdT then H1500 = H298 + Integral from 298 to 1500 of Cp dT
Cp = 2Cp (NO) - Cp (N2) - Cp(O2)
We can approximate Cp with a Taylor expansion.
Cp (O2)/R = 3.0673 + 1.6371x10-3 T - 5.118x10-7 T 2
Cp (N2)/R = 3.2454 + 0.7108x10-3 T - 0.406-7 T 2
Cp (NO)/R = 3.5326 - 0.186x10-3 T - 12.81x10-7 T 2 - 0.547x10-9 T 3
Cp /R = 0.7525 - 2.7199x10-3 T + 26.5448x10-7 T 2 - 1.094x10-9 T 3
1500
-3
2
2

Cp/R
dT

0.7525(150
0
298)
1/2(2.7199
x10
)(1500
298
)

298
1/3(26.544 8x10 -7 )(1500 3 - 2983 ) - 1/4(1.094x 10-9 )(1500 4 - 2984 )
Copyright © 2013 R. R.
Dickerson & Z.Q. Li
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What is H1500 ?
C p | 1500
298  - 454.176 R (K  cal/moleK)
 - 903.81 (cal/mole)
H1500  H 298 - 0.904  43.41 - 0.904   42.506 (kcal/mole )
Almost no change; H is strongly positive and
nearly independen t of temperatu re.
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Dickerson & Z.Q. Li
28
Room Temperature 9/3/2015
Separate
6
5
Mean = 25.2oC
Stdev = 2.04 oC
Frequency
4
3
2
1
0
18
19
20
21
22
23
24
25
26
27
28
29
Temperature (C)
Copyright © 2015 R. R. Dickerson
29
Room Temperature 9/3/2015
Together
7
Mean = 22.3 oC
Stdev = 1.14 oC
6
Frequency
5
4
3
2
1
0
18
19
20
21
22
23
24
25
26
27
28
29
Temperature (C)
Copyright © 2015 R. R. Dickerson
30
What is Your estimated uncertainty?
+/- oC
1.0, 1, 1, 0.2, 0.5, 2.0, 0.5, 0.1, 0.5, 0.6, 5.0, 1.0, “not”
with median 1.0K.
If the data are Gaussian then
95% CI = +/- 2 s
= +/- 2.2 oC
= +/- 4.1 oC (~7oF)
if we use separate measurements; +/-2.3 together.
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