CHEMICAL THERMODYNAMICS

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CHEMICAL THERMODYNAMICS

• A. ENTHALPY OF FORMATION AND COMBUSTION
• (In search of the Criterion of Feasibility)
• 1. First Law of Thermodynamics (Joule 1843 - 48)
• dU dQ dW
• or
• dQ dU dW
• 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 dQ - dW PdV VdP

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For example, consider the burning of graphitic
carbon in oxygen, the change in enthalpy is
?Ho Cgraph O2 ? CO2 -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.
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• At constant pressure and if the only work is done
  against the atmosphere, i.e. PdV work, then
• dW PdV
• dHp dQp
• and dQ is now an exact differential - that is
  independent of path.

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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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Heat capacity The amount of heat required to
produce a one degree change in temp in a given
substance. C dQ/dT Cp (?Q/?T)p
(?H/?T)p Cv (?Q/?T)v (?U/?T)v Because dQp
dH and dQv 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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• 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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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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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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For one mole of an ideal gas, P/T R/V At
constant volume, dU Cv dT Thus d? Cv
dT/T RdV/V Integrating ? Cv ln(T) R
ln(V) ?o Where ?o is the residual entropy.
This equation lets you calculate entropy for an
ideal gas at a known T and V.
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GIBBS FREE ENERGY The Second Law states that
for a reversible reaction d? dQ/T For an
irreversible reaction, d? gt dQ/T At constant
temperature and pressure for reversible and
irreversible reactions dU dQ - PdV - VdP dU
dQ - PdV - VdP dU - Td? PdV 0 Because dP
and dT are zero we can add VdP and ?dT to the
equation. dU - Td? - ? dT PdV VdP 0 d(U
PV - T?) 0 We define G as (U PV - T?) or
(H - T?)
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dG dH - Td? ?G ?H - T?? ?G is the Gibbs
free energy, 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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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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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 the
b.d.e. for either O-H bond.
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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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FREE ENERGY We have a problem, neither energy
(U) nor enthalpy (H) is the "criterion of
feasibility". Chemical systems generally tend
toward the minimum in U and H, but not
always. EXAMPLE 1. Everyday experience tells us
that water evaporates at room temperature, but
this is uphill in terms of the total energy,
U. H2O(l) ? H2O(g) P 10 torr, T 25 C ?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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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 ?P2 ? dG ? nRT/P dP ?G
nRTln(P2/P1) ?1 ?P1
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Consider the following reaction, aA bB ? cC
dD Where small case letters represent
coefficients. ?G nRT ln(P2/P1) ?G nRT
ln (PA)a (PB)b/ (PC)c (PD)d
-nRT ln (PC)c (PD)d / (PA)a (PB)b
Remember what an equilibrium constant is Keq
(PC)c (PD)d / (PA)a (PB)b thus ?G
-nRT ln (Keq)
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IIA. ENTHALPY (HEAT) 1.
FORMATION Definition The enthalpy of formation.
?Hfo is the amount of heat produced or required
to form a substance from its elemental
constituents. The standard conditions,
represented by a super "o", are a little
different from those for the Ideal Gas Law 25oC
(not 0oC), 1.0 atm. and the most stable form of
elements. The standard heat of formation is zero
for elements. This quantity is very useful for
calculating the temperature dependence of
equilibrium constants and maximum allowed rate
constants. It was thought for a long time that ?H
was the criterion of feasibility. Although ?H
tends toward a minimum, it is not the criterion.
Things usually tend toward minimum in ?H, but not
always. Examples are the expansion of a gas into
a vacuum, and the mixing of two fluids. Which
is hotter, an oxygen-acetylene flame or an
oxygen-methane flame? REACTIONS C2H2 2.5O2 ?
2CO2 H2O CH4 2O2 ? CO2 2H2O Note melting
point iron 1535 C. IIB. EXAMPLE 2. The
formation of nitric oxide from nitrogen and
oxygen occurs at combustion temperature. N2 O2
? 2NO We know that ?H gtgt0 at room temperature,
but we know that combustion produces large
quantities of NO. Therefore, these reactions
must be driven by some force other than internal
energy or heat. Note There is little change in
?H with T. You can prove it by integrating ?Cp
(Cp of the products - Cp of the reactants) from
298 to 1500 K. At room temperature ?H298
43.14 kcal/mole ?H1500 42.506 kcal/mole
ENTROPY AND THE SECOND LAW OF THERMODYNAMICS DFN
dS DQ/T For a reversible reaction the change
in entropy, ?S, is a function of state of system
only. To get a feel for what entropy is, let us
derive an expression for the entropy of an ideal
gas. GENERAL RELATIONS dE DQ - PdV dS
DQ/T dS dE/T PdV/T or, for one
mole Keq exp (- ?G/RT ) NOTE Keq 1
if and only if ?G 0.00 If a b is not the
same as c d, we can get into trouble trying to
take the log of an expression with units. For
this type of reaction, the Gibbs free energy is
the sum of the ?G for the chemical reaction and
the ?G for the change in pressure. Assuming that
the reactants start at 1.0 atm and go to an
equilibrium pressure and assuming that the
products finish at 1.0 atm. aA(PA 1) ?
aA(PA) ?G A aRTln (PA/1) Similarly for
B. For the products cC(PC) ? cC(PC 1) ?G C
cRTln (1/ PC) Similarly for D. ?Go
?Grxn ?GA ?GB ?GC ?GD I ?Grxn
aRTln(PA /1) bRTln(PB /1) cRTln(1/PC)
dRTln(1/PD) Combining the log terms, ?Go
?Grxn RT ln (PA)a (PB)b/(PC)c (PD)d For
the equilibrium partial pressures where ?Grxn
0, ?Go RT ln (PA)a (PB)b/(PC)c (PD)d
-RT ln (PC)c (PD)d /(PA)a (PB)b If
you remember that each of the partial pressures
was a ratio with the initial or final pressure
taken as 1.0 and that the ln(1) 0 are left out
you can see that Keq is always dimensionless.
Gibbs Free Energy and Equilibrium
aA bB ? cC dD ?G -nRT ln (Keq) This
holds only for reactants that start at standard
conditions and products that finish 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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Example Lightning
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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? Lets 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 ( 41.4E3/1.98298) 3.4 E-31
PNO2/(PN2PO2)
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?G 2(20.7) 41.4 kcal mole-1 Keq
PNO2/PN2PO2 exp (-?G/RT) 3.4 E-31 Assume
PN2 0.8 atm PO2 0.2 atm PNO vKeq0.80.2
2.3E-31 atm (pretty small) Lets try again at
a higher temperature (2000 K). Remember ?H and ?S
are independent of temperature. ?GT ?H
T?f 41.4 43.2 298 ?f ?f 6.04E-3 kcal
mole-1 K-1 ?GT 43.2 2000 6.04E-3 31.12
kcal mole-1
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?G(2000) 31.12 kcal mole-1 Keq
PNO2/PN2PO2 exp (-?G/RT) exp
(-31.12E3/1.982000) 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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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,
143-149, 1977. Rakov V., and M. A. Uman,
Lightning Physics and Effects, University Press,
Cambridge, 2003.
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Mean 23.4 /- 1.24oC Median 23.3oC
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Mean 23.3 /- 2.1C Median 23.4C
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Keeling Curve Linear growth, exponential
growth CO2o 35 ppm, 60 ppm?? (The
preindustrial mixing ratio was 280
ppm.) Unlabeled axes. 2 said Temp f(CO2) 4 were
perfect.
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Keeling Curve

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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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• 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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Example 2.

• The formation of nitric oxide from nitrogen and
  oxygen occurs at combustion temperature.
• We know that ?H gtgt0 at room temperature, but
  what about at combustion temperature? We can
  calculate ?H as a function of temperature with
  heat capacities, Cp, found in tables. Remember
  that
• R 1.99 cal/moleK and dH Cp dT
• N2 O2 ? 2NO

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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.

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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

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What is ?H1500 ?
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HENRYS LAW EXAMPLE

• What would be the pH of pure rain water in
  Washington, D.C. today? Assume that the
  atmosphere contains only N2, O2, and CO2 and that
  rain in equilibrium with CO2.
• Remember
• H2O H? OH?
• H?OH? 1 x 10?¹4
• pH -logH?
• In pure H2O pH 7.0
• We can measure
• CO2 ca. 370 ppm

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• Todays barometric pressure is 993 hPa 993/1013
  atm 0.98 atm. Thus the partial pressure of CO2
  is
• In water CO2 reacts slightly, but H2CO3 remains
  constant as long as the partial pressure of CO2
  remains constant.

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• We know that
• and
• THUS
• H 2.3x10-6 ? pH -log(2.3x10-6) 5.6
• EXAMPLE 2
• If fog water contains enough nitric acid (HNO3)
  to have a pH of 4.7, can any appreciable amount
  nitric acid vapor return to the atmosphere?
  Another way to ask this question is to ask what
  partial pressure of HNO3 is in equilibrium with
  typical acid rain i.e. water at pH 4.7? We
  will have to assume that HNO3 is 50 ionized.

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• This is equivalent to 90 ppt, a small amount for
  a polluted environment, but the actual HNO3
  would be even lower because nitric acid ionized
  in solution. In other words, once nitric acid is
  in solution, it wont come back out again unless
  the droplet evaporates conversely any
  vapor-phase nitric acid will be quickly absorbed
  into the aqueous-phase in the presence of cloud
  or fog water.
• Which pollutants can be rained out?
• See also Finlayson-Pitts Chapt. 8 and Seinfeld
  Chapt. 7.

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• We want to calculate the ratio of the aqueous
  phase to the gas phase concentration of a
  pollutant in a cloud. The units can be anything
  , but they must be the same. We will assume that
  the gas and aqueous phases are in equilibrium.
  We need the following
• Henrys Law Coefficient H (M/atm)
• Cloud liquid water content LWC (gm?³)
• Total pressure (atm)
• Ambient temperature T (K)
• LET
• be the concentration of X in the aqueous phase
  in moles/m³
• be the concentration of X in the gas phase in
  moles/m³
• Where is the aqueous concentration in M,
  and is the partial pressure expressed in
  atm. We can find the partial pressure from the
  mixing ratio and total pressure.

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• For the aqueous-phase concentration
• units moles/m³ moles/L(water) x
  g(water)/m³(air) x L/g
• For the gaseous content
• units moles/m³

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• Notice that the ratio is independent of pressure
  and concentration. For a species with a Henrys
  law coefficient of 400, only about 1 will go
  into a cloud with a LWC of 1 g/m³. This points
  out the need to consider aqueous reactions.

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• What is the possible pH of water in a high cloud
  (alt. ? 5km) that absorbed sulfur while in
  equilibrium with 100 ppb of SO2?
• In the next lecture we will show how to derive
  the pressure as a function of height. At 5km the
  ambient pressure is 0.54 atm.
• This SO2 will not stay as SO2H2O, but
  participate in a aqueous phase reaction, that is
  it will dissociate.

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• The concentration of SO2H2O, however, remains
  constant because more SO2 is entrained as SO2H2O
  dissociates. The extent of dissociation depends
  on H? and thus pH, but the concentration of
  SO2H2O will stay constant as long as the gaseous
  SO2 concentration stays constant. Whats the pH
  for our mixture?
• If most of the H? comes from SO2H2O
  dissociation, then
• Note that there about 400 times as much S in the
  form of HOSO2? as in the form H2OSO2. HOSO2? is
  a very weak acid, ant the reaction stops here.
  The pH of cloudwater in contact with 100 ppb of
  SO2 will be 4.5

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• Because SO2 participates in aqueous-phase
  reactions, Eq. (I) above will give the correct
  H2OSO2, but will underestimate the total
  sulfur in solution. Taken together all the forms
  of S in this oxidation state are called sulfur
  four, or S(IV).
• If all the S(IV) in the cloud water turns to
  S(VI) (sulfate) then the hydrogen ion
  concentration will approximately double because
  both protons come off H2OSO4, in other words
  HSO4? is a strong acid.
• This is fairly acidic, but we started with a
  very high concentration of SO2, one that is
  characteristic of urban air. In more rural areas
  of the eastern US an SO2 mixing ratio of a 1-5
  ppb is more common. As SO2H2O is oxidized to
  H2OSO4, more SO2 is drawn into the cloud water,
  and the acidity continue to rise. Hydrogen
  peroxide is the most common oxidant for forming
  sulfuric acid in solution we will discuss H2O2
  later.