Chemical Thermodynamics
2013/2014
5th Lecture:Thermochemistry
Valentim M B Nunes, UD de Engenharia
Thermochemistry
The study oh heat required or produced by chemical transformations is
called thermochemistry. Its an application of the first Law.
Remember: qp = ΔH and qV = ΔU
Thermochemistry is a branch
of thermodynamics. We can
measure (indirectly, for
instance work or temperature)
the energy a reaction produces
as heat, and identify q,
depending on the conditions,
with a change in internal
energy or enthalpy.
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Standard enthalpy changes
Changes in enthalpy when a system undergoes a physical or chemical
change are normally reported to a set of standard conditions. We will
consider the standard enthalpy change, ΔH°, the change of enthalpy
for a process in which the initial and final states are in their
standard states.
Standard states: The standard state of a substance is its pure more
stable form at the pressure of p° = 1 bar, and specified
temperature
“°” means
standard state!
Normally values are tabled at given temperatures, usually 298.15 K
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Examples: phase transitions
Standard enthalpy of vaporization, ΔH°vap
H2O(l)  H2O(g): ΔH°vap(373.15 K) = 44.7 kJ.mol-1
Standard enthalpy of fusion, ΔH°fus
H2O(s)  H2O(l): ΔH°fus(273.15 K) = 6.01 kJ.mol-1
Standard enthalpy of sublimation, ΔH°subl
C(s, graphite)  C(g): ΔH°subl(298 K) = 716.7 kJ.mol-1
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Standard enthalpy change in chemical reactions
The standard reaction enthalpy, ΔH°r ,at a given temperature T, is
the enthalpy of reaction when all reagents and products of reaction
are in their standard states.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l): ΔH°r = - 890 kJ.mol-1
The “-” signal means the process is
EXOTHERMIC!
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Standard enthalpy of formation
The standard enthalpy of formation, ΔH°f , of a substance is the
standard enthalpy for the reaction of its formation from its
constituent elements in their reference states.
6 C(s, graphite) + 2 H2(g)  C6H6(l): ΔH°f (benzene, l) = 49 kJ.mol-1
As a consequence, by definition, the standard enthalpy of formation
of any element in their reference state is zero at all temperatures
ΔH°f (C, graphite) = 0
ΔH°f (O2, g) = 0
…...
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Standard enthalpy of formation
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Hess's Law
We can now combine standard enthalpies of individual reactions to
obtain the enthalpy of another reaction. This is an application of the
first Law, resulting from the fact that enthalpy is a state function, is
called the Hess’s Law:
The enthalpy of an overall reaction is the sum of reaction enthalpies
of individual reactions into which a reaction may be divided.
As a consequence, we may calculate the enthalpy of a chemical
reaction, if we now the standard enthalpy of formation of all the
substances (see for instance table of slide 7)
H   i H
º
r
º
f ,i
products   i H reagents
i
º
f .i
i
νi are the stoichiometric coefficients
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Exothermic and Endothermic processes
At constant pressure, if ΔHr < 0, qp < 0,
and heat flows from the reaction to the
surroundings and the process is exothermic.
If ΔHr > 0, qp > 0 , and heat flows into
the reaction from the surroundings and
the process is endothermic.
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Combustion reactions
Combustion reactions are of particular importance in thermodynamics
and applications (thermoelectric power stations, motors,…). The
standard enthalpy of combustion, ΔH°c is the standard reaction
enthalpy for the complete oxidation of organic compounds to CO2(g)
and H2O(l) and N2 if nitrogen is also present.
Take for instance the combustion of natural gas (mainly methane):
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Zero!
Hcº  2  H ºf H2O  H ºf CO2   H ºf CH4   2  H ºf O2 
Hcº  2  (285.8)  (393.5)  (74.87)  890kJ
If we obtain water vapor (instead of liquid):
Hc  890 2  44  802kJ
All combustions are
extremely exothermic!
Vaporization of two moles of
water!
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Calorimetry
Constant volume
Adiabatic!
qV = (qwater + qpump + qr)
= 0
qr = -(qwater + qpump)
qwater = mwater  4.184 J.g-1.K-1 T
qpump = Cpump  t
qr = ΔU
ΔH = ΔU + Δ(pV)
Assume only contributions
from gases
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Calorimetry
Constant pressure
qp = ΔH
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The variation of enthalpy with temperature
The majority of industrial or biological reaction with interest occurs
at temperatures different from 298 K. Knowing the ΔHr at one
temperature, how to calculate it at other temperatures?
Consider the following thermodynamic cycle:
Reagents(T)
ΔHT
ΔHR
Reagents(298 K)
Products(T)
ΔHp
ΔH298
H R  H 298  H P  HT  0
 H r (T )
H r (T )  H 298  H R  H P
Products(298 K)
Tables!
?
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The variation of enthalpy with temperature
Recall
 H 
Cp  

 T  p
or
dH  C p dT
Integrating over two temperatures, T1 and T2:
T2
H (T2 )  H T1    C p dT
T1
This equation is especially simple when Cp is independent of T:
H T2   H T1   CpT2  T1 
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The variation of enthalpy with temperature
Since the previous equation applies to each substance in the
reaction, the standard reaction enthalpy at temperature T2 is:
T2
H rº T2   H rº T1    C p dT
T1
This equation is known as Kirchhoff’s law. ΔCp is calculated by;
C p   iC p,i Products   i C p,i Reagents
i
i
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Temperature dependence of heat capacities
If heat capacity is temperature independent previous equations are
very simple. If not, the temperature dependence of heat capacity is
taken into account by writing:
For organic compounds:
C p T   a  bT  cT 2
For inorganic gases:
C p T   a´b´T 
c´
T2
Characteristic
coefficients for
each substance!
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Adiabatic flame temperature
For a combustion process that takes place adiabatically with no shaft
work, temperature of the products is referred to as the adiabatic
flame temperature. This is the maximum temperature that can be
achieved for given reactants.
The maximum temperature for a given fuel and oxidizer combination
occurs with a stoichiometric mixture.
 H a  0  H 1  H 2
H 1  H cº isot hermicprocess
Tf
H 2   C p dT
Ti
Tf
_
Cp
 C dT

T  T 
p
Ti
f
i
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Adiabatic flame temperature
Adiabatic Flame Temperature (K)
Fuel
Oxygen as oxidizer
Air as oxidizer
Hydrogen, H2
3079
2384
Methane, CH4
3054
2227
Propane, C3H8
3095
2268
Octane, C8H18
3108
2277
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