Thermochemistry
- Energy of Chemical Reactions
Contents:
• heat, work, forms of energy
• specific heat and energies of phase changes
• enthalpy changes in chemical reactions
• standard enthalpies of formation
• Hess’s law
• estimating enthalpies of reaction from Bond Energies
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CHEMICAL ENERGY
Chemical bonds are a source of energy
• BOND BREAKING - requires energy
• BOND MAKING - releases energy
In a chemical reaction:
• if more energy is released in forming bonds than is used in
breaking bonds then . . . reaction is EXOTHERMIC
Energy is released as HEAT, LIGHT, SOUND, WORK
• if more energy is used in breaking bonds than is released in
forming bonds then . . . reaction is ENDOTHERMIC
Energy can be provided by
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- LIGHT - photochemistry
- WORK - electrochemistry
- COOLING of surroundings
THERMOCHEMISTRY (Ch. 6)
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Energy and Chemistry
ENERGY is the capacity to do work or transfer heat.
HEAT is the form of energy that flows between 2
samples because of a difference in temperature.
WORK is the form of energy that results in a
macroscopic displacement of matter such as gas
expansion or motion of an object (force x distance)
Other forms of energy —
• Chemical
• light
• gravitational potential
• electrical
• electrostatic potential
• kinetic
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Specific Heat Capacity
Thermochemistry is the science of heat (energy) flow.
A difference in temperature leads to energy transfer.
The heat “lost” or “gained” is related to
a) sample mass
b) change in T, and
c) specific heat capacity by
Specific heat capacity
=
heat lost or gained by substance (J)
(mass, g) (T change, K)
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Specific Heat Capacity
Substance
H2O
Al
glass
Spec. Heat (J/g•K)
4.184
0.902
0.84
Water
Aluminum
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Specific Heat Capacity - an example
If 25.0 g of Al cool from 310 oC to
37 oC, how many joules of heat
energy are lost by the Al?
Specific heat capacity
=
= 0.902 J/g.K
heat lost or gained by substance (J)
(mass, g)(T change, K)
heat gain/lost = q = (specific heat)(mass)(DT)
where DT = Tfinal - Tinitial = 37 - 310 = -273 K
q = (0.902 J/g•K)(25.0 g)(-273 K)
q = -6160 J
negative sign of q  heat is “lost by” or transferred from Al
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Heat Transfer and
Changes of State
Changes of state involve energy
Ice Water
333 J/g
(Heat of Fusion)
Water Vapor
2260 J/g
(Heat of vaporization)
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Heating/Cooling Curve for Water
3
Evaporate water
4
Heat water
1 2 Melt ice
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CHEMICAL REACTIVITY
• What drives chemical reactions? How do they occur?
The first is answered by THERMODYNAMICS and
the second by KINETICS.
• In Ch. 4 we saw a number of “driving forces” for
reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
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CHEMICAL REACTIVITY
• Energy transfer also allows us to predict
reactivity.
• In general, reactions that transfer
energy to their surroundings are
“product-favored”.
• How do we describe heat transfer in
chemical processes ?
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Heat Energy Transfer in
Physical & Chemical Processes
• CO2 (s, -78 oC) ---> CO2 (g, -78 oC)
Heat flows into the SYSTEM (solid CO2) from the
SURROUNDINGS in an ENDOTHERMIC
process.
Surroundings
System
heat
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• ENERGY is the capacity to do work or transfer heat.
• HEAT is the form of energy that flows between
2 samples because of a difference in temperature.
• WORK is the form of energy that results in a
macroscopic displacement of matter such as
gas expansion or motion of an object (force x distance)
In CO2 sublimation & expansion, the same amount
of ENERGY flows from surroundings to system
If expanding gas is enclosed, part of the energy transfer
appears in the form of WORK OF EXPANSION
wexp = - PDV
(for an ideal gas)
If expanding gas is not enclosed, the energy transfer appears
only as HEAT (CO2 gas gets warm).
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FIRST LAW OF THERMODYNAMICS
heat energy transferred
q = DE - w
Energy change
OR
work done
by the
system
DE = q + w
NB - q and w positive
when they are
transferred
FROM surroundings
TO system
Surroundings
Heat
Energy is conserved!
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System
qsys > 0
THERMOCHEMISTRY (Ch. 6)
Work
wsys > 0
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ENTHALPY
Most chemical reactions occur at constant P, so
Heat transferred at constant P is called qp with
qp = DH = DE - w
= DE + P DV = D(E+PV)
where H
H is defined as (E + PV)
= enthalpy
DH = heat transferred at constant P
DH = change in heat content of the system
DH = Hfinal - Hinitial
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ENTHALPY
DH = Hfinal - Hinitial
If Hfinal > Hinitial then DH is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then DH is negative
Process is EXOTHERMIC
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Endo- and Exothermic
Surroundings
Surroundings
System
Heat
qsys < 0
Heat
qsys > 0
ENDOTHERMIC
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System
EXOTHERMIC
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USING ENTHALPY
Consider the combustion of H2 to form water . .
H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJ
Exothermic reaction — heat is a “product”. DH = -242 kJ.
This is spontaneous and proceeds readily once initiated.
But the reverse reaction, the decomposition of water :
H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g)
Endothermic reaction — heat is a “reactant”, DH = +242 kJ.
This does not occur spontaneously.
BUT . . . Decomposition of water can be made to occur
by coupling to another, spontaneous process . . .
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LIGHT
How can we
make H2 gas ?
w ire
H2
O2
H2 O
S e m ic o n d u c t o r
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Me t a l
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N. Lewis,
American Scientist,
Nov. 1995, page 534.
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Making H2 from liquid H2O involves two steps.
H2O(liq) + 44 kJ  H2O(g)
H2O(g) + 242 kJ H2(g) + 1/2 O2(g)
--------------------------------------------------H2O(liq) + 286 kJ H2(g) + 1/2 O2(g)
This is an example of HESS’S LAW —
If a reaction is the sum of 2 or more others,
the net DH is the sum of the DH’s of the other rxns.
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Hess’s Law - a second example :
Calc. DHrxn for S(s) + 3/2 O2(g) --> SO3(g)
knowing that
S(s) + O2(g) --> SO2(g)
DH1 = -320.5 kJ
SO2(g) + 1/2 O2(g) --> SO3(g) DH2 = -75.2 kJ
S(s) + 3/2 O2(g) --> SO3(g)
DH3 = -395.7 kJ
The two rxns. add to give the desired rxn.,
so DHrxn = DH1 + DH2 = -395.7 kJ
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energy
S solid
direct path
+ 3/2 O
2
DH3 =
-395.7 kJ
SO3 gas
+O2
DH1 =
-320.5 kJ
SO2 gas
+ 1/2 O 2
DH2 = -75.2 kJ
DH3 = -395.7
DH(2+3) = -320.5 + -75.2 = -395.7
S DH along one path = S DH along another path
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SDH along one path =
SDH along another path
• This equation is valid because
DH is a STATE FUNCTION
• These depend only on the state of
the system and not how it got there.
• Other state functions include:
V, T, P, energy . .
— and your bank account!
• Unlike V, T, and P, one cannot
measure absolute H. Can only measure DH.
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Standard Enthalpy Values
Most DH values are labeled DHo
o
•
•
•
•
means measured under standard conditions
P = 1 atmosphere ( = 760 torr = 101.3 kPa)
Concentration = 1 mol/L
T = usually 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas
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DHof = standard molar enthalpy of formation
- the enthalpy change when 1 mol of
compound is formed from elements under
standard conditions.
Values: Kotz, Table 6.2 and Appendix K
By definition, DHof = 0 for elements in
their standard states.
H2(g) + 1/2 O2(g) --> H2O(g)
DHof = -241.8 kJ/mol
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Using Standard Enthalpy Values
Calculate DH
of reaction?
In general, when ALL
enthalpies of formation
are known,
DHorxn =
S DHof (products)
- SDHof (reactants)
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Example: Calculate the heat of combustion of
ethanol, i.e., DHorxn for
C2H5OH(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(g)
DHorxn = S DHof (prod) - S DHof (react)
DHorxn = { 2 DHof (CO2) + 3 DHof (H2O) }
- {7/2 DHof (O2) + DHof (C2H5OH)}
= { 2 (-393.5 kJ) + 3 (-241.8 kJ) }
- {7/2 (0 kJ) + (-235.1 kJ)}
DHorxn = -1035.5 kJ per mol of ethanol
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Bond Energies (Kotz, sect. 9.4, pp 418-422)
• Given by D - the bond dissociation energy
D = energy required to break a bond in a gas phase
molecule under standard conditions
• D can be derived from DHrxn for atomization . . .
H
e.g. CH4 (g) C (g) + 4 H (g)
DHrxn = -1664 kJ = 4 * D(C-H)
H
D(C-H) = 416 kJ per mole of C-H bonds
H
H
• D (C-H) (kJ/mol) varies slightly among compounds :
CH4 416 C2H6 392
C3H8 380
C2H4 432 C2H2 445
C6H6 448
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Bond Energies
• D is similar for same bond in different molecules
• Average values over many compounds are tabulated
• Bond energy depends on bond order
BOND
D (kJ/mol) (Bond Energy)
H—H
436
C—C
347
C=C
611
CC
837
N—N
159
NN
946
see table 9.5 for Dissociation Energies of other bonds.
The GREATER the number of bonds (bond order) the
HIGHER the bond dissociation energy
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Using Bond Energies
• Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
Net energy = DHrxn
= energy required to break bonds
- energy evolved when bonds are made
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Estimating DHrxn for H—H + Cl—Cl  2 H—Cl
H—H = 436 kJ/mol
Cl—Cl = 243 kJ/mol
H—Cl = 431 kJ/mol
• Sum of H-H + Cl-Cl bond energies =
436 kJ + 243 kJ = +679 kJ
• 2 mol H-Cl bond energies = 862 kJ
• Net = DH = +679 kJ - 862 kJ = -183 kJ
THEREFORE,
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EXAMPLE 2:
Estimate the energy of the reaction
2 H—O—O—H ----> O=O + 2 H—O—H
• Is the reaction exo- or endothermic?
• Which is larger: energy req’d to break bonds . . .
or energy evolved on making bonds?
Energy for bond breaking:
4 mol O—H bonds = 4 (464 kJ)
2 mol O—O bonds = 2 (138 kJ)
TOTAL = 2132 kJ
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Energy from bond making :
1 mol O=O bonds = 498 kJ
4 mol O—H bonds = 4 (464 kJ)
TOTAL = 2354 kJ
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2 H—O—O—H ---->
O=O + 2 H—O—H
Net energy = +2132 kJ - 2354 kJ = - 222 kJ
The reaction is exothermic!
More energy is evolved on
making bonds than is
expended in breaking
bonds.
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Enthalpies of Reaction from Bond Energies
ENDOTHERMIC
EXOTHERMIC
Gaseous
Atoms
Gaseous
Atoms
REACTANTS
PRODUCTS
PRODUCTS
REACTANTS
Bond Breaking costs more
Bond Making releases more E
than is gained by Bond Making than required for Bond Breaking
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Key Concepts from Chapter 6: Thermochemistry
• heat transfer - specific heat
• phase transitions - heats of fusion, vaporization, etc
• First law of thermodynamics DE = q - w
• endothermic versus exothermic reactions
• enthalpy change in chemical reactions
• Hess’s law
• standard molar enthalpies of formation
DHrxn = S DHf(products) - S DHf (reactants)
• bond energies

DHrxn = S D(bonds broken) - S D(bonds made)
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