Enthalpy and Hess Law

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Enthalpy and Hess Law

• Yep, its a law not a rule . . . meaning it
  always works!

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Enthalpy (DH), Lets Review

• The thermodynamic variable used to describe the
  heat of a reaction at constant pressure, qP
• The potential thermodynamic energy of a reacting
  system
• The potential energy stored (as heat) in
  chemical bonds
• Exothermic reactions have negative DHrxn values
• Typically (but not always) spontaneous reactions
  have negative values of DHrxn (the heat term is
  added to the products side)
• We express enthalpy for a chemical reaction
  (DHrxn) as a stoichiometric component in a
  thermochemical equation
• 2H2O(l) ? 2H2(g) O2(g) 572 kJ
• Enthalpy is a state function which means it is
  independent of path
• ? If we do not know the enthalpy for a reaction
  we can calculate it using the known enthalpies
  for smaller reactions

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Enthalpy (DH), Lets Review a Little More

• When we discuss enthalpy we like to express it
  in terms of molar enthalpy
• ? The molar enthalpy for the transformation of
  water into its constituent elements is 286 kJ/mol
  (not 572 kJ as written in the thermochemical
  equation . . . why?)
• Enthalpy data is often expressed in tables in a
  standard state, DHo, this is 101.3 kPa, 298 K
  (and 1 mol/L for solutions)
• We can determine DHrxn by measuring the
  temperature change of a pure substance to which
  the heat (energy) of the reaction has been
  transferred
• Process known as calorimetry

In a calorimeter Heat absorbed (heat given off
by rxn) (mcDT)liquid (mcDT)calorimeter The
calorimeter can absorb some heat so choose your
material wisely!!!
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Enthalpy and Heat . . . Some Problems

• How much heat energy is required to increase the
  temperature of 10 g of nickel (specific heat
  capacity 440 J kg-1 K-1) from 323 K to 343 K?
• The enthalpy of combustion (DHcomb) of ethanol
  (C2H5OH) is 1370 kJ/mol. How much heat is
  released when 0.20 moles of ethanol undergo
  complete combustion?
• Consider the following reaction
• H2(g) ½ O2(g) ? H2O(l) 286 kJ
• DHrxn -286 kJ/mol, what mass of O2(g) must be
  consumed to produce 1144 kJ of energy?

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Enthalpy Diagrams
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Hess Law
Because enthalpy (DH) and energy (DE) are state
functions, we do not concern ourselves with the
reaction pathway when determining these values.
They are said to be independent of path.
Therefore when we cannot measure a reaction to
determine its enthalpy change (DH) we can use
literature data to calculate these energetic
quantities. For a given reaction with an unknown
energy term we can calculate DH by assembling the
desired reaction from several smaller
well-defined ones.
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Hess Law and the Formation of Ethylene (C2H4), A
Polymer Precursor
Ethylene or ethene is the monomer unit of
polyethylene commonly known to us as plastic. The
overall formation reaction is as
follows 2C(graphite) 2H2(g) ? C2H4(g)
DHrxn ? Remember a formation reaction is
the formation of a molecule from its elements in
their most stable form (H2 not H). By definition
the heat of formation of an element DHform
0. How do we find DHrxn for this reaction?
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Hess Law and the Formation of Ethylene (C2H4), A
Polymer Precursor
The overall formation reaction is as
follows (a) 2C(graphite) 2H2(g) ? C2H4(g)
DHrxn ? We are given the following known
data (b) C(graphite) O2(g) ? CO2(g) DH
-393.5 kJ (c) C2H4(g) 3O2(g) ? 2CO2(g)
2H2O(g) DH -1410.9 kJ (d) H2(g) ½ O2(g) ?
H2O(l) DH -285.8 kJ
Lets begin by finding an equation that will
place 2 moles of C(graphite) on the left hand
side. We need to multiply (b) by 2 to accomplish
this 2(b) 2C(graphite) 2O2(g) ? 2CO2(g) DH
-787 kJ
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Hess Law and the Formation of Ethylene
(a) 2C(graphite) 2H2(g) ? C2H4(g)
DHrxn ? (b) C(graphite) O2(g) ? CO2(g) DH
-393.5 kJ (c) C2H4(g) 3O2(g) ? 2CO2(g)
2H2O(g) DH -1410.9 kJ (d) H2(g) ½ O2(g) ?
H2O(l) DH -285.8 kJ
Next, lets identify an equation that will place
the C2H4(g) term on the right hand side. We can
achieve this by reversing (c) and the changing
the sign of DH. -(c) 2CO2(g) 2H2O(g) ?
2C2H4(g) 3O2(g) DH 1410.9 kJ
Finally, equation (a) has a 2H2 term so we need
to double (d) to obtain this. 2(d) 2H2(g) O2(g)
? 2H2O(l) DH -571.6 kJ
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Hess Law and the Formation of Ethylene (C2H4), A
Polymer Precursor
We can now sum (perform a summation) of the 3
known equations and obtain (a), which will yield
an new energy term. 2(b) 2C(graphite) 2O2(g) ?
2CO2(g) DH -787 kJ -(c) 2CO2(g) 2H2O(g) ?
2C2H4(g) 3O2(g) DH 1410.9 kJ 2(d) 2H2(g)
O2(g) ? 2H2O(l) DH -571.6 kJ

• 2C(graphite) 2H2(g) ? C2H4(g) DH (-787
  1410.0 571.6) kJ
• DH 52.3 kJ

Yielding the thermochemical equation 2C(graphite)
2H2(g) 52.3 kJ ? C2H4(g) The formation of
ethylene is an endothermic reaction.
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Enthalpy and Hess Law
This completes our work in chapter 6 of our
textbook and roughly our work with enthalpy as
the sole energy term for a reacting
system. Please review your notes from this
section and read chapter 6 if you have not done
so.
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Why do molecules form the way they do?

• Bond Enthalpies, Hess Law, The Born-Haber Cycle,
  and Heats of Reaction

Textbook Reference Chapter 6 with parts from
Chapter 9
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Molecular Compounds
Why does oxygen form O2 rather than O8 (more
accurately 4O2 rather than O8)? ? We know that
oxygen is a diatomic, but this is not a reason
this is merely an observation of trend. ? We
need to consider DHBDE (Bond Dissociation Energy)
which is the energy required to cleave a
covalent bond.
BDE O2 498 kJ/mol BDE 4O2 4 x 498 kJ 1992
kJ Meaning 1992 kJ is required to break 4 moles
of O2 OR 1992 kJ of energy is given off when we
form 4 moles of O2 from O atoms.
BDE OO 146 kJ/mol BDE O8 8 x 146 kJ 1168
kJ Meaning only 1168 kJ is given off when we form
8 OO single bonds in O8. O2 is energetically
favored.
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Some other elements to consider
Why is phosphorus P4 rather than 2P2? ? P4
(white phosphorus) is tetrahedral ? PP BDE
209 kJ ? PP BDE 490 kJ
Why is sulfur S8 rather than S2? ? This is the
converse of oxygen which prefers O2. ? SS BDE
266 kJ ? SS BDE 427 kJ
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The Ionic Lattice . . . One More Time
The find the lattice energy (DHlatt)of an ionic
compound we can use the following formula, known
as the Born-Lande Equation DHlatt
(-LA)(z)(z-)(e2)(1 1/n)

4per Where L 6.022 x 1023 A
Madelung Constant z summation of charges on the
ions e electron charge 1.6 x 10-19 C e
permittivity in a vacuum 8 x 10-12 F/m r
distance between the ions n Born constant
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Lattice Energy there has to be an easier way . .
.
(this would be a pretty lousy slide if there
wasnt)
We use whats called the Born-Haber cycle, which
makes use of some specific heats of reaction
(DHrxn). DHf the standard heat of formation of
a compound from its elements DHsub the heat of
sublimation (solid ? gas) DHBDE the Bond
Dissociation Energy for a covalent bond DHI1
first ionization energy (neutral atom losing an
e-, always positive) DHI2 second ionization
energy (1 to 2, large and positive) DHEA
electron affinity (always a negative term except
Be and N) DHlatt lattice energy (always
negative, usually quite large)
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Formation of NaCl(s)
Na(g) Cl(g) e-
DHEA -349 kJ
DHI1 496 kJ
Na(g) Cl-(g)
Na(g) Cl(g)
DHlatt -787 kJ
DHBDE 122 kJ
Na(g) ½ Cl2(g)
DHsub 107 kJ
Na(s) ½ Cl2(g)
DHf ???
NaCl(s)
BDE Cl2(g) 244 kJ, so ½ (244 kJ) 122 kJ
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How do we calculate DHf from the Born-Haber
Cycle?
From our work with Hess Law we know that
energies are additive. ? Therefore we can add up
all of the components from the cycle which yield
the overall formation reaction (from the
elements). DHf NaCl 107 122 496 (-349)
(-787) -411 kJ/mol of NaCl
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Formation of NaCl(s)
Na(g) Cl(g) e-
DHEA -349 kJ
DHI1 496 kJ
Na(g) Cl-(g)
Na(g) Cl(g)
DHlatt -787 kJ
DHBDE 122 kJ
Na(g) ½ Cl2(g)
DHsub 107 kJ
Na(s) ½ Cl2(g)
DHf -411 kJ/mol
NaCl(s)
DHf NaCl 107 122 496 (-349) (-787)
-411 kJ/mol of NaCl
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Determine the lattice energy of MgF2(s)
DHsub Mg(s) ? Mg(g) 146 kJ/mol DHI1 Mg(g) ?
Mg(g) 738 kJ/mol DHI2 Mg(g) ? Mg2(g) 1451
kJ/mol DHBDE F2(g) 159 kJ/mol of F2 DHEA F
-328 kJ/mol of F DHform MgF2(s) -1124 kJ/mol
(this is a DHf)
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Lattice Energy of MgF2(s)
Mg2(g) 2F(g) 2e-
DHEA -656 kJ (2 x -328 kJ)
DHI2 1451 kJ
Mg2(g) 2F-(g)
Mg(g) 2F(g) e-
DHI1 738 kJ
Mg(g) 2F(g)
DHlatt ?
DHBDE 159 kJ
Mg(g) F2(g)
DHsub 146 kJ
Mg(s) F2(g)
DHf -1124 kJ/mol
MgF2(s)
DHlatt DHf (DHsub DHBDE DHI1 DHI2
DHEA) -2962 kJ/mol MgF2(s)
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Lets leave it here as far as new material . . .
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Your Assignment (and no not if you choose to
accept it, just accept it)

• Using your notes and the textbook suggest
  possible reasons why some reactions are
  exothermic and some are endothermic (5.4.2) in
  terms of average bond energy/enthalpy.
• The combustion of methane is represented by the
  equation CH4(g) 2O2(g) ?
  CO2(g) 2H2O(l) 890.3 kJ.
• a) what mass of CH4(g) must be burned to give
  off 1.00 x 105 kJ of heat?
• b) how much heat is produced when 2.78 moles of
  CO2(g) are generated?
• 3) Using standard enthalpies of formation from
  Appendix B in your textbook calculate the
  standard enthalpy change for the following
  reactions
• a) NH3(g) HCl(g) ? NH4Cl(s)
• b) 3C2H2(g) ? C6H6(l)
• c) FeO(s) CO(g) ? Fe(s) CO2(g)

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• When burning a Dorito you find that the
  temperature of 150 g of water in an aluminum
  (mass 12 g) can is raised by 64 K. What amount
  of energy was released by the Dorito? You may
  assume that no heat was lost to the surrounding
  and it was completely transferred to the can and
  water.
• Use the following 2 reactions calculate the DHrxn
  for 2NO2(g) ? N2O4(g). N2(g) 2O2(g) ?
  N2O4(g) DH 9.2 kJ and N2(g) 2O2(g) ?
  2NO2(g) DH 33.2 kJ
• Calculate the enthalpy of reaction
• BrCl(g) ? Br(g) Cl(g) DHrxn ?
• Using the following data
• Br2(l) ? Br2(g) DH 30.91 kJ
• Br2(g) ? 2Br(g) DH 192.9 kJ
• Cl2(g) ? 2Cl(g) DH 243.4 kJ
• Br2(l) Cl2(g) ? 2BrCl(g) DH 29.2 kJ
• 7) Question 9.33 from your textbook. Using a
  Born-Haber cycle for KF to calculate DHEA of
  fluorine.

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Tying Up Some Loose Ends . . .
The brocolli must die!

• Enthalpy Cycles, Calculation of DHrxn and
  Variations in Lattice Energy

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Multiple Representations of Enthalpy
So far we have looked at 2 ways of representing
the energy/enthalpy term for a chemical
reaction 1) as a component in a thermochemical
equation 2) as a term outside the equation,
calculated from a formula such as q
mcDT 3) as an enthalpy diagram ? a graphical
way to show the change in enthalpy rather than
relying solely on equations ? this in NOT a
commonly used method and it will be addressed
purely as the presentation of a third option
to enthalpy problems, I strongly recommend you
use equations and Hess Law to solve enthalpy
problems
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What is an Enthalpy Diagram?
A diagram that shows the overall and net reaction
steps with their corresponding energy terms for a
chemical reaction. Consider below and the
combustion of methane.
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What is an Enthalpy Cycle?
Lets consider an alternative route to finding
the DHcomb of methane.
?H1 enthalpy change for bond breaking 4 x
E(C-H)     2 x E(OO) 4 x 413 2 x 498
2648 kJ mol-1
?H2 enthalpy change for bond making  - 2 x
E(CO) 4 x E(O-H)   - 2 x 805 4 x 464
  - 3466 kJ mol-1
DHrxn DH1 DH2 -818 kJ/mol
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What is an Enthalpy Cycle?
DH2
A
C
DH1
DH3
DH1 DH2 DH3
B
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Variations in Lattice Energy
Sometimes when we calculate a lattice energy and
then compare it to our experimental value, there
is a difference. If this difference is large
enough between the theoretical and experimental
values this indicates that rather than the
lattice being totally ionic, there is a
significant degree of covalent character to each
interaction. ? Yes this is a return to the
bonding continuum! Consider the following
AgCl has significant covalent character! Is
there another way we could have identified this
phenomenon?
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Time for Questions
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