Download Energy

a.
b.


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1
TWO Trends in Nature
• Order  Disorder


• High energy  Low energy

Energy
The capacity to do work
or to produce heat.
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3
Law of Conservation
of Energy
Energy can be converted from one form to
another but can neither be created nor
destroyed.
(Euniverse is constant)
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4
The Two Types of Energy
Potential: due to position or composition can be converted to work
Kinetic: due to motion of the object
KE = 1/2 mv2
(m = mass, v = velocity)
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5
(a) In the initial
positions, ball A has
a higher potential
energy than ball B.
(b) After A has
rolled down the hill,
the potential energy
lost by A has been
converted to what
other forms?
Frictional heating
Of the hill
Increase PE of Ball B
Temperature v. Heat
Temperature reflects random motions of
particles, therefore related to kinetic energy
of the system.
Heat involves a transfer of energy between
2 objects due to a temperature difference
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Due to motion, or thermal energy of particles.
Energy will move spontaneously from high to low
concentration.
This infrared photo
of a house shows
where energy leaks
occur. The more red
the color, the more
energy (heat) is
leaving the house.
Work: force acting over a distance
Ball A does work on Ball B
What is the hill is bumpy and
ball A comes to a rest before
hitting ball B?
No work done. Only
transfer of HEAT through
frictional heating
Thus…
there are two ways to transfer energy
work (generally movement) and heat
(generally chemical reactions/particles)
9
The way the energy transfer is
divided between work and heat
depends on ..
Pathway
If all energy goes to friction in the
hill and Ball B does not move….
no work is done.
Therefore, Work and heat are
pathway dependent
10
HOWEVER,…….Regardless of the pathway
the actual energy change remains constant.
Total energy transfer is independent of the
pathway
Energy is a state function.
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11
State Function
(state property)
Depends only on the present state of the
system - not how it arrived there.
It is independent of pathway.
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12
Chemical Energy
• Same applies for chemical systems
Example:
CH4 + 2O2  CO2 + H2O + energy
Actual reaction may follow different modes of
changes (different intermediate steps) but
net change in Energy is the same.
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13
System and Surroundings
System: chemical reaction
Surroundings: Everything else in the universe
Universe = System + Surroundings
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14
Exo and Endothermic
Heat exchange accompanies chemical reactions.
Exothermic: Heat flows out of the system (to the
surroundings). [product] usually represent as
Reactant  Products + ∆. Indicates output of
energy
Endothermic: Heat flows into the system (from the
surroundings). [reactant] usually represent as
∆
Reacants  Products. Indicates input of energy
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15
Figure 6.2: The combustion of methane
releases the quantity of energy change ∆(PE)
to the surroundings via heat flow. This is an
exothermic process.
16
Figure 6.3: The energy diagram for the
reaction of nitrogen and oxygen to form nitric
oxide. This is an endothermic process: Heat
{equal in magnitude to ∆(PE)} flows into the
system from the surroundings.
17
Concept Check
Classify each process as exothermic or
endothermic. Explain. The system is
underlined in each example.
Exo
a)
Endo b)
Endo c)
Exo
d)
Endo e)
Your hand gets cold when you touch ice.
The ice gets warmer when you touch it.
Water boils in a kettle being heated on a
stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
18
Law of Conservation of Energy is
also called:
First Law of Thermodynamics:
Which states that the energy of
the universe is constant.
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19
First Law
E = q + w
E = change in system’s internal energy
q = heat
w = work
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20
Exothermic versus endothermic.
Endothermic * products have Higher PE
Exothermic * reactants have Higher PE
(Higher PE = weaker bonds)
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21
First Law cont.
Sign is the direction of the flow. (system point of view)
q = + endothermic
system E increase
q = - exothermic
system E decrease
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22
Also applies to work..
System does work on surroundings: w is –
Surroundings do work on system: w is +
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23
JOULE
• SI unit for energy:
kg  m
J
2
s
2
Internal
energy
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24
Common type of work in chemistry
• Done by a gas (through expansion)
• Done to a gas (through compression)
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25
How does a car engine work?
• Combustion results in production of a gas:
•
expansion in cylinder: Endo or exothermic
•
pushes piston: q (– or +) (∆E to surroundings)
•
motion of car: w (- or +) (work done to surr.)
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26
Figure 6.4: (a) The
piston, moving a
distance ∆h against a
pressure P, does work
on the surroundings.
(b) Since the volume of
a cylinder is the area of
the base times its
height, the change in
volume of the gas is
given by ∆h x A = ∆ V.
F
P
A
Force per unit area
Work is a force applied over a distance.
Work= force x distance = F x Δh
F
SINCE P 
,
A
F=PxA
Work = P x A x Δh
ΔV = final V – initial (piston moving)
28
Since V = (A x ∆h) (area of the piston x height)
∆V = Vfinal - Vinitial = A x ∆h
substitute into work:
Work = P x (A x Δ h) = PΔV
For a gas expanding w is (-) due to work flowing
out of the system.
W = -P ΔV
29
PV Work
• Calculate the work associated with the
expansion of a gas from 46.L to 64 L at a
constant external pressure of 15atm.
For a gas at constant P,
w= -P∆V
30
P = 15 atm
And ∆V = 46-64 = -18 L
W = 15atm x (-18 L) = -270 L∙atm
31
Just a note…
• For an ideal gas, work can occur ONLY
when its V changes.
Thus, if a gas is heated at constant
volume, the pressure increases but no
work occurs.
32
A balloon is being inflated to its full extent by
heating the air inside it. In the final stages
of this process, the volume of the balloon
changes form 4.00 x 106 L to 4.50 x 106 L
by the addition of 1.3 x 108 J of energy as
heat. Assuming that the balloon expands
against a constant pressure of 1.0 atm,
calculate ∆E for the process
(1L atm = 101.3 J)
8 x 107 J
33
Enthalpy, H
•What is Enthalpy?
–a thermodynamic quantity equivalent to the total
heat content of a system.
–equal to the internal energy of the system plus
the product of pressure and volume.
•Enthalpy = H = E + PV
–Since it’s rather difficult to measure heat content
directly, we are more interested in the change in
Enthalpy or ΔH
–State function
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34
Determining Enthalpy
Recall: At constant pressure, the only work allowed is PV
qP = defined as heat at constant pressure, and is calculated from
E = qP + w = qP  PV
qP = E + PV= H
Therefore, H = energy flow as heat (at constant
pressure)
**Heat of reaction and change in enthalpy are
interchangeable**
35
Enthalpy (H) is used to quantify the heat flow into (or out of) a
system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0 exotherm
Hproducts > Hreactants
H > 0 endotherm6.4
Hess’s Law
Enthalpy
(state function)
Reactants  Products
The change in enthalpy is the same whether
the reaction takes place in one step or a
series of steps.
37
Figure 6.7: The principle of Hess's law. The same change in
enthalpy occurs when nitrogen and oxygen react to form
nitrogen dioxide, regardless of whether the reaction occurs in
one (red) or two (blue) steps.
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38
Oxidation of N2 to produce NO2
N2(g) + 2O2(g)
2NO2(g)
ΔH1=68kJ
N2(g) + O2(g)
2NO(g) + O2(g)
N2(g) + 2O2(g)
Or..
2NO (g)
2NO2(g)
2NO2(g)
ΔH2=180kJ
ΔH3=-112kJ
ΔH=68kJ
39
Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g)
2NO(g)  N2(g) + O2(g)
2.
H = 180 kJ
H = 180 kJ
If the coefficients of a reaction are multiplied by
an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g)
H = 540 kJ
40
3. Fractional equations
**Coefficients do not have to be whole numbers,
as long as ALL values are adjusted appropriately
Example: Determine enthalpy change if half mole
of nitrogen monoxide gas is produced, given the
equation:
N2(g) + O2(g)  2NO(g) H = 180 kJ
0.25 N2(g) + 0.25 O2(g)  0.50 NO(g) H = 45
kJ
Two forms of carbon are graphite, the soft, black,
slippery material used in “lead” pencils and as
a lubricant for locks, and diamond, the
brilliant, hard gemstone. Using the enthalpies
of combustion for graphite (-394 kJ/mol) and
diamond (-396kJ/mol), calculate ∆H for the
conversion of graphite to diamond:
Cgraphite(s)  Cdiamond(s)
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Cgraphite(s)  Cdiamond(s)
Cgraphite(s) +O2 (g)  CO2 (g)
Cdiamond(s) + O2 (g)  CO2 (g)
∆H=-394 kJ
∆H=-396 kJ
Cgraphite(s) +O2 (g)  CO2 (g) ∆H=-394 kJ
CO2  Cdiamond(s) + O2 (g)
∆H=-(-396 kJ)
Cgraphite(s)  Cdiamond(s)
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43
Diborane (B2H6) is a highly reactive boron hydride,
which was once considered as a possible rocket
fuel for the US space program. Calculate ∆H for
the synthesis of diborane from its elements,
according to the equation
2B(s) + 3H2 (g)  B2H6(g)
Using the following data:
Reactions:
2B(s)
+ 3/2 O2(g)
B2H6(g) + 3O2(g)
H2(g)
+ 1/2O2(g)
H2O(l)




B2O3(s)
B2O3(s) + 3H2O(g)
H2O(l)
H2O(g)
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ΔH
ΔH
ΔH
ΔH
-1273 kJ
-2035 kJ
-286 kJ
44kJ
44
Hess’s Law
Enthalpy
(state function)
Reactants  Products
The change in enthalpy is the same whether
the reaction takes place in one step or a
series of steps.
45
Oxidation of N2 to produce NO2
ΔH1=68kJ
N2(g) + 2O2(g)
Or..
2NO2(g)
N2(g) + 2O2(g)
2NO(g) + O2(g)
2NO (g)
ΔH2=180kJ
2NO2(g) ΔH3=-112kJ
N2(g) + 2O2(g)
2NO2(g)
ΔH=68kJ
46
Figure 6.7: The principle of Hess's law. The same
change in enthalpy occurs when nitrogen and oxygen
react to form nitrogen dioxide, regardless of whether
the reaction occurs in one (red) or two (blue) steps.
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47
Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g)
2NO(g)  N2(g) + O2(g)
2.
H = 180 kJ
H = 180 kJ
If the coefficients of a reaction are multiplied
by an integer, H is multiplied by that same
integer.
6NO(g)  3N2(g) + 3O2(g)
H = 540 kJ
48
Two forms of carbon are graphite, the soft,
black, slippery material used in “lead”
pencils and as a lubricant for locks, and
diamond, the brilliant, hard gemstone.
Using the enthalpies of combustion for
graphite (-394 kJ/mol) and diamond (396kJ/mol), calculate ∆H for the conversion
of graphite to diamond:
Cgraphite(s)  Cdiamond(s)
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49
Cgraphite(s)  Cdiamond(s)
Cgraphite(s) +O2 (g)  CO2 (g)
Cdiamond(s) + O2 (g)  CO2 (g)
∆H=-394 kJ
∆H=-396 kJ
Cgraphite(s) +O2 (g)  CO2 (g) ∆H=-394 kJ
CO2  Cdiamond(s) + O2 (g)
∆H=-(-396 kJ)
Cgraphite(s)  Cdiamond(s)
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50
Diborane (B2H6) is a highly reactive boron
hydride, which was once considered as a
possible rocket fuel for the US space
program. Calculate ∆H for the synthesis of
diborane from its elements, according to the
equation
2B(s) + 3H2 (g)  B2H6(g)
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51
3
2B(s)  O 2 (g) B2 O3 (g)
  1273kJ
2
B2 H 6 (g)  3O 2 (g) B2 O3 (g)  3H 2 O(g)   2035kJ
1
H 2 (g)  O 2 (g) H 2 O(l)   286kJ
2
H 2 O(g) H 2 O(l)   44kJ
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52
Standard Enthalpy of Formation
• Defined as the CHANGE IN ENTHALPY associated with
the FORMATION of ONE MOLE of a COMPOUND
from its COMPONENT ELEMENTS with ALL
substances in their STANDARD STATES.
– Symbol: ΔHof, where the degree o symbol indicates that all
substances are in their STANDARD STATES, and subscript f
indicates “formation”
– Examples:
• 1/2 N2(g) + O2(g)  NO2(g)
• C(s) + 2H2(g) + 1/2 O2(g)  CH3OH(l)
ΔHof
ΔHof
= 34 kJ/mol
= -239 kJ/mol
**Manipulate equation in such as way to form EXACTLY 1 mol
product only
• Standard Enthalpy of Formation of elements is ALWAYS
ZERO “0”. WHY?
– because elements are NOT formed through chemical reactions.
• Sometimes difficult to determine directly.
– Example: Formation of DIAMOND from
graphite. ΔHof CANNOT be obtained directly.
Process is too slow.
• ΔHof can be obtained from OTHER KNOWN
processes, such as ENTHALPY OF COMBUSTION
– **unlike standard Enthalpy of Formation, Enthalpy of
Combustion is given for EXACTLY 1 mol of the substance
combusting.
» WHY?
» Combustion products are variable. How can you keep
track of which one to count/measure. Easier to consider
substance actually undergoing combustion.
Factors to Consider
• Conventional Definitions of Standard States
– For a Compound
• The standard state of a gaseous substance (ex: NO2) is a
pressure of exactly 1 ATMOSPHERE
• For a pure substance in a condensed state (liquid or solid), the
standard state is the PURE LIQUID OR SOLID .
• For a substance present in solution, the standard state is a
CONCENTRATION OF EXACTLY 1M
– For an Element
• The standard state of an element is the form in which the
element exists under condition of 1 ATM AND 25oC.
– Ex: Standard state of Oxygen: O2(g) at a pressure of 1 ATM and 25oC
– Ex: Standard state of Sodium: Na(s)
– Ex: Standard state of Mercury: Hg(l)
• Rules to Follow
– When a reaction is reversed, the magnitude of ΔH
remains the SAME, but its SIGN CHANGES
– When the balanced equation for a reaction is multiplied
by an integer (or fraction), the value of ΔH for that
reaction MUST ALSO BE MULTIPLIED by the SAME
FACTOR
– The change in enthalpy for a given reaction can be
calculated from the enthalpies of formation of the
reactants and products
• ΔHorxn
=
Σ np ΔHof (products)
- Σ nr ΔHof (reactants)
where n is # moles
– Elements in their standards states are NOT INCLUDED
in the ΔHoreaction calculations, because ΔHof for an
element in its standard state is ZERO.
Example 1: Determine the standard enthalpy of reaction for
the combustion of methane
Pathway for combustion of methane.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Reactants are first taken apart in (a) and (b) then used to
assemble the products in reactions (c) and (d).
57
Figure 6.9: A schematic diagram of the energy changes for the reaction
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l).
Reverse of the formation
58
Althernately, apply the equation for change of enthalpy
given the standard enthalpy of formation of each
substance, given
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l).
(a) formation of CH4
(b) Formation of O2
(c) formation of CO2
(d) formation of H2O(l)
ΔHorxn
ΔHorxn
=
ΔHof = -75kJ/mol
ΔHºf = 0
(b/c already an element)
ΔHºf = -394kJ/mol
ΔHºf = -286kJ/mol
Σ np ΔHof (products) - Σ nr ΔHof (reactants)
where n is # moles
= (1 mol CO2(-394 kJ/mol) + 2 mol H2O(-286 kJ/mol))
–
(1 mol CH4(-75 kJ/mol ) + 2 mol O2(0 kJ/mol))
59
NOTE: reverse sign for methane, use 2 moles water, note water (g) vs. (l)
• Example 2: Using a standard enthalpy of
formation reference table, determine the
standard enthalpy change for the overall
reaction that occurs when ammonia is
burned in air to form nitrogen dioxide and
water. This is the first step in the formation
of nitric acid
• Using enthalpies of formation, determine the standard
change in enthalpy for the thermite reacation, which
occurs when a mixture of powdered aluminum and iron
III oxide is ignited with a magnesium fuse.
• Methanol (CH3OH) is often used as a fuel in high
performance engines in race cards. Use your reference
table to compare the standard enthalpy of combustion
per gram of methanol with the combustion per gram of
gasoline. Gasoline is actually a mixture of compounds,
but assume for this problem that gasoline is a pure
substance (C8H18)
Specific Heat Capacity
• Specific Heat Capacity, c
– Amount of energy required to raise the
temperature of 1 g of a substance by 1 oC
• Unique value/intrinsic property
• Translation:
– Transfer of identical amounts of energy to equal masses of
two different substances will NOT result in the same
temperature change of the samples.
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63
Calculating energy transfer, q
• Amount of energy transferred, q, is related to
temperature change ∆T by the following equation,
where c = specific heat capacity
q = m c ∆T
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Terms to Know
– C = Heat capacity
• is the heat capacity of a given sample of a substance
• Units: J/oC
or
J/oK
– c = specific heat capacity
• Heat capacity per gram of substance
• Units: J/oC·g or
J/oK·g
– Molar heat capactiy
• Heat capacity per mole of substance
• J/oC·mol
or
J/oK·mol
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65
Practice Problems
• Determine the amount of energy required to
change the temperature of a 50.0 g sample of
water from 25.0 oC to 94.0 oC, given the specific
heat capacity of water is 4.184 J/g·oC
• Given that the combustion of methane is an
exothermic reaction with
ΔHcomb CH4 = -891 kJ/mol
and assuming 100% efficiency, determine the final
temperature of a 50.0 g sample of water, originally
at 28oC, if 8.00 g methane undergoes complete
combustion.
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Heating and Cooling Curves
• Explain energy transitions of a given
substance from one phase to another when
pressure is kept constant and energy is
added or removed.
– temperature is changed,
– phase is changed
OR
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• Starting with a solid below its melting point, the following
effects can be observed, when energy is added
1. The temperature of solid increases at a constant rate until it
begins to melt (melting point)
2. When melting begins, the temperature is constant until solid has
all turned to liquid
3. The temperature of the liquid increases at a constant rate until it
begins to boil (boiling point)
4. When boiling begins, the temperature is constant until the liquid
is all turned to gas
5. The temperature of the gas increases at a constant rate.
•
In summary, energy is being used to change the
temperature but not the phase, or it is being used to change
the phase but not the temperature.
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68
Heating Curve
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• In the regions where the temp of the
solid, liquid or gas is being increased,
amount of energy added is,
q = m c ∆T
where, q = energy, m = mass, c = specific heat
capacity of the substance and ∆T = change in
temperature
**BE CAREFUL, c is generally different for solids,
liquids and gases
• Where the solid is melting, the amount of energy being added is
q = (∆Hfusion)(moles)
where ∆Hfusion is the molar enthalpy of fusion and is the energy absorbed
when 1 mole of a solid melts
• Where the liquid is vaporizing, the amount of energy being
added is
q = (∆Hvaporization)(moles)
where ∆Hvaporization is the molar enthalpy of vaporization and is the energy
absorbed when 1 mole of a liquid vaporizes
Cooling Curve
• A cooling curve shows the same processes as a
heating curve, only in reverse, where the energy is
released, rather than absorbed.
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Calorimetry
• Experimental technique used to measure
energy changes in a chemical system.
• Process involves
– Chemical reaction (or a phase change)
– Thermal contact with heat bath (usually water)
• Heat capacity of the heat bath must be known.
– Ideally no heat loss or gain with the universe
• Only heat transfer is between system (reaction/phase
change) and surrounding (heat bath)
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• As the chemical reaction (or phase change) takes
place
– Energy is transferred between reaction/phase change
and heat bath
– Since specific heat capacity of heat bath is known,
we can calculate amount of energy transferred
(gained or lost) by applying
q = m c ∆T
• If temperature of heat bath goes up, the chemical reaction (or phase
change) must have released energy (i.e, the reaction was exothermic).
• If the temperature of heat bath goes down, the chemical reaction (or
phase change) must have absorbed energy (i.e. the reaction was
endothermic)
Therefore, for exothermic reaction
- q (energy lost by system) = + q (energy gained by surroundings)
-(m c ∆T)system = + (m c ∆T)surroundings
OR
For an endothermic reaction
- q (energy lost by surrounding) = + q (energy gained by system)
-(m c ∆T)surroundings = + (m c ∆T)system
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• Applying the Law of Conservation of
Energy (and assuming that there is no
energy losses), we can conclude
– Magnitude of the energy lost or gained by the
chemical reaction (or phase change, must be
equal to the magnitude of the energy gained or
lost y the heat bath.
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75
Calorimetry and Hess’s Law
• By measuring the energy changes in various
reactions, and applying Hess’s Law (the
final energy change is a sum total of
individual energy changes), we can
indirectly determine the energy change
associated with a multistep process
• Tomorrow’s lab
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76
Born Haber Process
• Describes the thermochemistry of ionic bonds
– Process of ionic bond formation can be broken
down into several stages.
– Ex: two possible routes for formation of sodium
chloride
• Single step process (Enthalpy change = standard
enthalpy of formation of NaCl)
Na(s) + ½ Cl2(g) 
NaCl(s)
• Recall: standard enthalpy of formation is the enthalpy
change when one mole of a substance is formed from
its component elements in their standard states. 77
• A multi-step process involving five separate changes
1. Atomization of sodium
Na(s)

Na(g)
∆Ha(M) = standard enthalpy of atomization is the energy
required to form one mole of gaseous atoms from the
element under standard conditions (endothermic)
2. Ionization of sodium
Na(g)

Na+1(g)
+
e∆Hi(1) = First ionization energy is the energy required to
remove one mole electrons from one mole of atoms in
the gaseous phase (endothermic)
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3. Dissociation of chlorine molecules
½ Cl2(g)  Cl(g)
∆Ha(NM) =Also considered as atomization
of Chlorine molecules, equal to 1/2
standard bond dissociation energy of
chlorine (since only ½ mole Cl2 is required
to form 1 mol NaCl
The standard enthalpy of bond dissociation
is the energy required to dissociate one
mole of molecules into atoms.
(endothermic)
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4. Formation of gaseous choride ions from gaseous
chlorine atoms
Cl(g) +
e
Cl-1(g)
∆HEa =The first electron affinity is the enthalpy
change when one mole of gaseous atoms gains an
electron to form a mole of gaseous ions.
(exothermic)
4. Bring together the gaseous ions
Na+1(g) Cl-1(g) 
NaCl(s)
∆HLE =The standard lattice enthalpy is the enthalpy
change when one mole of a solid is formed from its
constituents gaseous ions (exothermic)
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80
Born-Haber Cycle Diagram
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• RECALL: Hess’s Law states that the energy change
for a reaction is independent of the route taken,
therefore,
ΔHf = ΔHa(Me) + ΔHi(1) + ΔHa(NM) + ΔHEa + ΔHLE
• Standard conventions
– Arrows pointing upwards represent endothermic changes
(ΔH = +)
– Arrows pointing downwards represent exothermic
changes (ΔH = -)
82
Practice
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83
Practice
• Which ions are present in MgO(s)?
• Using the data given, draw a schematic
diagram representing each process in the
formation of MgO.
• Calculate the 2nd electron affinity of Oxygen
• Is the 2nd electron affinity exothermic or
endothermic? Explain.
ΔHa(O)
ΔHa(Mg)
ΔHi(1)(Mg)
ΔHi(2)(Mg)
ΔHEa(1)(O)
ΔHLE(MgO)
ΔHf(MgO)
+249 kJ/mol
+148 kJ/mol
+738 kJ/mol
+1451 kJ/mol
-141 kJ/mol
-3791 kJ/mol
-548 kJ/mol
• The actual ΔHf for this reaction is 602 kJ/mol. What does this
information indicate about the ionic
characteristic of this compound?
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84
Considerations
• When considering the relative attractions of
ions for one another, it can be useful to
consider charge density.
– Small, highly charged ions have HIGH charge
densities
• Tend to attract one another to greater degree
– Leads to higher melting points and higher lattice energies
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• Discrepencies between calculated and
experimental lattice enthalpy
– Assumption-a compound is essentially 100%
ionic
• Possible to calculate theoretical value of lattice
enthalpy (Born Haber Process)
• If theoretical values agree closely to experimental
values, bond is considered to be essentially ionic
86
Compound
Theoretical Lattice
Experimental
Formation
Lattice Formation
Enthalpy (kJ/mol) Enthalpy (kJ/mol)
Agreement?
NaCl
-766
-781
Good Match
ZnS
-3427
-3565
Poor Match
• If theoretical values do not agree closely, the
assumption of 100% ionic bond is incorrect.
– Differences are due to polarization
– Ionic bond is said to take on a degree of covalent
character
» The greater the discrepancy, the greater the
covalent character.
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87
CONGRATULATIONS.
YOU’VE JUST
FINISHED
THERMOCHEMISTRY
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88