Download Unit5_Chemical Thermodynamics

Heat
 Defined as transfer of energy
 RECALL: Kelvin temperature is a measure of kinetic
energy
 The higher the kelvin temperature of a sample of
particles, the more they move.

Vectors can be used to represent individual and average
velocities of particles at two different temperatures.
Notice that the sum of the
vectors (length/magnitude)
representing particles at a
higher temperature is greater
than the sum of the particles
at a lower temperature.
HIGH TEMP
LOW TEMP
 Example: a 500 mL sample of water at 25 oC has less
energy than the same volume of water at 35 0C.
 Doubling the Kelvin temperature of a system, also
doubles the kinetic energy as represented by the
equation below.
KE per molecule = 1 mv2
2
 This also must mean a change in the velocity of the
particles

RECALL (from gas laws): that change in velocity however, is
NOT directly proportionate.
 Rather, the change in velocity is dependent on the mass of the
particle. More massive particles will have a less significant
change in velocity (root mean square velocity)
Heat: Energy transfer
 When two systems that are at different temperatures
come into thermal contact (i.e. the particles collide)
 Energy is transferred from the hotter one to the cooler
one, until equal temperatures to each other is achieved.
 Average kinetic energy of the particles in each system
becomes the same

A single, intermediate temperature is achieved.
 Transfer of energy is called HEAT or HEAT
TRANSFER.
 Heat is NOT a substance
 When the intermediate, same temperature is achieved,
there is no further transfer of energy and THERMAL
EQUILIBRIUM is achieved.
Specific Heat Capacity
 Defined as the amount of energy required to raise the
temperature of a 1 g sample of a substance by 1 oC or 1 K.
 Different for different substances (intrinsic property)
 Implications
 Transfer of an identical amount of energy to the SAME mass of
two different substances will NOT result in the same temperature
change of the two substances.
 Ex: when you heat a pot of water on the stove
 Is it possible that the mass of water is LESS than the mass of the
pot?
 Which gets hotter first-the pot or the water? Why?
 The amount of energy transferred (q) can be related to the
temperature change of a substance (∆T) by the equation below,
where m = mass and c = specific heat capacity (unique value, must
be given)
q = m c ∆T
cwater = 4.186 J/g ·oC
Practice
 Given that the specific heat of water is 4.186 J/g ·oC, determine
the amount of energy required to raise the temperature of 125 g
sample from room temperature to 37.5 oC
 A 200. g sample of water requires 4186 J energy to raise the
temperature by 5 oC. Given the specific heat values below,
predict which sample will have the greatest increase in
temperature, if each sample is the same mass as the water
sample and the same amount of energy is applied to each.
0.386 J/g ·oC
 Al:
0.900 J/g ·oC
 Steel:
0.49 J/g ·oC
 Calculate the temperature change to each vessel and compare to
your prediction.
 Cu:
Heating and Cooling Curves
 It is possible to investigate the changes in a substance as energy
is either added or removed at constant pressure.
 In these cases, heating and cooling curves result.
 Starting with a solid below its melting point, the following effects
can be observed





The T (temperature) of the solid increases at a constant rate until it begins
to melt (i.e. reaches its m.p at the given pressure)
 **RECALL: normal melting and boiling points are the temperature at
which these changes occur at 1 atm of pressure)
When melting begins, the temperature is constant until the solid has all
turned to a liquid
The T of the liquid increases at a constant rate unit it begins to boil (i.e
reaches it b.p)
When boiling begins, the temperature is constant until the liquid has all
turned to gas
The temperature of the gas increases at a constant rate.
 In summary, energy is either being used to increase temperature,
OR change phase, but not both.
 In regions where the temperature of the solid, liquid or gas is being increased, the
amount of energy being added is q = m c ∆T
 Where q = energy, m = mass, c = specific heat of the substance (in that state) and ∆T =
Tfinal – Tinitial, i.e. the temperature change
 Where the solid is melting, the amount of energy being added is q = (∆Hfusion) (moles)
 Where ∆Hfusion is the molar enthalpy of fusion and representes the energy absorbed
when 1 mole of a solid melts
 Where the liquid is boiling, the amount of energy being added is q = (∆Hvaporization)
(moles)
 Where ∆Hvaporization is the molar enthalpy of vaporization and represents the energy
absorbed when 1 mole of a liquid vaporizes.
 **Keep in mind/common points of mistake
 Conventionally, q has units of Joules (J) and ∆H has units of kiloJoules (kJ) (must convert
one or another to ensure proper magnitude and units.
 temperature change of solid or liquid uses mass, ∆H uses moles (again, must convert one
or the other to ensure proper magnitude and units.
 In a heating curve, energy is always positive (INTER bonds are being broken)
Energy
 The capacity to do work or generate heat.
 The FIRST LAW OF THERMODYNAMICS is also
called the Law of Conservation of Energy, and it states
that the total energy of the universe is constant
 Universe is composed of the system and its
surroundings.
 We ALWAYS want to consider energy change from the
system’s perspective.

Energy can be gained by the system (+) in which case the
surrounding must lose energy (-) or vice versa
 Energy transfer can occur through HEAT or WORK
Heat vs. Work
 Ball A and ball B at initial
position have some internal
energy (potential).
•Thus,
there are their
two ways to transfer
 At their
final positions,
potential
energies have
energy—
changed. •through work (generally movement)
 Consider the energy transitions
(generally chemical
between•through
the initialheat
and final
reaction/particle changes)
positions
 Ball A moves and two energy
transitions occur


Ball A does work on ball B
Ball A loses some energy to the
hill due to frictional heating.
•What if the hill is bumpy and Ball A
comes to rest before hitting ball B.
•Potential energy of Ball A is
still reduced
•ALL of the energy transfer is
due to frictional heating.
•No work is done
 Therefore, the way energy is transferred is PATHWAY
dependent
 RECALL: whether Ball A hits and Ball B moves or not
determines how the energy was transferred, not how
much.
 However, regardless of the pathway, the total energy
of the system and surroundings remains constant
 The total energy gained and lost remains the same
 Total energy of the system is INDEPENDENT of the
pathway.

**Energy is a STATE FUNCTION (property)
 Depends ONLY on the present state of the system, not how it
got there (pathway independent)
Energy transitions in chemical
systems
 Consider the products of a hydrocarbon combustion
reaction
CH4
C2H6 + O2  CO2 + H2O + energy
C3H8
 Carbon dioxide and water are produced in all
combustion of hydrocarbon reactions.


However, regardless of HOW CO2 and H2O are produced, the
internal energy (bond energies) of CO2 and H2O remains the
same
Actual reactions may require different mechanisms, but net
energy change for the formation of CO2 and H2O is the same
 but, in the three examples above, the actual NET change will
be different for the OVERALL reactions
Classification of energy transitions
 Energy transitions (heat transfer) in chemical
reactions are always classified with respect to the
system, as
 exothermic (net output): potential energy of products is
less than the reactants, OR
 endothermic (net input): potential energy of the
products is greater than the reactants
 The opposite change must occur to the surrounding
 The overall energy of the universe will remain constant
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.
Work
 Work: scientific (physical) definition—use of force, F,
to move an object over some distance.
 Specifically in chemistry, work is considered only in
terms of expansion or compression (contraction) of
gases.

Energy transfer can occur through work.
 Consider a gas inside a vessel with a movable piston
 As the gas expands, the particles collide with the piston and
energy is transferred from the gas to the piston (and the
piston moves away)
 Work is done BY the gas
 Energy is lost
Quantitative analysis of work
 Work can by quantitatively analyzed by considering the
following
 change in volume of the gas, and
 External pressure, P, that the gas is working against

Via equation
w = -P∆V
(see attachment for full derivation)
 Because of law of conservation of energy, any energy lost (-
w) by one system(in this case the gas), must be gained (+w)
in equal magnitude by another (in this case the piston).
 In this work scenario, energy flows from one system (gas) to the
other (piston)
Practice
 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
P = 15 atm
And ∆V = 46-64 = -18 L
W = 15atm x (-18 L) = -270 L·atm
Note that….
 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.
First Law
From the calculations of heat (q) and work
(w), we can calculate change in internal
energy of system.
E = q + w
Where E = change in system’s internal
energy
 Sign is the direction of the flow. (system point of view)
q = + endothermic system E increases
q = - exothermic system E decreases
AND
w = + gas compressed
w = - gas expands
system E increases
system E decreases
Joules
 Energy is measured in Joules
kg  m
J
2
s
 Work gives units of L·atm
 Conversion to J
1 L·atm = 101.3 J
2
Practice
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)
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)
26
 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)
 Applying the Law of Conservation of Energy (and
assuming that there are 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.
28
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
29
Practice
 Propane is a gas that is commonly used in gas grills. A
sample of propane with a mass of 44.0 g is completely
burned in oxygen and in the process it releases 2002 kJ of
energy. This chemical reaction is brought in contact with a
water bath, and the transfer of energy from the reaction to
the water takes place.
Assuming the specific heat capacity of water is 4.184 J/g·K,
and that in such an experiment 100% of the energy
generated is transferred to the water causing the water
temperature to increase, answer the questions below
 In which direction does the energy flow?
 Calculate the change in temperature of 20.00 kg of water
 Is the combustion (burning) of propane an exothermic or
endothermic process? Explain your answer.
Practice
 Consider the following
data
Specific heat
capacity of ice
2.05 J/gK
Specific heat
capacity of water
4.184 J/gK
Specific heat
capacity of steam
2.08 J/gK
Molar heat of
fusion for H2O
6.02 kJ/mol
Molar heat of
vaporization for
H2O
40.7 kJ/mol
1.
Determine the energy change
when 1.00 mol (18.0 g) of water
at 20.0 oC is frozen, and then
cooled to a temperature of
-2.00 oC
a. Is the process exothermic
or endothermic
2. Considering ONLY the energy
change in the phase change
above, what would be the
temperature change observed
in a heat bath made of gold
with a mass of 0.502 kg? The
specific heat capacity of gold is
0.129 J/gK
a. Is the process exothermic
or endothermic?
Practice
 Example: If a 35.0 gram block of Aluminum at 375 K is
added to 200. g water at 300 K, calculate final
temperature, assuming no heat is lost. Specific heat of
Al: 0.92 J/gK
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
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
Hproducts > Hreactants
exotherm
H > 0
endotherm
6.4
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 that EXACTLY, ONLY 1 mol
product is formed
 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)
Determining Enthalpy of Reaction
from Standard Enthalpy Values
 Enthalpy values can be used for various types of calculations
 Enthalpy of reaction: energy transitions in a given reaction
 Standard enthalpy of combustion
 Hess’s Law
 Others
 Rules to Follow to Calculate Enthalpy of Reaction (and generally)
 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

Fractional equations are allowed
 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).
41
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
42
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
Σ np ΔHof (products)
ΔHof = -75kJ/mol
ΔHºf = 0
(b/c already an element)
ΔHºf = -394kJ/mol
ΔHºf = -286kJ/mol
- 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))
NOTE: reverse sign for methane, use 2 moles water, note water (g) vs. (l) 43
 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)
Standard Enthalpy of Combustion
 Heat of reaction for the combustion of exactly 1 mole
of a substance (commonly hydrocarbon)
 Practice:
 Determine ∆Hcombustion for each of the following
hydrocarbons, using standard enthalpy of formation
values.



Methane (CH4)
Propane (C3H6)
Ethanol (C2H5OH)
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.
47
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-right) or two (blue-left) steps.
Copyright©2000 by Houghton Mifflin Company.
All rights reserved.
48
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
49
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
50
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) Hrxn = 180 kJ
0.25 N2(g) + 0.25 O2(g)  0.50 NO(g) Hrxn = 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)
Copyright©2000 by Houghton Mifflin Company.
All rights reserved.
52
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)
Copyright©2000 by Houghton Mifflin Company.
All rights reserved.
53
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)
Copyright©2000 by Houghton Mifflin Company.
All rights reserved.
ΔH
ΔH
ΔH
ΔH
-1273 kJ
-2035 kJ
-286 kJ
44kJ
54
BOND ENERGIES and ENTHALPIES
 RECALL: Atoms are attracted to each other when the outer
electrons of one atom are electrostatically attracted to the nuclei of
another atom
 Attraction between the two atoms makes them increasingly stable,
giving lower and lower potential energies
 However, if atoms continue to approach one another, and get
increasingly close, there comes a point at which the two nuclei
(and/or the more densely packed core electrons) will start to repel
each other
 As they start to repel, potential energy is raised, and the atoms
become less stable
 Happy medium (i.e. bond formed) at a distance where the force of
attraction and repulsion results in the LOWEST potential energy
 Distance between the nuclei at this point is the bond length
 Potential energy at this point is the bond strength
Plotting potential energy vs. bond
length
 Typical plot of potential energy
vs. bond length appears as
follows
 Ex: Hydrogen, H2
bond length: 74 pm
bond strength: 436 kJ/mol
 Since forces of attraction
stabilize atoms, the greater the
number of electrons involved,
the stronger the attraction
 Result: triple bonds tend to be
stronger than double bonds, and
double bonds are stronger than
single bonds
 Shorter bonds also tend to be
stronger than longer bonds
Practice
 Assume that the potential energy vs. bond length plot
represents Oxygen, draw a curve representing
Nitrogen.
Bond dissociation energy
 RECALL: Bond breaking is generally an endothermic
process (ΔH is +) and bond making is generally an
exothermic process (ΔH is -) (exceptions exist)
 The strength of a bond can be measured by measuring the
amount of energy required to break a bond
 Ex: H2 
2H
Bond Dissociation Energy (ΔH) = +436 kJ
 Consider a polyatomic molecule, such as methane, CH4
 Composed of 4 identical C-H bonds.
 However, each bond dissociation energy is actually quite
different
CH4(g)
CH3(g)
CH2(g)
CH(g)
 CH3(g)
 CH2(g)
 CH (g)
 C(g)
+
+
+
+
TOTAL:
H(g)
H(g)
H(g)
H(g)
435 kJ/mol
453 kJ/mol
425 kJ/mol
339 kJ/mol
1652 kJ/mol
 TO overcome complexity of accounting for each individual
bond dissociation energy, we define bond energy as the
AVERAGE of each of the four, separate C-H bonds. Therefore

1652/4 = 413 kJ/mol
Determining ΔH values based on
Bond strength
 The energy change in a reaction can be calculated by
 Summing up the energy required to break each of the bonds in the
reactants (a positive value)
 Summing up the energy released by making each of the bonds in the
products (a negative value)
 Summing the totals of the two values above
 If the total energy needed to break the reactant bonds is greater than
the total energy released in making the product bonds, then the
reaction (system) is endothermic
 Energy is absorbed from surrounding
 Temperature of surrounding decrease
 If the total energy needed to break the reactant bonds is less than
the totoal energy released in making the product bonds then the
reaction (system) is exothermic
 Energy is released to the surrounding
 Temperature of surrounding rises
Common Bond Enthalpy Values
Practice
 Calculate the standard enthalpy change of reaction below
CH3CH=CH2 +
H2

CH3CH2CH3
2. CH3CH=CH2 +
Br2 
CH2BrCHBrCH3
3. CH4 + 2Cl2 + 2F2  CF2Cl2 + 2HF + 2HCl
4. CH3CH2OH + CH3COOH  CH3CO2CH2CH3 + H2O
(solve WITHOUT any calculations)
1.
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.
63
 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)
64
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)
65
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)
66
Born-Haber Cycle Diagram
67
 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 = -)
68
Practice
69
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.
• The actual ΔHf for this reaction is -602
kJ/mol. What does this information
indicate about the ionic characteristic of
this compound?
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
70
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
71
 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)
Compare to experimental value.
If theoretical values agree closely to experimental
values, bond is considered to be essentially ionic
72
Compound
Theoretical
Lattice
Formation
Enthalpy
(kJ/mol)
Experimental
Lattice
Formation
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.
73
CONGRATULATIONS.
YOU’VE JUST
FINISHED
TOPIC 5C of
THERMOCHEMISTRY
74
Review Unit 2: Chemical Bonding
 Recall: 3 types of INTERmolecular bonds
 London dispersion forces




Present in ALL substances
Weakest
Increases with increasing polarizability
 Increase in amount of electrons. Generally can be directly related to increasing molar
mass.
Ex: noble gases, halogens (reason why Cl2 is liquid and I2 is solid, even though individual
molecules are entirely non-polar covalent). Substances that are gases at room temp almost
always display LDF as primary forces, alkanes, alkenes, alkynes
 Dipole-dipole forces
 Present in polar covalent molecules
 Stronger
 Strength increases with increasing polarity,
 Ex: esters, HCl, HBr
 Hydrogen bonding
 H connected to F, O, N
AND
 Attracted to lone pair of electrons on F, O or N in another molecule
 Ex: HF, H2O, NH3, alcohols, carboxylic acids
Effects of intermolecular forces
 Amount AND relative strengths of intermolecular bonds
affect physical properties of covalently bonded compounds
such as boiling point and melting point
 For molecules of SIMILAR molar mass, primary factor affecting
b.p./m.p. is relative strength of type of INTER bond
 For molecules of similar type (ex: comparing ALKANES ONLY)
primary factor affecting b.p./m.p. is relative molar mass

Although pure alkanes display ONLY LDFs as primary inter bond,
starting with 6C (C6H14-hexane) exists as liquid and with greater mass
as solid due to INCREASING polarizability (greater number of
electrons).
 Conclusion: greater molar mass = greater polarizability = stronger
overall combined LDF
 When comparing across both type and mass, pay extra careful
attention to information given.

Ex: alkane X is a solid, and alcohol Y is a liquid. Given this information,
you can conclude that the combined LDF forces in alkane X MUST BE
stronger than the hydrogen bonds in the alcohol
 Dissolution in water
 Hydration (dissociation) of ionic substances involve
three steps



Separation of ion (input of energy-endothermic)
Separation of water molecules (input of energy-endothermic)
Ions being surrounded by water (attractive forces-exothermic)
Predicting whether a reaction will occur
 RECALL: spontaneous reaction are those that proceed to the
product side without external influence.
 RECALL: particles with LOWER potential energy are more stable
than particles with higher potential energy.
 By this definition, one may assume that ALL exothermic reaction
always proceed since a negative ∆H implies that products have lower
potential energy.
 However, this is not necessarily true.


For ex: if a reaction has a very high activation energy, then it will not occur
and is described as kinetically stable or under kinetic control.
 In these circumstances, it is possible that a reaction that one might predict
as being highly likely, produces little or no product.
Ex: oxidation of ammonia by oxygen to produce nitrogen monoxide and water
has an enthalpy change value of -909 kJ/mol. However, the reaction does not
proceed due to the high activation energy.
 Similarly, one may assume that an endothermic reaction will never occur,
because by definition the products have higher potential energy than
reactants
 However, many endothermic reactions (ex: dissolution of KCl (+19.2
kJ/mol) or acetic acid and sodium bicarbonate (XXXXXXXXX) do proceed
to completion.
 Since one can assume that endothermic reactions must also have some
required activation energy that must be obtained before the reaction will
proceed, we can conclude that ∆H and activation energy cannot be the
only factors that determine whether a chemical reaction will or will not
occur.
 Further, careful examination of these reactions shows that there is a
change from a relatively ordered (organized) to relatively less ordered
(disorganized) state.
 KCl(s)  K+(aq) + Cl-(aq)
 1 mole  2 mole
 Solid to aqueous
 CH3COOH(aq) + NaHCO3(s)  NaC2H3O2(aq) + CO2(g) + H2O(l)
 2 moles  3 moles
 Solid/aqueous  aqueous/liquid/gas
 We call this degree of disorder or dispersal of energy ENTROPY
Entropy
 Defined as degree of disorder or disarray in a system.
 A measure of the randomness of a system.
 Symbol, S
 Units, (J / K·mol)
 RECALL: reactions are either reversible or irreversible.
 By definition the entropy of the universe
 Increases for spontaneous (thermodynamically favorable)
processes (+ ∆S)
 Does not change for reversible processes (= ∆S)
 **A system however can display an increase (+) or decrease () in entropy. Ex: gas  liquid, system is more organized, and
the system has a decrease in entropy (-)
Perfect Crystal
 An element is considered to be a pure, perfect crystal at 0
K.
 It is completely organized or ordered.
 RECALL: By definition, at absolute zero, all motion stops.
 The absolute entropies of substances in the real world
(not at 0 K) are measured relative to a perfect crystal
 As such all substances in he real world must have entropy
values greater than 0.


Reasonable, as we recognize that all particles have motion; the
greater the motion, the greater the entropy.
The state, temperature, number of particles, bonds, molar mass
and volume of gases are all important factors in entropy.
Comparing Entropies
 Absolute entropy of
 Solids
<
Liquids
 Small number of particle
 Gases of smaller volume
 Solids
<
<
<
<
Gases
large number of particles
gases of larger volume
(larger volume has more
space to be disordered)
aqueous solution
 Increases with increasing temperature since particles move
around more and are more dispersed
Practice
 Predict the sign of ∆So in these reactions
 MgO(s) +
H2O(l) 
Mg(OH)2(s)
 Na2CO3(s)

Na2O +
CO2(g)
 Which of the following has the greater entropy?
 A metal at 273 K or the same metal at 400 K?
 A flexible soft metal like lead, or a rigid solid like
diamond?
 Two samples of the same gas, at the same temperature,
but one at a pressure of 1 atm and the other at a pressure
of 0.5 atm
Calculating Entropy of a Reaction
 The change in Entropy (∆So) can be calculated by
subtracting the sum of the absolute entropies of the
reactants from the sum of the absolute entropies of the
products
 This is analogous to calculating change in Enthalpy.
∆Sorxn = Σ moles Soproducts - Σ moles Soreactants
Example
 Calculate the entropy change, Sorxn for the formation of
ammonia from its component elements, given the
following information

N2(g) + 3H2(g)

 Absolute entropies values
 N2(g)
192 J/(K mol)
 H2(g)
131 J/(K mol)
 NH3(g)
193 J/(K mol)
2 NH3(g)
∆Sorxn=[(2 moles NH3 * 193 J/(K mol)]
–[(1 mole N2 * 192 J/(K mol)+(3 mol H2 * 131 J/(K mol)]
-199 J/(K molrxn)
 Is the calculated value reasonable?
 Yes—negative value implies that the system has become
more ordered (entropy has decreased) since four moles of
gas molecules have converted to two moles of gas molecules
Gibbs Free Energy
 RECALL: we established that given that exothermic
reactions don’t always occur and that many
endothermic reactions do occur, enthalpy change and
activation energy are not the only factors affecting
favorability of a reaction. The second law of
thermodynamics (entropy) must also be considered.
 Enthalpy and entropy are brought together in the
Gibbs-Helmholtz free energy equation
∆Go = ∆Ho - T ∆So
∆Go = ∆Ho - T ∆So
 Gibbs Free Energy allows us to predict the spontaneity or
thermodynamic favorability of a reaction occuring.
 A reaction is said to be thermodyamically favorable, and
therefore will occur, when the value of ∆G is negative



Considering the relationship above, indicates that a negative ∆Go is
favored by a negative ∆Ho and a positive ∆So
Other combinations may also yield a negative ∆Go value
**Know that relationships below-common topic of AP questions.
Additional Considerations
 In order to apply the Gibbs Free Energy equation, all
reactants and products in an equilibrium mixture (Unit 6)
must be present in their standard states (standard states of
elements, gases at 1.0 atm, solutios at 1.0 M). Temperature
should be stated, and is usually (but not always) 298 K
 Entropy values tend to be given in units that involve Joules,
J, whereas enthalpy values are given in kiloJoules, kJ.
 This means that when performing calculations that involve
both entropy and enthalpy, appropriate conversions must be
applied.
 Thermodynamica favorability does NOT indicate any
information about the kinetics (rate, speed) of a reaction.
 A reaction with very negative ∆Go value (spontaneous) may
still be very slow.
Calculating Gibbs Free Energy of a
Reaction
 Gibbs Free Energy (∆Go) can be calculated by subtracting
the sum of the ∆Go values of the reactants from the sum of
the ∆Go values of the products
 This is analogous to calculating change in Enthalpy and
Entropy.
∆Gorxn = Σ moles Goproducts - Σ moles Goreactants
 If ∆Go is positive, then the reaction is NOT
thermodynamically favorable.
 This usually means that the reverse reaction IS
thermodynamically favorable.
 If ∆Go is zero, then the reaction is favored EQUALLY in both
the forward and backward directions
 System is at equilibrium
Free Energy Changes
At temperature of 25oC, use appendix 4
At temperatures other than 25°C,
G° = H  TS
How does G change with temperature?
Free Energy and Temperature
How does G change with temperature?
 There are two parts to the free energy equation:
 H: the enthalpy term
 TS : the entropy term
 The temperature dependence of free energy, then
comes from the entropy term.
1.
2.
3.
4.
H = TS
H < TS
H > TS
We cannot determine without additional
information.
1.
2.
3.
4.
It indicates the process is spontaneous under
standard conditions.
It indicates the process has taken place under
standard conditions.
It indicates the process has taken place at 273K and 1
barr.
It indicates the process has taken place at 1 atm and
0K.
1.
2.
3.
4.
Entropy of universe increases and free energy of the
system decreases.
Entropy of system decreases and free energy of the
universe increases.
Entropy of system increases and free energy of the
universe decreases.
Entropy of universe decreases and free energy of the
system increases.
1.
2.
3.
always increase
always decrease
sometimes increase and sometimes decrease,
depending on the process