THERMOCHEMISTRY

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THERMOCHEMISTRY

• CHAPTER 5

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WHY ENERGY?

• Life requires energy
• From plants photosynthesis
• To equipment electricity
• Vehicles fossil fuels
• To animals - food

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ENERGY FROM WHERE?

• Most of our energy as humans
• From food
• i.e. from chemical reactions
• In the world
• From chemical reactions

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WHAT IS THERMOCHEMISTRY?

• Quantitative study of chemical reactions
  involving heat changes

Consume energy
Produce energy
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Examples Food we eat - provides us with
energy Burning coal- to produce electrical
energy Energy from the sun - responsible for
chemical reactions in plants which make them grow
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RECALL

• FORCE (F) push/pull exerted on an object
• WORK (w) energy used to cause an object to
  move against a force (w Fd)
• HEAT (q) energy transfer as a result of a
  temperature difference.
• Heat always flows from a hotter to a colder
  object until they have the same temperature
• ENERGY (E) capacity to do work or to transfer
  heat

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HOT STUFF?

• Can something transfer cold?

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• Does ice make a drink cold?

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• When you blow hot soup do you make it cold?

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SO IN FACT

• In theory stirring soup long enough and fast
  enough could make it boil

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Heat always flows from a hotter to a colder
object until they have the same temperature
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BUT I KNOW ITS COLD!!

• Heat from our body is transferred,
• making something
• feel cold because we
• lose heat from the area
• touching the cold object

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NATURE OF ENERGY
2 types of energy
Kinetic energy
Potential energy
Energy due to position of object relative to
others (Stored energy)
Energy due to motion of object
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KINETIC ENERGY
Note Ek increases as velocity increases and Ek
increases as mass increases.
Atoms and molecules have mass and are in motion,
thus they have kinetic energy!
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energy an object possesses because of its
temperature.
Thermal energy
Associated with kinetic energy of atoms and
molecules.
The higher the temperature, ? the faster
the atoms and molecules move ? the more
kinetic energy molecules and atoms have
? the greater the thermal energy of the
object.
The total thermal energy of the object is the sum
of the individual energies of the atoms or
molecules.
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MISSING KINETIC ENERGY, PLEASE HELP

• What is it to have no kinetic energy?
• Absolute zero!
• Can we detect that?
• No!
• Why?!
• Because in order to detect it we would have to
  interact with it which would heat it

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POTENTIAL ENERGY
We know Ep mgh
Forces other than gravity that lead to potential
energy ELECTROSTATIC forces
Attractions and repulsions due to oppositely
charged objects (e.g. positive and negative ions)
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Recall
Eel electrostatic energy k Coulomb
constant Q charge d distance between
charges
Describes the electrostatic energy between 2
charges.
UNITS OF ENERGY SI unit J (Joule) 1 J 1
kg.m2.s-2  (Others 1 cal (calorie) 4.184 J)
Ek0.5mv2 (kg)(m.s-1)2
(The amount of heat energy to raise temp of 1g
H2O by 1oC)
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A BRAINTEASER

• If everything has kinetic energy
• And chemical potential energy
• Why is everything not moving around constantly?
• Why are solids not felt to vibrate?

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SYSTEM AND SURROUNDINGS
Everything else
Portion singled out for study
System
Surroundings
System surroundings universe
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Open system can exchange both energy and matter
with the surroundings
Closed system exchanges energy but NOT matter
with the surroundings
Isolated system neither energy nor matter can
be exchanged with the surroundings
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A QUOTE NOT TO LIVE BY

• Thermodynamics is a funny subject. The first
  time you go through it, you don't understand it
  at all. The second time you go through it, you
  think you understand it, except for one or two
  small points. The third time you go through it,
  you know you don't understand it, but by that
  time you are so used to it, it doesn't bother you
  anymore.
• http//www.eoht.info/page/Thermodynamicshumor

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FIRST LAW OF THERMODYNAMICS
Energy is neither created nor destroyed, it is
merely converted from one form into another.
Energy is transferred between the system and the
surroundings in the form of work and heat.
The total energy of the universe remains
constant.
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INTERNAL ENERGY (U)
Also given as E
Internal energy (of the system) sum of all Ek
and Ep of all components of the system
Symbol for change ? ? Change in internal
energy of the system ? ?U
?U UFinal - Uinitial
Number Unit ? Magnitude Sign ? Direction (in
which energy is transferred)
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?U UFinal - Uinitial
?Usyst gt 0 ? Uf gt Ui ? GAINED energy from
surroundings
?Usyst lt 0 ? Uf lt Ui ? LOST energy to
surroundings
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RELATING ?U TO WORK AND HEAT
General ways to change ENERGY of a closed system
WORK Done by/on system
HEAT lost/gained by system
When a system undergoes a chemical/physical change
Relationship ?U q w
We cant measure U, but we can determine ?U.
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SIGN CONVENTION
qgt0 Heat transferred TO system
qlt0 Heat transferred FROM system
SYSTEM
wgt0 Work done ON system
wlt0 Work done BY system
The sign of ?U will depend on the sign and
magnitude of q and w since ?U q w.
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Example
Calculate the change in internal energy for a
process in which the system absorbs 120 J of heat
from the surroundings and does 64 J of work on
the surroundings.
120 J heat
64 J work
?U q w
(120 J) (- 64 J)
56 J
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SOME DEFINITIONS
Extensive property dependent on the amount of
matter in the system. E.g. mass, volume etc
Intensive Property NOT dependent on the amount
of matter in the system. E.g. density,
temperature etc
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State function a function that depends on the
state or conditions of the system and NOT on the
details of how it came to be in that state.
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DEFINITIONS APPLIED TO INTERNAL ENERGY
Recall Total internal energy of a system sum
of all Ek and Ep of all components of the
system Thus total internal energy of a system
? total quantity of matter in system
?U is an extensive property
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Internal energy depends on the state or
conditions of the system (e.g. pressure,
temperature, location)
Does not depend on how it came to be in that
state. ? state function
??U only depends on Ui and Uf and not how the
change occurred.
NOTE Heat and work are not state functions!
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ENTHALPY (H)
We cannot measure enthalpy (H), we can only
measure change in enthalpy (?H).
Change in enthalpy (?H) is the heat gained or
lost by the system when a process occurs under
constant pressure.
E.g. atm pressure
?H Hfinal - Hinitial qp
?H is a state function and an extensive property
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NOTE At constant pressure, most of the energy
lost / gained is in the form of heat.
Very little work is done for the expansion /
contraction against the atmospheric force,
especially if the reaction does not involve gases.
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qpgt0
qplt0
SYSTEM
?Hgt0
?Hlt0
Exothermic process system evolves heat
Endothermic process - system absorbs heat
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Example of an exothermic reaction
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Example of an endothermic reaction
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FINDING THE RELATIONSHIP BETWEEN ?U AND ?H
CONSIDER PV WORK
We know that ?U q w
Assume we do PV work only
(e.g. expanding gases in cylinder of a car does
PV work on the piston)
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Gas expands and moves piston distance d
W Fd
? W PAd
But F PA
W P?V
Sign system is doing work on the piston W
-P?V
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?U q w and w -P?V
? ?U q - P?V
At constant volume, ?V 0
? ?U qv
At constant pressure
?U qp - P?V
But ?H qp
? ?U ?H - P?V
The volume change in many reactions is very
small, thus P?V is very small and hence the
difference between ?U and ?H is small.
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ENTHALPIES OF REACTION
enthalpy change that accompanies a reaction
(heat of reaction)
?Hrxn ?H(products) ?H(reactants)
final
initial
Thermochemical equation 2H2(g) O2(g) ?
2H2O(g) ?H - 483.6 kJ
exothermic
Balanced equation
?H associated with the reaction
Coefficients in balanced equation no. of moles
of reactant/product producing associated ?H
(extensive)
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2H2(g) O2(g) ? 2H2O(g) ?H -483.6 kJ
Note Since enthalpy is an extensive property, ?H
depends on the amount of reactant consumed.
4H2(g) 2O2(g) ? 4H2O(g) ?H -967.2 kJ
Also, ?H for a reaction is equal in magnitude,
but opposite in sign to ?H for the reverse
reaction. 2H2O(g) ? 2H2(g) O2(g) ?H
483.6 kJ
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Notes are available on
http//hobbes.gh.wits.ac.za/chemnotes/fmain-chem10
33-35.htm Go to the section for Mrs C
Billing Lectures Chapter 7, Chapter 8 and
Chapter 9 are this work We are busy with Chapter
7 at the moment
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Example
Calculate the heat needed to convert 25 ml water
to steam at atmospheric pressure if the enthalpy
change is 44 kJ/mol. (Assume the density of
water is 1.0 kg/l)
1.0 g/ml
H2O(l) ? H2O(g)
?H 44 kJ/mol
25 ml
m ?v (1.0 g/ml)(25 ml) 25 g
qp ?H 44 kJ/mol x 1.4 mol 62 kJ
Check - endothermic
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Example
Hydrogen peroxide can decompose to water and
oxygen by the reaction 2H2O2 (l) ? 2H2O (l)
O2 (g) ?H -196 kJ Calculate the heat
produced when 2.50 g H2O2 decomposes at constant
pressure.
qp ?H -196 kJ for 2 mol H2O2
-7.2 kJ for 0.0735 mol
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Example burning money

• Ethanol burns in air to give water vapour and
  carbon dioxide
• Calculate the enthalpy of reaction given enthalpy
  of formations of ethanol (l) -277.7 kJ/mol
• water (l) -285.8 kJ/mol carbon dioxide
  (g) -393.5 kJ/mol
• 2. Calculate the minimum amount of water needed
  to prevent the paper from burning after be soaked
  in 1g of ethanol

First find the balanced chemical equation
C2H5OH (l) 3O2? 3H2O (l) 2CO2 (g)
Enthalpy of reaction
?H ?Hf(products) - ?Hf(reagents)
2(-393.5) 3(-285.8) (-277.7 3x0) kJ/mol
-1365.8 kJ/mol
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Example burning money
Ethanol burns in air to give water vapour and
carbon dioxide 2. Calculate the minimum amount of
water needed to prevent the paper from burning
after be soaked in 1.00g of ethanol
Find the heat released by burning 1.00 g of
ethanol
qp ?H -1365.8 kJ/mol x 0.0217 mol -29.6 kJ
qp
m nM (0.67 mol)(18.0148 g/mol) 12 g
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CALORIMETRY
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Heat Capacity
? The amount of heat required to raise the
temperature by 1K.
Recall a change of 1K a change of 1oC
Units J.K-1 Extensive property
Molar Heat Capacity
? The heat capacity of 1 mol of substance.
Units J.K-1mol-1
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Example 25oC ? 30oC
?T 30oC - 25oC 5oC
Add 273 to convert oC to K 298 K ? 303 K
?T 303 K 298 K 5 K
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Specific Heat Capacity (C) ? heat capacity of 1 g
of substance Units J.K-1g-1
? determined by measuring the change in
temperature that a known mass of substance
undergoes when it gains/loses a specific quantity
of heat.
Specific Heats for Some Substs at 298K H2O
(l) 4.18 J.K-1g-1 N2 (g) 1.04 J.K-1g-1 Al
(s) 0.90 J.K-1g-1 Fe (s) 0.45 J.K-1g-1 Hg
(l) 0.14 J.K-1g-1
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Example

• a) How much heat is needed to warm 1.00 L of
  water from 23.0oC to 98.0oC?
• (Assume the density of water is 1.00 kg/L).
• What is the molar heat capacity of water?

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a) How much heat is needed to warm 1.00 L of
water from 23.0oC to 98.0oC? (Assume the
density of water is 1.00 kg/L).
m ?v (1.00 L)(1.00 kg.L-1) 1.00x103 g
?T 98.0 23.0 75.0oC
75.0 K
q (4.18 J.g-1.K-1)(1.00x103 g)(75.0 K) q
3.14x105 J
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b) What is the molar heat capacity of water?
C 4.18 J.g-1.K-1
MH2O 18.016 g.mol-1
75.3 J.mol-1.K-1
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CONSTANT PRESSURE CALORIMETRY
Any calorimeter at atmospheric pressure has a
constant pressure, e.g. coffee-cup calorimeter.
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The heat produced by a reaction is entirely
absorbed by the solution at constant
pressure. i.e. heat does not escape the
calorimeter
Recall ?H qp
Can calculate qsoln Cm?T, but we want qrxn
Since heat given off by the reaction is absorbed
by the solution qrxn -qsoln
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Example
When 4.25 g solid ammonium nitrate dissolves in
60.0 g water in a coffee-cup calorimeter, the
temperature drops from 22.0oC to 16.9oC.
Calculate ?H (in kJ/mol NH4NO3) for the
dissolution process. Assume the specific heat of
solution is the same as that for pure water.
NH4NO3(s) ? NH4(aq) NO3-(aq)
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When 4.25 g solid ammonium nitrate dissolves in
60.0 g water in a coffee-cup calorimeter, the
temperature drops from 22.0oC to 16.9oC.
Calculate ?H (in kJ/mol NH4NO3) for the
dissolution process.
?T 16.9 22.0 -5.1oC
-5.1 K
endothermic
m 60.0 4.25 64.3 g
assumption
qrxn -Cm?T qrxn -(4.18 J.g-1.K-1)(64.3
g)(-5.1 K) qrxn 1.4x103 J
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When 4.25 g solid ammonium nitrate dissolves in
60.0 g water in a coffee-cup calorimeter, the
temperature drops from 22.0oC to 16.9oC.
Calculate ?H (in kJ/mol NH4NO3) for the
dissolution process.
qrxn 1.4x103 J
MNH4NO3 80.052 g.mol-1
endothermic
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CONSTANT VOLUME CALORIMETRY BOMB CALORIMETER
Recall ?U qv
? At constant volume measure ?U rather than ?H,
but for most ?U ? ?H
A bomb calorimeter is used to study combustion
reactions etc. We calculate the heat of
combustion from the measured change in
temperature. We need to know the heat capacity
of the calorimeter (Ccal).
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qrxn -Ccal x ?T
NB This equation cannot be applied
blindly!!! Always check units for
Ccal. Always include units in calculations to
ensure they cancel.
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Bomb Calorimeter
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Example
A 2.200 g sample of quinone, C6H4O2, is burned in
a bomb calorimeter whose total heat capacity is
7.854 kJ/oC. The temperature of the calorimeter
increases from 23.44oC to 30.57oC. What is the
heat of combustion per gram of quinone? per mole
of quinone?
?T 30.57 23.44 7.13oC
exothermic
qrxn -Ccal?T qrxn -(7.854 kJ.oC-1)(7.13oC) qrx
n -56.0 kJ
exothermic
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A 2.200 g sample of quinone, C6H4O2, is burned in
a bomb calorimeter whose total heat capacity is
7.854 kJ/oC. The temperature of the calorimeter
increases from 23.44oC to 30.57oC. What is
the heat of combustion per gram of quinone? per
mole of quinone?
produced by
qrxn -56.0 kJ
2.200 g quinone
-25.5 kJ
1 g quinone
i.e. -25.5 kJ.g-1
-25.5 kJ.g-1 x 108.09 g.mol-1 -2.75 x 103
kJ.mol-1
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Example
50.0 g of water at 62.5oC was poured into a
calorimeter containing 50.0 g water at 18.7oC.
The final temperature was 35.0oC. How much heat
was lost to the surroundings during this process?
(Heat capacity of water 4.184 J.oC-1.g-1)
Ans -2.34 kJ
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50.0 g of water at 62.5oC was poured into a
calorimeter containing 50.0 g water at 18.7oC.
The final temperature was 35.0oC. How much heat
was lost to the surroundings during this process?
(Heat capacity of water 4.184 J.oC-1.g-1)
H2O 1 m 50.0 g
H2O 2 m 50.0 g
Ti 62.5oC Tf 35.0oC
Ti 18.7oC Tf 35.0oC
?T -27.5oC
?T 16.3oC
qsoln Cm?T
q1 (4.184)(50.0)(-27.5)
q2 (4.184)(50.0)(16.3)
q1 -5753 J
q2 3410 J
q1 q2 -2343 J -2.34 kJ
Negative q ? heat lost by soln to the surroundings
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HESSS LAW
If a reaction is carried out in a series of
steps, ?H for the reaction will be equal to the
sum of the enthalpy changes for the individual
steps.
A B ? C D ?H1 ?H2
?H for a reaction is calculated from ?H data of
other reactions.
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Recall that enthalpy is a state function ? ?H is
independent of the path followed.
It is useful to use Hesss Law in cases where ?H
cannot be measured.
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Example
Carbon occurs in two forms, graphite and diamond.
The enthalpy of combustion of graphite is 393.5
kJ/mol and that of diamond is 395.4 kJ/mol.
Calculate ?H for the conversion of graphite to
diamond.
C(s,graph) O2(g) ? CO2(g) ?H 393.5 kJ/mol
C(s,diam) O2(g) ? CO2(g) ?H 395.4 kJ/mol
C(s,graph) ? C(s,diam) ?H ?
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C(s,graph) O2(g) ? CO2(g) ?H 393.5 kJ/mol
C(s,diam) O2(g) ? CO2(g) ?H 395.4 kJ/mol
C(s,graph) ? C(s,diam) ?H ?
C(s,graph) O2(g) ? CO2(g) ?H
-393.5 kJ/mol
CO2(g) ? C(s,diam) O2(g) ?H
395.4 kJ/mol
C(s,graph) ? C(s,diam)
?H -393.5 -(-395.4) 1.9 kJ/mol
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Example
From the following enthalpies of reaction
H2(g) F2(g) ? 2HF(g) ?H -537 kJ C(s)
2F2(g) ? CF4(g) ?H -680 kJ 2C(s) 2H2(g)
? C2H4(g) ?H 52.3 kJ Calculate ?H for the
reaction of ethylene with F2 C2H4(g) 6F2(g)
? 2CF4(g) 4HF(g)
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H2(g) F2(g) ? 2HF(g) ?H -537 kJ
C(s) 2F2(g) ? CF4(g) ?H -680 kJ 2C(s)
2H2(g) ? C2H4(g) ?H 52.3 kJ C2H4(g) 6F2(g)
? 2CF4(g) 4HF(g)
C2H4(g) ? 2C(s) 2H2(g)
?H -(52.3 kJ)
2C(s) 4F2(g) ? 2CF4(g)
?H 2(-680 kJ)
2H2(g) 2F2(g) ? 4HF(g)
?H 2(-537 kJ)
C2H4(g) 6F2(g) ? 2CF4(g) 4HF(g)
?H -(52.3 kJ) 2(-680 kJ) 2(-537 kJ) ?H
-2.49x103 kJ
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STANDARD STATES
The magnitude of ?H depends on conditions of
temperature, pressure, and state of products and
reactants. In order to compare enthalpies, need
same set of conditions.
Standard state of a substance pure form at 1
atm and at the temperature of interest (usually
298 K (25oC))
Standard enthalpy (?Ho) ? when all products and
reactants are in their standard states
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STANDARD ENTHALPIES OF FORMATION
?Hfo? change in enthalpy for the reaction that
forms 1 mol of a substance from its elements in
their standard states. Units kJ/mol
Note If there is more than one form of the
element present under standard conditions, use
the most stable form.
?Hfo 0 kJ/mol for the most stable form of any
element
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Examples of most stable form
Oxygen(g) can exist as O2 or O3 (ozone) at 1
atm and 298 K
Most stable form O2
?Hfo(O2(g)) 0 kJ/mol
Carbon(s) can exist as graphite or diamond at 1
atm and 298 K
Most stable form graphite
?Hfo(C(s,graphite)) 0 kJ/mol
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Tabulated data of enthalpies of formation can be
used to calculate enthalpies of reaction as
follows
Stoichiometric coefficients
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Table of standard enthalpies of formation
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General Knowledge
Gray tin and white tin are two solid forms of
tin. The denser white metallic form is the most
stable phase above 13oC, and the powdery gray
form is more stable below 13oC. The formation of
gray tin is said to have contributed to
Napoleons defeat at Moscow, when his soldiers
buttons fell off their clothes at the low
temperatures they encountered there.
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Example
Nitroglycerine is a powerful explosive, giving
four different gases when detonated 2C3H5(NO3)3
(l) ? 3N2(g) 1/2O2(g) 6CO2(g)
5H2O(g) Given that the enthalpy of formation of
nitroglycerine is ?Hfo -364 kJ.mol-1, calculate
the enthalpy change when 10.0 g of nitroglycerine
is detonated.
?Hfo C3H5(NO3)3 (l) -364 kJ.mol-1 ?Hfo H2O
(g) -241.8 kJ.mol-1 ?Hfo CO2 (g) -393.5
kJ.mol-1
?Hfo O2 (g) ?Hfo N2 (g)
0 kJ.mol-1
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2C3H5(NO3)3 (l) ? 3N2(g) 1/2O2(g) 6CO2(g)
5H2O(g) Calculate the enthalpy change when 10.0 g
of nitroglycerine is detonated.
?Hfo C3H5(NO3)3 (l) -364 kJ.mol-1 ?Hfo H2O
(g) -241.8 kJ.mol-1 ?Hfo CO2 (g) -393.5
kJ.mol-1
?Horxn 3(0) 1/2(0) 6(-393.5)
5(-241.8) - 2(-364) ?Horxn
-3570--728 ?Horxn -2842 kJ
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2C3H5(NO3)3 (l) ? 3N2(g) 1/2O2(g) 6CO2(g)
5H2O(g) Calculate the enthalpy change when 10.0 g
of nitroglycerine is detonated.
?Horxn -2842 kJ
-2842 kJ for 2 mol NG
? x kJ for 0.0440 mol NG
x -62.5 kJ for 10.0 g
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Example For you to do!
Calculate ?Ho for the decomposition of
limestone CaCO3(s) ? CaO(s)
CO2(g) Hence calculate the heat required to
decompose 1 kg of limestone.
?Hfo CaCO3 (s) -1207.1 kJ.mol-1 ?Hfo CaO
(s) -635.5 kJ.mol-1 ?Hfo CO2 (g) -393.5
kJ.mol-1
Answer ?Ho 178.1 kJ for 1 mol qp ?Ho
1779 kJ for decomposition of 1 kg CaCO3
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BOND ENTHALPIES
The stability of a molecule is related to the
strengths of the covalent bonds it contains.
The strength of a covalent bond between two atoms
is determined by the energy required to break
that bond.
Bond enthalpy enthalpy change for
breaking a particular bond in a mole of
gaseous substance.
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Can determine approximate value for ?Horxn if we
have the values of average bond
enthalpies. ?Horxn is calculated by determining
the energy required to break all bonds minus the
energy evolved to form all bonds.
Reactants initial
Products final
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This is only an approximation because the
average bond enthalpies used as the bond
enthalpies are not exactly the same in all
molecules.
e.g. C-H bond energy in CH4 is slightly different
to that in C2H6 and so on.
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You must know these
O2 N2 CO CO2 CN H2CO
OO
N?N
C?O
OCO
C?N
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Example
Hydroiodic acid reacts with chlorine as
follows 2HI(aq) Cl2(g) ? 2HCl(aq)
I2(s) Approximate ?Ho for the reaction using
tabulated bond energies.
2HI(aq) Cl2(g) ? 2HCl(aq) I2(s)
H-I
Cl-Cl
H-Cl
I-I
H-Cl
H-I
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2HI(aq) Cl2(g) ? 2HCl(aq) I2(s)
Bonds Broken
Bond Energy / kJ.mol-1
2(297)
H-I
2 x
Cl-Cl
1 x
239
833
Bonds Formed
2 x
H-Cl
2(431)
1 x
I-I
149
1011
833 1011 -178 kJ.mol-1
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Example
Estimate ?H using bond enthalpies for the
following reaction C2H6(g) 7/2O2(g) ?
2CO2(g) 3H2O(g)
OCO
OO
6 x
C-H
OC
4 x
6 x
O-H
C-C
1 x
OO
7/2 x
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C2H6(g) 7/2O2(g) ? 2CO2(g) 3H2O(g)
Bonds Broken
Bond Energy / kJ.mol-1
6(413)
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7/2(494)
4555
Bonds Formed
4(707)
6(463)
5606
4555 5606 -1051 kJ.mol-1