Chemistry 100(02) Fall 2009
Webpage: http://blackboard.latech.edu/
Dr. Upali Siriwardane
CTH 311 Phone 257-4941
Upali@chem.latech.edu
Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 10:00
- 12:00.
Exams:
October 5,
2009 (Test 1): Chapter 1 & 2
October 28, 2009 (Test 3): Chapter 3 & 4
November 18, 2009 (Chapter 5 & 6)
November 19, 2009 (Make-up test) comprehensive:
Chapters 1 through 6 )
8:00-9:15 AM, CTH 328
CHEM 100, Fall 2009, LA TECH
6-1
Chapter 6. Energy and Chemical Reactions
6.1 The Nature of Energy Page 214
6.2 Conservation of Energy Page 217
6.3 Heat Capacity Page 222
6.4 Energy and Enthalpy Page 227
6.5 Thermochemical Expressions Page 233
6.6 Enthalpy Changes for Chemical Reactions Page 235
6.7 Where Does the Energy Come From? Page 240
6.8 Measuring Enthalpy Changes: Calorimetry Page 242
6.9 Hess's Law Page 246
6.10 Standard Molar Enthalpies of Formation Page 248
6.11 Chemical Fuels for Home and Industry Page 253
6.12 Foods: Fuels for Our Bodies Page 256
PORTRAIT OF A SCIENTIST: James P. Joule Page 215
ESTIMATION: Earth's Kinetic Energy Page 217
CHEM 100, Fall 2009, LA TECH
6-2
Chapter 6. KEY CONCEPTS
Energy and Chemical Reactions
•Kinetic energy and potential
energy
•Energy units and unit conversion
•Conservation of energy: heat and
work
•Thermodynamic terms: system,
surroundings, thermal
equilibrium, exothermic,
endothermic, and state function
•Heat capacity and calorimetry
•Internal energy and enthalpy
Thermochemical equations
Thermostoichiometric factors
CHEM 100, Fall 2009, LA TECH
•Standard enthalpy change for a
reaction, DHo
•Enthalpy change from bond
enthalpies
•Calorimetry and thermal
energy transferred during a
reaction
•Hess's law and enthalpy
•Standard molar enthalpies of
formation DHo of a reaction.
•Chemical fuels and heating
•Main components of food
•Caloric intake
6-3
Kinetic energy and potential energy
External or Macroscopic Energy
Potential Energy: Energy of an object as a result of
its position
Kinetic Energy: Energy of an object as a result of its
motion.
Internal or submicroscopic (nano-scale) Energy)
Potential Energy: Energy of an atoms or molecules
as a result of its position at nano-scale
Kinetic Energy: Energy of an object as a result of
motion of its atoms and molecules at nano-scale.
Temperature is directly proportional to kinetic
energy (thermal energy) of atoms and molecules
Total Energy = Kinetic + Potential
CHEM 100, Fall 2009, LA TECH
6-4
Energy Transformations
CHEM 100, Fall 2009, LA TECH
6-5
Thermal Energy
CHEM 100, Fall 2009, LA TECH
6-6
Forms of Energy
Energy - the ability to do work.
Work - when a force is applied to an object.
There are several types of energy:
Thermal - heat
Electrical
Radiant - including light
Chemical
Mechanical - like sound
Nuclear
CHEM 100, Fall 2009, LA TECH
6-7
Energy units
Kinetic energy was defined as:
1
2
2
kinetic energy = mv
m = mass and v = velocity.
Joule (J) - the energy required to move a 2 kg mass at
a speed of 1 m/s. It is a derived SI unit.
2
1
J = kinetic energy = 2 (2 kg) (1 m/s)
2 -2
= 1 kg m s
Volume expansion work; P DV
24.5 L atm x 101. 3 J = 2482 J
1 L atm
CHEM 100, Fall 2009, LA TECH
6-8
Energy units and unit
conversion
1J = 1 kg m2/sec2
1 cal = 4.184 J
1kcal = 1 Cal
thus
1 Cal = 1 kcal = 1000 cal = 4.184 kJ = 4184 J
CHEM 100, Fall 2009, LA TECH
6-9
Law of Conservation of Energy
“Energy cannot be created or destroyed in a
chemical reaction.”
During a reaction, energy can change from one form
to another.
Example. Combustion of natural gas.
Chemical bonds can be viewed as potential energy.
So during the reaction:
2CH4 (g) + 3O2 (g)
2CO2 (g) + 2H2O (l) + thermal energy + light
some potential energy is converted to thermal
energy and light.
CHEM 100, Fall 2009, LA TECH
6-10
First Law of Thermodynamics
the total amount of energy in the universe is a
constant
the amount of heat transferred into a system plus
the amount of work done on the system must
result in a corresponding increase of internal
energy in the system
CHEM 100, Fall 2009, LA TECH
6-11
Measuring Temperature
CHEM 100, Fall 2009, LA TECH
6-12
Hot and Cold Iron
CHEM 100, Fall 2009, LA TECH
6-13
Heat Transfer
Heat is always
transferred
from the hotter to
the cooler sample
CHEM 100, Fall 2009, LA TECH
6-14
State functions
Depend only on the initial and final states of a
system. They are independent of how the
system gets from one state to another.
State functions include:
Energy
Pressure
Volume
Temperature
Enthalpy (DH)
CHEM 100, Fall 2009, LA TECH
6-15
Path Independent Energy Changes
CHEM 100, Fall 2009, LA TECH
6-16
Enthalpy Diagram
CHEM 100, Fall 2009, LA TECH
6-17
Thermochemistry Terminology
Chemical energy – energy associated with a
chemical reaction
Thermochemistry – the quantitative study of
the heat changes accompanying chemical
reactions
Thermodynamics – the study of energy and
its transformations
CHEM 100, Fall 2009, LA TECH
6-18
Thermochemistry Terminology
State functions
 properties which depend only on the initial and
final states
 properties which are path independent
Non-state properties
 properties which are path dependent
state properties  E
non-state properties  q & w
CHEM 100, Fall 2009, LA TECH
6-19
Stepwise Energy Changes in Reactions
CHEM 100, Fall 2009, LA TECH
6-20
Bond Energy and DH
Energy associated with holding 2 atoms together
For example, if the reaction
HF(g) -> H(g) + F(g) ; DH = +565 kJ
then we can say that the HF bond energy is 565
kJ/mol.
CHEM 100, Fall 2009, LA TECH
6-21
Why is it necessary to divide Universe
into System and Surrounding
Universe = System + Surrounding
System  that part of the universe under investigation
Surroundings  the rest of the Universe
CHEM 100, Fall 2009, LA TECH
6-22
Internal Energy
The sum of the individual energies of all nano-scale
particles (atoms, ions, or molecules) in that
sample
Chemical Energy: Potential energy as stored in
bonds
Nuclear energy: E = 1/2mc2
Thermal Energy: Depends on the temperature
Total Internal Energy: Depends on the type of
particles, and how many of them there are in the
sample
CHEM 100, Fall 2009, LA TECH
6-23
Internal Energy, Heat, and Work
Sign Conventions: gain (+). Loss(-)
CHEM 100, Fall 2009, LA TECH
6-24
Internal Energy
CHEM 100, Fall 2009, LA TECH
6-25
What is the internal energy change (DE)
of a system?
DE is associated with changes in atoms,
molecules and subatomic particles
Total Energy = Eexternal + DEinternal
DE = work (w) or volume expansion + heat (kinetic
energy) + nuclear energy + chemical energy
DE = heat (q) + w (work)
DE = q + w
DE = q -P DV; w =- P DV
CHEM 100, Fall 2009, LA TECH
6-26
Volume Expansion Work
w = -p DV, Why is negative
sign?
Work has a sign: performed (- , loss) or done on
the system (+, gain)
volume expansion work: compression and
expansion
compression DV = Vf -Vi ;DV is negative
expansion DV = Vf -Vi ;DV is positive
compression: w = -p DV; DV = -, w is +
expansion: w = -pDV; DV = +, w is -
CHEM 100, Fall 2009, LA TECH
6-27
Volume Expansion Type Work
w = PDV
Expansion w is +
compression w is -
DV = Vinitial + Vincrease
P
Vincrease
P
qp = +2kJ
CHEM 100, Fall 2009, LA TECH
Vinitial
6-28
Calculations using DE = q + w and
DE = q - PDV
a) In a process in which 89 J of work is done
on a system, 567 J of heat is given off. What is
the DE of the system?
b) In a process in which a gas expands from
25 L to 50 L against a constant pressure of
0.980 atm and 650 J of heat is absorbed.
What is the DE of the system?
CHEM 100, Fall 2009, LA TECH
6-29
a) In a process in which 89 J of work is
done on a system, 567 J of heat is given
off. What is the DE of the system?
DE = q + w
DE = -567 J +89 J
DE = -478 J
Internal energy (DE) of a system is decreased or
loss by an amount of 478 J.
CHEM 100, Fall 2009, LA TECH
6-30
b) In a process in which a gas expands from 25
L to 50 L against a constant pressure of 0.980
atm and 650 J of heat is absorbed. What is the
DE of the system?
DV = 50 L- 25 L= 25 L
w = -pDV = -0.980 atm
x 25 L =- 24.5 L atm
to convert L atom to J use conversion factor
1 L atm = 101. 3 J
-24.5 L atm x 101. 3 J = - 2482 J
1 L atm
-pDV = -2482 J
q = 650
DE = q -pDV
DE = 650 -2482 J = -1832 J
Internal energy (DE) of a system is decreased or loss by an amount of
1832 J
CHEM 100, Fall 2009, LA TECH
6-31
Measuring thermal energy changes
Thermal energy cannot be directly measured.
We can only measure differences in energy.
To be able to observe energy changes, we must
be able to isolate our system from the rest of
the universe.
Calorimeter - a device that is used to measure
thermal energy changes and provide isolation
of our system.
CHEM 100, Fall 2009, LA TECH
6-32
Heat capacity vs Specific heat
Every material will contain thermal energy.
Identical masses of substances may contain
different amounts of thermal energy even if
at the same temperature.
Heat capacity. The quantity of thermal energy
required to raise the temperature of an
object by one degree.
Specific heat. The amount of thermal energy
required to raise the temperature of one
gram of a substance by one degree.
CHEM 100, Fall 2009, LA TECH
6-33
o
Specific Heats at 25 C, 1 atm
Substance
DH
Substance DH
Al(s)
0.90
Fe (s)
0.45
Br2 (l)
0.47
H2O (s)
2.09
C (diamond)
0.51
H2O (l)
4.18
C (graphite)
0.71
H2O (g)
1.86
CH3CH2OH (l)
2.42
N2 (g)
1.04
CH3(CH2)6CH3 (l)
2.23
O2 (g)
0.92
DH = specific heat, J g-1 oC-1
CHEM 100, Fall 2009, LA TECH
6-34
What basic ideas are used in
calorimetric calculations?
Heat gain = - heat loss (qlost = - qgained)
1st law of thermodynamics
Heat gain/loss = Specific heat x mass x Dt
Heat gain/loss = Heat capacity x Dt
Dt = final temperature – initial temperature
Dt = tf - ti
What unknown?
Spec. heat, heat cap, tf, mass, DH(coffee cup), DE
(bomb calorimeter)
CHEM 100, Fall 2009, LA TECH
6-35
EXAMPLE If 100. g of iron at 100.0oC is placed
in 200. g of water at 20.0oC in an insulated
container, what will the temperature, oC, of the iron
and water when both are at the same temperature?
The specific heat of iron is 0.106 cal/goC.
(100.g 0.106cal/goC  (Tf - 100.)oC) = qlost
- qgained = - (200.g  1.00cal/goC  (Tf - 20.0)oC)
10.6(Tf - 100.oC) = - 200.(Tf - 20.0oC)
10.6Tf - 1060oC = - 200.Tf + 4000oC
(10.6 + 200.)Tf = (1060 + 4000)oC
Tf = (5060/211.)oC = 24.0oC
CHEM 100, Fall 2009, LA TECH
6-36
What is it?
heat gain/loss?
Heat gain= [Specific heat x mass x Dt]
Heat loss = - [Specific heat x mass x Dt]
specific heat?
amount of heat needed to increase
temperature of 1g of a material by 1oC
heat capacity?
heat capacity = [Specific heat x mass]
temperature drop/gain (Dt)?
Dt = tfinal - tinitial
CHEM 100, Fall 2009, LA TECH
6-37
Two ways of measuring heat changes
Constant pressure?
qp (open to atmosphere-coffee cup calorimeter)
Constant volume?
qv (bomb calorimeter).
CHEM 100, Fall 2009, LA TECH
6-38
Bomb Calorimeter
CHEM 100, Fall 2009, LA TECH
6-39
Coffee cup calorimeter
Conduct a reaction
and look at the
temperature change.
CHEM 100, Fall 2009, LA TECH
6-40
What exactly is DH?
Heat measured at constant pressure qp
Chemical reactions are exposed to
atmosphere and are held at a constant
pressure.
Volume of materials or gases produced can
change. ie: work = -PDV
DE = qp + w
qp = DE + PDV;
w = -PDV
DH = DE + PDV;
qp = DH(enthalpy )
CHEM 100, Fall 2009, LA TECH
6-41
When 1.00 mole of HCl is reacted with
1.00 mole of NaOH in 1.0 L of water, the
temperature of water increases by 13.7
oC. Calculate the heat of the reaction for
the following thermochemical equation.
HCl(aq) + NaOH (aq) ---> NaCl (aq) + H2O(l); DH= ?
CHEM 100, Fall 2009, LA TECH
6-42
How do you measure DE?
Heat measured at constant volume qv
Chemical reactions takes place inside a
bomb.
Volume of materials or gases produced
can not change. ie: work = -PDV= 0
DE = qv + w
qv = DE + 0; w = 0
DE = qv = DE(internal energy )
CHEM 100, Fall 2009, LA TECH
6-43
Bomb calorimeter(constant volume)
CHEM 100, Fall 2009, LA TECH
6-44
DE and DH from First Law of
Thermodynamics
DE = q + w
at constant V, wexpansion = 0
DE = qv + 0
at constant P, wexpansion = PDV
DE = qp + PDV
qp = DH = DE - PDV
CHEM 100, Fall 2009, LA TECH
6-45
EXAMPLE A 1.000g sample of a
compound was burned in an oxygen
bomb calorimeter. It produced 12.0 kJ
of heat. The temperature of the
calorimeter and 2000 g of water was
raised 4.645oC. How much heat is
gained by the calorimeter?
heat gained = – heat lost
heatcalorimeter + heatwater = heatreaction
heatcalorimeter = heatreaction - heatwater
CHEM 100, Fall 2009, LA TECH
6-46
EXAMPLE A 1.000g sample of a
compound was burned in an oxygen
bomb calorimeter. It produced 42.0 kJ
of heat. The temperature of the
calorimeter and 2000 g of water was
raised 4.645oC. How much heat is
gained by the calorimeter?
heatcalorimeter = heatreaction - heatwater
heat = 42.0 kJ ((2.000kg)(4.645oC)(4.184kJ/kgoC))
= 3.13 kJ
CHEM 100, Fall 2009, LA TECH
6-47
Example
What is the mass of water equivalent of the heat
absorbed by the calorimeter?
#g = (3.13 kJ/4.645oC)(1.00kg  C/4.184kJ)
= 1.61 x 102 g
CHEM 100, Fall 2009, LA TECH
6-48
What is it?
heat of fusion
Heat absorbed in this process of breaking
intermolecular forces in a solid to become
a liquid
heat of evaporation
Heat absorbed in this process of breaking
intermolecular forces in a liquid to
become a gas
CHEM 100, Fall 2009, LA TECH
6-49
Example A 1.000 g sample of ethanol
(MM 46.07) was burned in the sealed
bomb calorimeter described above.
The temperature of the water rose from
24.284oC to 27.559oC. Determine the
heat for the reaction.
m = (2000 + ”161")g H2O
Dt = 27.559oC - 24.284oC = 3.275oC
q = m  s.h.  Dt
q= (2161g)(4.184J/goC)(3.275oC) = 29.61 kJ
q = (29.61 kJ/1.000g)(46.07 g/mol) = 1364
kJ/mol
CHEM 100, Fall 2009, LA TECH
6-50
Freezing and Melting
CHEM 100, Fall 2009, LA TECH
6-51
EXAMPLE: How much heat is required to
heat 10.0 g of ice at -15.0oC to steam at
127.0oC?
q ice = m  s.h.  Dtice
2.09J/goC
qwater = m  s.h.  Dtwater
4.18J/goC
qsteam = m  s.h.  Dtsteam
2.03J/goC
qfusion = m  heat of fusion
333J/g
qboil. = m  heat of vaporization 2260J/g
qoverall = qice + qfusion + qwater + qboil. + qsteam
CHEM 100, Fall 2009, LA TECH
6-52
EXAMPLE: How much heat is required to heat
10.0 g of ice at -15.0oC to steam at 127.0oC?
qoverall = qice + qfusion + qwater + qboil. + qsteam
q = (10.0g  2.09J/goC  15.0oC)
+ (10.0g  333J/g)
+ (10.0g  4.18J/goC  100.0oC)
+ (10.0g  2260J/g)
+ (10.0g  2.03J/goC  27.0oC)
q = (314 + 3.33X103 + 4.18X103 + 2.26X104 +
548)J
= 23.3 kJ
CHEM 100, Fall 2009, LA TECH
6-53
Heating, Temperature Change, and
Phase Change
CHEM 100, Fall 2009, LA TECH
6-54
Vaporization and Condensation
CHEM 100, Fall 2009, LA TECH
6-55
Thermochemical Equation
Thermochemistry: deals with heat changes
during chemical reactions
Thermochemical equation. E.g.
2H2 (g) + O2 (g) ---> 2H2O(l) DH = - 285 kJ;
DH is called the enthalpy of reaction.
if DH is + reaction is called endothermic
if DH is - reaction is called exothermic
The magnitude of DH is directly proportional to
the amount of reactant or product.
Sign of DH change from + to - for the reverse
reaction.
CHEM 100, Fall 2009, LA TECH
6-56
Exothermic Reaction
Energy
Reactants
Products
Since excess energy is released,
the products are more stable.
CHEM 100, Fall 2009, LA TECH
6-57
Endothermic Reaction
Energy
Products
Reactants
Additional energy is required
because the products are less stable.
CHEM 100, Fall 2009, LA TECH
6-58
Thermostoichiometry
Standard State Enthalpy: DH°
enthalpy at thermodynamic standard conditions of
298 K, 1 atm, and 1molar for solutions
How do you calculate the heat given out or absorbed
in a reaction ?
• Get the thermochemical equation
• Get the grams of materials
• Convert grams to moles
• Do a ratio calculation using stoichiometric
coefficients and DH in the thermochemical
equation
CHEM 100, Fall 2009, LA TECH
6-59
Calculate the amount of heat given out by
burning 50g of methanol (CH3OH) in excess
oxygen using following the thermochemical
equation:
2 CH3OH(l) + 3O2(g) ----> 2CO2(g) + 4H2O(l) ; D H = 1691.73 kJ
CHEM 100, Fall 2009, LA TECH
6-60
CHEM 100, Fall 2009, LA TECH
6-61
Stoichiometric Calculations
Determine the thermal energy released when 50.0
grams of methane is burned in an excess of
oxygen.
First, determine the number of moles of methane
(MM = 16.043 u).
mol CH4 = (50.0 g) / (16.043 g/mol)
= 3.12 mol CH4
CHEM 100, Fall 2009, LA TECH
6-62
Stoichiometric Calculations
Now look at the balanced thermochemical equation.
CH4 (g) + 2O2 (g)
CO2 (g) + H2O (l) DHrxn = -890.32
kJ
DHrxn = -890.32 kJ / mol CH4 so:
Thermal energy released
= (3.12 mol CH4(g)) (-890.32 kJ / mol CH4 )
= - 2.78 x 103 kJ
CHEM 100, Fall 2009, LA TECH
6-63
What is Hess's Law of
Summation of Heat?
To get DH for new reactions.
Two methods?
• 1st method: new DH is calculated by adding
DHs of several other reactions.
• 2nd method: Where DHf ( DH of formation) of
reactants and products are used to calculate
DH of a reaction.
CHEM 100, Fall 2009, LA TECH
6-64
Hess’s law
The value of H for the reaction is the same whether it occurs
directly or in a series of steps.
DHoverall = DH1 + DH2 + DH3 + · · ·
A+B
A+B
energy
F
E+
C
CHEM 100, Fall 2009, LA TECH
D+B
C
6-65
EXAMPLE
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
CH4(g)  C(s) + 2 H2(g)
DH1
2 O2(g)  2 O2(g)
DH2
C(s) + O2(g)  CO2(g)
DH3
2 H2(g) + O2(g)  2 H2O(l)
DH4
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
DHoverall = DH1 + DH2 + DH3 + DH4
CHEM 100, Fall 2009, LA TECH
6-66
Hess’s law
The thermal energy given off or absorbed in a
given change is the same whether it takes place
in a single step or several steps.
This is just another way of stating the law of
conservation of energy.
If the net change in energy were to differ based on
the steps taken, then it would be possible to
create energy -- this cannot happen!
CHEM 100, Fall 2009, LA TECH
6-67
Calculating enthalpies: 1st method
Thermochemical equations can be combined to
calculate DHrxn.
Example. 2C(graphite) + O2 (g)
2CO (g)
This cannot be directly determined because CO2 is
always formed.
However, we can measure the following:
C(graphite) + O2 (g)
2CO (g) + O2 (g)
CHEM 100, Fall 2009, LA TECH
CO2 (g)
2CO2 (g)
DHrxn= -393.51 kJ
DHrxn= -565.98 kJ
6-68
Calculating enthalpies
By combining the two equations, we can
determine the DHrxn we want.
2 [ C(graphite) + O2 (g)
CO2 (g) ] DHrxn= -787.02 kJ
2CO2 (g)
2CO (g) + O2 (g)
DHrxn= +565.98 kJ
Note.
Because we need 2 moles of CO2 to be produced
in the top reaction, the equation and its DHrxn
were doubled.
CHEM 100, Fall 2009, LA TECH
6-69
Calculating enthalpies
Now all we need to do is to add the two equations together.
2C(graphite) + 2O2 (g)
2CO2 (g)
DHrxn= -787.02 kJ
2CO2 (g)
2 CO (g) + O2 (g)
2 C(graphite) + O2 (g)
2 CO (g)
DHrxn= +565.98 kJ
DHrxn= -221.04 kJ
Note.
The 2CO2 cancel out, as does one of the O2 on the right-hand
side.
CHEM 100, Fall 2009, LA TECH
6-70
Method 1: Calculate DH for the reaction:
so2(g) +
1/2 O2(g) + H2O(g) -----> H2SO4(l) DH = ?
Other reactions:
SO2(g) ------> S(s) + O2(g)
DH = 297kJ
H2SO4(l)------> H2(g) + S(s) + 2O2(g) DH = 814 kJ
H2(g) +1/2O2(g) -----> H2O(g)
DH = -242 kJ
CHEM 100, Fall 2009, LA TECH
6-71
SO2(g) ------> S(s) + O2(g);DH1 =
297 kJ - 1
H2(g) + S(s) + 2O2(g) ------> H2SO4(l)
DH2 = -814 kJ - 2
H2O(g) ----->H2(g) + 1/2 O2(g)
DH3 = +242 kJ - 3
______________________________________
SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l)
DH = DH1 + DH2 + DH3
DH = +297 - 814 + 242
DH = -275 kJ
CHEM 100, Fall 2009, LA TECH
6-72
Calculating enthalpies
2nd method
The real problem with using Hess’s law is figuring
out what equations to combine.
The most often used equations are those for
formation reactions.
Formation reactions
Reactions in which compounds are formed from
elements.
2 H2 (g) + O2 (g)
2 H2O (l) DHrxn = -571.66 kJ
CHEM 100, Fall 2009, LA TECH
6-73
Calculation of DH
DH = Sc 
o
CHEM 100, Fall 2009, LA TECH
o
o
DHf products
– Sc 
o
DHf reactants
6-74
Example
What is the value of DHrx for the reaction:
2 C6H6(l) + 15 O2(g)  12 CO2(g) + 6 H2O(g)
from Appendix J Text
D Hrx
C6H6(l) DHfo = + 49.0 kJ/mol
O2(g) DHfo = 0
CO2(g) DHfo = - 393.5
H2O(g) DHfo = - 241.8
=[S c  DHfo]product – [S c  DHfo]reactants
CHEM 100, Fall 2009, LA TECH
6-75
Example
What is the value of DHrx for the reaction:
2 C6H6(l) + 15 O2(g)  12 CO2(g) + 6 H2O(g)
from Appendix J Text
C6H6(l) DHfo = + 49.0 kJ/mol; O2(g) DHfo = 0
CO2(g) DHfo = - 393.5; H2O(g) DHfo = - 241.8
D Hrx =[S c DHfo]product - [S c  DHfo]reactants
D Hrx =[12(- 393.5) + 6(- 241.8)]product
- [2(+ 49.0 ) + 15(0)]reactants kJ/mol
= - 6.2708  103 kJ
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Standard enthalpy of formation
o
DHf
Enthalpy change that results from one mole of a
substance being formed from its elements.
All elements are at their standard states.
The DHfo of an element in its standard state has a
value of zero.
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Standard enthalpies of
Substance DH , kJ/mol
formation
CaCO
-1206.92
f
o
3 (s)
Standard enthalpy of
formation values are
available for a wide
range of substances.
In addition, separate
values for a substance in
different states will also
be given where
appropriate.
CHEM 100, Fall 2009, LA TECH
CaO (s)
CH4 (g)
C2H6 (g)
CH3OH (l)
CH3OH (g)
CO (g)
CO2 (g)
HCl (g)
H2O (l)
H2O (g)
NaCl (s)
SO2 (g)
-635.09
-74.85
-84.67
-238.64
-201.2
-110.52
-393.51
-92.31
-285.83
-238.92
-411.12
-296.83
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Why is DHof of elements is zero?
DHof, Heat formations are only for compounds
Note: DHof of elements is zero
Note: If the element is not at standard state (25o C
and 1 atm ) it’s DHof, is not zero. O indicate
standard state.
Eg. Br2(g) DHof is not zero because at 25o C and 1
atm Br2 is a liquid: Br2(l).
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Calculate Heat (enthalpy) of
Combustion: 2nd method
C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(g) ;DH = ?
DHf (C7H16) = -198.8 kJ/mol
DHf (CO2) = -393.5 kJ/mol
DHf (H2O) = -285.9 kJ/mol
DHf O2(g) = 0 (zero)
What method?
2nd method
CHEM 100, Fall 2009, LA TECH
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C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(g) ; DH = ?
DH = [S (DHof) Products] - [S (DHof) reactants]
DH = [ 7(-393.5 + 8 (-285.9)] - [-198.8 + 11 (0)]
= [-2754.5 - 2287.2] - [-198.8]
= -5041.7 + 198.8
= -4842.9 kJ = -4843 kJ
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Phase changes Affect DHof
We can use DHof values to determine the energy
required to change from one phase to another.
Example. Conversion of methanol from a liquid to a
solid.
kJ
C (s) + 2 H2 + O1 2 (g)
CH3OH (g) DHorxn = -201.2
2
C (s) + 2 H2 +
CHEM 100, Fall 2009, LA TECH
O1 2 (g)
2
CH3OH (l) DHorxn = -238.6
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Phase change
kJ
C (s) + 2H2 + O2 (g)
CH3OH (l)
CH3OH (l)
1
2
CH3OH (g)
DHorxn = -201.2
C (s) + 2H2 (g) + 12 O2 (g) DHorxn = +238.6
CH3OH (g) DHorxn = +37.4
This is not DHovap because the values are at 25 oC.
DHovap would be the thermal energy required at the
boiling point of methanol.
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Fossil Fuels
coal
petroleum
natural gas
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Energy Resources in the U.S.
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Energy units
Calorie (cal)
Originally defined as the quantity of heat
required to heat of one gram from 15 to 16oC.
It is now defined as: 1 cal = 4.184 J
Dietary Calorie
This is what you see listed on food products.
It is actually a kilocalorie.
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Caloric Value of Some Foods
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