Chapter 5: Thermochemistry
1. Thermodynamics
2. Energy
3. Specific Heat
4. Enthalpy
5. Enthalpies of Reactions
6. Hess’s Law
7. State Functions
8. Standard Enthalpies of Formation
9. Determining Enthalpies of Reactions
5.1 Thermodynamics
“The science of heat and work”
of energy transfer
therme – heat
dynamis – power (Greek)
ENERGY
Energy - Capacity to Do Work (w) or
Transfer Heat (q)
Kinetic Energy - Energy of Motion
Potential Energy - Stored Energy
Molecular Kinetic Energy
-Translational energy
Ek, translational = 1/2mv2
-Rotational energy
Ek, rotaional = 1/2Iw2
moment of inertia
frequency
I=
w = rotational
-Vibrational energy
Ek, vibrational = 1/2Kx2
K = Hooke’s constant
x = displacement from equilbrium
Molecular Potential Energy
Eatom –coulombic attraction
of e- to nucleus
Ebond – coulombic forces of
covalent and ionic bonds
Enucleus – strong and weak
nuclear forces holding
nucleus together
Law of Conservation of Energy
The total energy of the universe is
constant
Energy can not be created or destroyed,
only transformed
Units of Energy
Calorie (cal) - Amount of Energy (Heat)
Required to Raise 1 Gram of Water 1
Degree Celsius (14.5 to 15.5o C)
Modern Definition is
Based on The Joule
1 Calorie = 4.184 Joule
Note Dietary Calorie (Cal) = 1 Kcal
Units of Energy
Two Common Units of Energy
Joule –work based definition
Calorie –heat based definition
Joule – SI derived unit, the amount of
work required to accelerate a 1 kg
object 1 m/s2 a distance of 1 meter
1J  1kg  m / sec
2
2
JP Joule
(1818-1889)
Thermal Energy
Depends on the Temperature, # and Type
of Atoms Present
The Hotter an Object Is, the More Kinetic
Energy It’s Constituent Atoms Posses
Spontaneous Thermal Energy
Transfer
Heat (q) Flows From Hot to Cold
5.2 Heat Capacity (C)
q = CDT
-The heat required to induce a
temperature change in a substance
C = q/DT -The capacity of a substance to
absorb heat as its temperature
changes
*Note: Heat capacity describes an object and has
units in terms of energy per degree temperature
Is Heat an Intensive or
Extensive Property?
Extensive: It requires twice as
much heat to raise 2 grams of
water 1oC than it does 1 gram.
Specific Heat Capacity
-the heat capacity per gram of substance
q = mcDT
q = Heat or Energy (cal or joule)
m = Mass (in grams)
c = Specific heat, the heat required to raise one
gram of a substance by one degree Celsius
DT = Change in Temperature (Tfinal - Tinitial)
Specific Heat Capacity
c=q
mDT
•Specific heat capacity describes a type of
substance and not a specific object.
•Has Units of J/goC or cal/ goC
• To know how much heat capacity an object has
you need to know how much of it you have (it’s
mass times it’s specific heat capacity).
Is Specific Heat Capacity an
Intensive or Extensive
Property?
Intensive: The amount of heat
required to raise the temperature of
substance 1oC on a per gram basis
is the same, no matter how many
grams the substance has.
What Is the Specific
Heat of Water?
c(H2O) = 1cal/(goC)
(original definition of the calorie)
Specific Heat of Common Materials
Substance
Aluminum
Graphite
glass
gold
water(l)
water(s)
water(g)
wood
c(J/g. K)
0.902
0.72
0.888
0.128
4.184
2.06
2.01
1.76
Metals have small
specific heat
capacities, so it does
not take much heat to
raise the temperature
of a metal
Specific Heat of Common Materials
Substance
Aluminum
Graphite
glass
gold
water(l)
water(s)
water(g)
wood
c(J/g. K)
0.902
0.72
0.888
0.128
4.184
2.06
2.01
1.76
The Heat Capacity of a
Substance Depends on
it’s Phase
Difference between C and c:
C: Heat Capacity describes an object, like a
10 g piece of aluminum has a heat
capacity of 9.02 J/K.
c: Specific Heat Capacity describes a
material, aluminum has a specific heat
capacity of 0.902 J/(g-K).
Calculating Energy Requirements
How much Energy is Required to raise 1.000
gallon of water from 0.000 deg C to 100.0 deg
C?
dwater = 1 g/ml, 1 gal = 3.7854 L
Q=mcDT
3.7854l 1000ml 1g
1cal
Q  1gal (
)(
)( )(
)(100 deg C  0 deg C )
gal
l
ml g  deg C
Q=379 kcal
Heat Problem
What would be the final temperature if a 250 g piece
of aluminum at 20.oC absorbed 1.5 kJ of energy?
q  mcDT  mc T f  Ti 
q  mcT f  mcTi
mcT f  q  mcTi
q
Tf 
 Ti
mc
1,500 J
Tf 
 20.o C  27 0 C


250 g  0.902 J 0 
g C

Chapter 6 Interactive Quizzes 1-3
http://www.ualr.edu/rebelford/chem1402/q1402/chem1402QP.htm
First Law of Thermodynamics
-Total amount of energy of the universe
is constant
DE = q + w
The change in the internal energy of a system
is equal to the heat transfer and work done
on/by the system with the surroundings.
Internal Energy
surroundings
system
Q
(reaction)
W
DE = Ef - Ei = Q + W
DE > 0, the system gains energy
DE < 0, the system loses energy
Heat of Reaction
Endothermic Reaction - heat is added
to the system (q is positive and the
surroundings lose the heat)
Exothermic Reaction - heat is lost from
the system (q is negative and
surroundings gain the heat)
5.3 Energy & Changes of State
What Are the Three Effects That Adding
or Subtracting Heat Can Have on a
Substance.
1.
Decomposition the destruction of intramolecular
bonds (chemical change)
2.
Temperature Change within a phase
(physical change)
3.
Phase Change - Changes in the intermolecular
forces (physical change)
Heat & Changes of State
6 Phase Changes
V
L
S
ENDO
6 Phase Changes
V
EXO
L
S
6 Phase Changes
V
EXO
L
S
ENDO
How Do We Quantitatively Determine the Heat
Associated With Vaporization or Fusion Change?
With the Molar Heat of Fusion
or Vaporization
Define the Molar Heat of Fusion: DHf (kJ/mol)
-Heat required to melt one mole of a
substance at it’s melting pt
Define the Molar Heat of Vaporization: DHv(kJ/mol)
-Heat required to vaporize one mole
of a substance at it’s boiling pt
For a Given Substance,
Which Is Greater, the
Molar Heat of Vaporization
or the Molar Heat of
Fusion?
Heating Diagram
Calculating Energy Changes
How much energy is required raise 100. g
of ice from - 100C to 110oC?
Given:
DHf=6.02 kJ/mol
DHv=40.6 kJ/mol
cl=4.184 J/g.oC
cv=2.01 J/g.oC
cs=2.06 J/g.oC
T
e
m
p
e
r
a
t
u
r
e
(oC)
Heat Added
Calculating Energy Changes
qtotal  qDTice  qmelt ice  qDTwater  qboil water  qDTsteam
qtotal  mi ci DTi  nDH fus  mwcwDTw  nDH vap  ms cs DTs

qtotal  100 g 2.06
J
g C o

 mol H 2O 
KJ
 0 C  10 C   100 g  18g  6.02 mol


o
o


100 g  2.01  110 C  100 C 
100 g 4.186
J
g C o
J
g C o
 mol H 2O 
KJ
100 C  0 C   100 g  18g  40.6 mol


o
o
o
o
Calculating Energy Changes
qtotal  qDTice  qmelt ice  qDTwater  qboil water  qDTsteam
qtotal  mi ci DTi  nDH fus  mwcwDTw  nDH vap  ms cs DTs
qtotal  2.06kJ  33.4 KJ  41.86 KJ  225kJ  2.01kJ
 304kJ
Work
-Work occurs when something
moves against an opposing force
- Lets investigate the work of expansion at
constant pressure
W = FDX
F = Force, X = distance
P = F/A, F = P.A
V = X.A, X = V/A
W = (P.A)(DV/A)
W = -PDV
Why is this negative?
Enthalpy
DH = Enthalpy change, the heat transfer in/out
of a system at constant P
DH = qp
DE = q + w
w = -PDV
DE = q -PDV
DH = qP = DE + PDV
State Functions
Enthalpy & Internal Energy are State
Functions - their values are “path
independent” and only depend on their
current states, not how they were attained
Is Work a State Function?
Sign Conventions of Energy
Transfer
Endothermic Reactions - absorb heat
from the surroundings (DH > 0)
Exothermic Reactions - release heat
to the surroundings (DH < 0)
5.5 DH and Chemical Reactions
Reactants + Heat --> Products
-endothermic, heat was added to
the reaction
Reactants --> Products + Heat
-exothermic, heat was released by
the reaction
DHrxn is associated with a
chemical equation
C(s) + 2H2(g) --> CH4(g) DH = -74.8kJ
What do the units of the DH in the
above reaction mean?
74.8 kJ are released for every mole of
carbon or for every 2 moles of
hydrogen consumed
Is the Following Reaction
Endothermic or Exothermic?
CaO(s) + CO2(g) --> CaCO3(s)
DH = -178 kJ
DH Is Negative and Energy Is Lost
From the System So It Is Exothermic
Would this reaction heat the
surroundings?
DHForward = - DHreverse
CaO(s) + CO2(g) --> CaCO3(s)
DH = -178 kJ
What is DH For the reaction:
CaCO3(s) --> CaO(s) + CO2(g) DH =+178 kJ
It is endothermic by the same order of
magnitude the first was exothermic
Calorimetry
What happens when a hot object comes
into thermal contact with a cold object?
Insulated Jacket
Calorimetry
What is the final temperature if 2g of gold at
100oC is dropped into 25 ml of water at 30oC in
an ideal calorimeter? cAu = 0.128J/g.oC.
Heat lost by Au = Heat gained by water
-Qhot = Qcold
(First Law)
-Qhot = Qcold
-mHcH(Tf - TH) = mCcC(Tf - TC)
TF = mCcCTC + mHcHTH
mCcC + mHcH
TF 


J
0.128J
(25ml H 2O)( 1mlg HH 22OO )( 4g.184
)(
30
.
0
C
)

(
2
g
Au
)(
)(
100
C)
 C
g  C
(25ml H 2O)(
1 g H 2O
ml H 2O
J
0.128J
)( 4g.184
)

(
2
g
Au
)(
)
 C
g  C
TF= 30.17oC
Calorimetry
What is the final temperature if 2g of gold at
100oC is dropped into 25 ml of water at 30oC in
a real calorimeter? cAu = 0.128J/g.oC, Ccal =
36J/oC
Heat lost by Au = Heat gained by water + calorimeter
-Qhot = Qcold
(First Law)
-Qhot = Qcold
-QAu = Qwater +Q calorimeter
-mHcH(Tf - TH) = mCcC(Tf - TC) + Ccal (Tf - TC)
-mHcH(Tf - TH) = (mCcC+ Ccal )(Tf - TC)
TF = (mCcC+ Ccal ) TC + mHcHTH
(mCcC+ Ccal ) + mHcH
(25ml H 2O)( 1mlg HH2OO )( 4.184 J )  36 oJ  (30.0 C )  (2 g Au)( 0.128 J )(100 C )
g C
C
g C
2

TF 
(25ml H 2O)( 1mlg HH2OO )( 4.184 J )  36 oJ   (2 g Au)( 0.128 J )
g C
C
g C
2

-Qhot = Qcold
-QAu = Qwater +Q calorimeter
TF = (mCcC+ Ccal ) TC + mHcHTH
(mCcC+ Ccal ) + mHcH
(25ml H 2O)( 1mlg HH2OO )( 4.184 J )  36 oJ  (30.0 C )  (2 g Au)( 0.128 J )(100 C )
g C
C
g C
2
TF  
(25ml H 2O)( 1mlg HH2OO )( 4.184 J )  36 oJ   (2 g Au)( 0.128 J )
g C
C
g C
2

TF= 30.12oC
6.7 Hess’s Law
Conservation of Energy
If the products of one reaction are
consumed by another, the two equations
can be coupled into a third equation.
The energy of the third reaction will be
the sum of the energies of the first 2
reactions
Hess’s Law
Calculate DH for the reaction
2C(s) + O2(g) --> 2CO(g)
From the reactions
2C(s) + 2O2(g) --> 2CO2(g)
2CO2(g) --> 2CO (g) + O2(g)
DH= -787kJ
DH= 566kJ
2C(s) + O2(g) --> 2CO(g)
DH= -221kJ
Hess’s Law
2C(s) + 2O2(g)
DH= -221kJ
E
2CO (g) + O2(g)
DH= 566kJ
DH= -787kJ
1
2
2CO2(g)
2C(s) + 2O2(g) --> 2CO2(g)
2CO2(g) --> 2CO (g) + O2(g)
2C(s) + O2(g) --> 2CO(g)
DH= -787kJ
DH= 566kJ
DH= -221kJ
2 Step Path
Bomb Calorimeter &
Combustion Reactions
Hess’s Law
Calculate DH for the reaction
2S(s) + 3O2(g) --> 2SO3(g)
From the reactions
S(s) + O2(g) --> SO2(g)
DH=-296.8kJ
2SO2(g) + O2(g) --> 2SO3(g) DH= -197.0kJ
Note, to cancel SO2, multiply first eq by 2
S(s) + O2(g) --> SO2(g)
DH= -296.8kJ
2SO2(g) + O2(g) --> 2SO3(g) DH= -197.0kJ
DH (kJ)
2S(s) + 2O2(g) --> 2SO2(g) -593.6
2SO2(g) + O2(g) --> 2SO3(g) -197.0
2S(s) + 3O2(g) --> 2SO3(g) -790.6
Interactive Quiz 6-11
http://www.ualr.edu/rebelford/chem1402/q1402/chem1402QP.htm
Why are Enthalpies Additive?
-Enthalpy is a state function
-Its value is only dependent on the state
of the system, not the path
-It’s a consequence of the conservation
of energy
For a Chemical Reaction
DH = Hproducts-Hreactants
Is Not Dependent on the Steps the
Reaction Involves
Standard State Enthalpies
DHo = Standard State Enthalpy
Change of Reaction
Standard State is the most stable
form of a substance as it exists
at 1 atm and 25oC.
For a Solution this is at a
Concentration of 1M
Standard Molar Enthalpy of
Formation
DHof = Standard Molar Enthalpy
of Formation
The Enthalpy Change Associated With the
Formation of 1 Mol of a Substance From It’s
Elements in Their Standard States
H2(g) +1/2O2(g) --> H2O(l) DHof = - 285.8kJ/mol
DHof Listed in Appendix L of Text
Exercises
Write the DHof equation for the following
1. NH3(g)
1/2N2(g) + 3/2H2(g) --> NH4(g)
2. H2O(l)
1/2O2(g) + H2(g) --> H2O(l)
3. O2(l)
O2(g) --> O2(l)
4. O2(g)
No reaction
5. O3(g)
3/2O2(g) --> O3(g)
Enthalpy of Formation
Calculate the Enthalpy of Formation of Acetylene
(C2H2) from the following combustion data
1. 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
-2600KJ
2.
C(s) + O2(g)  CO2(g)
-394KJ
3.
2H2(g) + O2(g)  2H2O(l)
-572KJ
2C(s) + H2(g)  C2H2(g)
Enthalpy of Formation
Calculate the Enthalpy of Formation of Acetylene
(C2H2) from the following combustion data
2CO2(g) + H2O(l)  C2H2(g) + 5/2O2(g)
2C(s) + 2O2(g)  2CO2(g)
H2(g) + 1/2O2(g)  H2O(l)
2C(s) + H2(g)  C2H2(g)
- ½ (-2600KJ)
2(-394KJ)
½(-572KJ)
226KJ
Standard Enthalpies of Reaction
DHo rxn can be determined from DHof
o
DH rxn
  n [DH of ( products )]   m[DH of (reactants )
Where n & m are the stoichiometric
coefficients of the products & reactants
Explain This in Terms of Enthalpy as a State Function
and Hess’s Law for the following Reaction:
8Al(s) + 3Fe3O4 --> 4Al2O3(s) + 9Fe(s)
DHo rxn Exercises
8Al(s) + 3Fe3O4 --> 4Al2O3(s) + 9Fe(s)
From Appendix L:
DHo f (Fe3O4 ) = -1118.4 kJ/mol
DHo f (Al2O3 ) = -1675.7 kJ/mol
3Fe(s) + 2O2(g)  Fe3O4(s)
-1118.4kJ
2Al(s) + 3/2O2(g)  Al3O3(s)
-1675.5kJ
DHo rxn Exercises
8Al(s) + 3Fe3O4 --> 4Al2O3(s) + 9Fe(s)
3(Fe3O4(s)  3Fe(s) + 2O2(g))
-3(-1118.4kJ)
4(2Al(s) +3/2O2(g)  Al2O3(s))
4(-1675.5kJ)
8Al(s) + 3Fe3O4  4Al2O3(s) + 9Fe(s) -3363.3kJ
In 1 Step, Using the Eq:
o
DH rxn
  n [DH of ( products )]   m[DH of (reactants )
DHo rxn Exercises
8Al(s) + 3Fe3O4  4Al2O3(s) + 9Fe(s)
From Appendix L:
DHo f (Fe3O4 ) = -1118.4 kJ/mol
DHo f (Al2O3 ) = -1675.7 kJ/mol
o
DH rxn
  n [DH of ( products )]   m[DH of (reactants )
DHo rxn =[4(-1675.7 kJ/mol) + 8(0)]
-[3(-1118.4 kJ/mol) + 9(0)] = -3363.6 kJ
Standard Enthalpies of Reaction
DHo rxn can be determined from DHof
o
DH rxn
  n [DH of ( products )]   m[DH of (reactants )
Interactive Quiz 6-1 & 6-2
http://www.ualr.edu/rebelford/chem1402/q1402/chem1402QP.htm
Enthalpy of Vaporization for Water
from Enthalpies of Formation
Calculate DHVap(H2O)
From Appendix 4:
DHo f [H2O(l)] = -286 kJ/mol
DHo f [H2O(g)] = -242 kJ/mol
What is the Equation that
Describes Vaporization of Water?
Enthalpy of Vaporization for Water
from Enthalpies of Formation
From Appendix 4:
DHo f [H2O(l)] = -286 kJ/mol
DHo f [H2O(g)] = -242 kJ/mol
H2O(l)  H2O(g)
o
DH rxn
  n [DH of ( products )]   m[DH of (reactants )
-242kJ/mol – [-286kJ/mol]
= 44 kJ/mol
Enthalpies of Reaction and
Stoichiometric Problems
8Al(s) + 3Fe3O4  4Al2O3(s) + 9Fe(s)
Determine the standard state enthalpy change
when 15 g of Al react with 30.0 g of Fe3O4.
3 Steps
• Balance Eq. & Identify Limiting Reagent
• Calculate DHo rxn for Balanced Eq.
• Calculate DHo rxn for this reaction based on
complete consumption of the limiting reagent
Enthalpies of Reaction and
Stoichiometric Problems
8Al(s) + 3Fe3O4  4Al2O3(s) + 9Fe(s)
1. Calculate DHo rxn.for balanced Equation
This was done in previous
problem, DHo rxn = -3363.6 kJ
Enthalpies of Reaction and
Stoichiometric Problems
8Al(s) + 3Fe3O4  4Al2O3(s) + 9Fe(s)
DHo rxn = -3363.6 kJ
2. Determine Limiting Reagent

 mol Al  
1
15gAl 
 0.069



 26.98 g   8mol Al 

 mol Fe3O4  
1
30 gFe3O4 

  0.043

 231.54 g   3mol Fe3O4 
Fe3O4 is Limiting Reagent
Enthalpies of Reaction and
Stoichiometric Problems
8Al(s) + 3Fe3O4  4Al2O3(s) + 9Fe(s)
DHo rxn = -3363.6 kJ
3. Determine DHo rxn based on complete
consumption of limiting reagent
 mol Fe3O4   3363kJ 
30 gFe3O4 

  145kJ

 231.54 g   3mol Fe3O4 
“Key Step” Correlates Ratio of the Enthalpy of
Reaction to the Coefficient of a Chemical Species