Chapter 6 — Thermochemistry
Objectives
1
2
• Definitions of energy, heat, work and how they
interrelate
• 1st Law of Thermodynamics
• Heat Capacity
• Energy Transfer
• Define Universe, System, and Surroundings
• Enthalpy
• Hess’s Law
• Calorimetry
Thermochemistry
CHAPTER 6
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© 2009 Brooks/Cole - Cengage
3
Geothermal power —Wairakei
North Island, New Zealand
Ch. 6.1 - Energy & Chemistry
4
• Burning peanuts
supply sufficient
energy to boil a cup
of water.
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• Burning sugar
(sugar reacts with
KClO3, a strong
oxidizing agent)
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Energy & Chemistry
Energy & Chemistry
5
6
ENERGY is the capacity to
do work or transfer heat.
HEAT is the form of energy
• These reactions are PRODUCT
FAVORED
• They proceed almost completely from
reactants to products, perhaps with
some outside assistance.
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that flows between 2
objects because of their
difference in temperature.
Other forms of energy —
• light
• electrical
• kinetic and potential
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1
Chapter 6 — Thermochemistry
Potential Energy
on the Atomic Scale
7
Potential & Kinetic Energy
Potential energy —
energy a motionless
body has by virtue of
its position.
• Positive and
negative particles
(ions) attract one
another.
• Two atoms can
bond
• As the particles
attract they have a
lower potential
energy
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NaCl — composed of
Na+ and Cl- ions.
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Potential Energy
on the Atomic Scale
9
Potential & Kinetic Energy
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Potential & Kinetic Energy
Kinetic energy —
energy of
motion.
10
Kinetic energy
— energy of
motion
• Translation
• Positive and
negative particles
(ions) attract one
another.
• Two atoms can
bond
• As the particles
attract they have a
lower potential
energy
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12
Kinetic and Potential Energy
Summary
• Chemical Energy is due to the potential
energy stored in the arrangements of
the atoms in a substance.
• If you run a reaction that releases heat,
you are releasing the energy stored in
the bonds.
• Energy because of temperature is
associated with the kinetic (movement)
energy of the atoms and molecules.
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2
Chapter 6 — Thermochemistry
Internal Energy (E)*
13
Internal Energy (E)
14
PE + KE = Internal energy (E)
• PE + KE = Internal energy (E)
(1st Law of Thermodynamics)
• Internal energy of a chemical
system depends on
• number of particles
• type of particles
• temperature
*Note: Internal energy is sometimes
symbolized by U
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Internal Energy (E)
Thermodynamics
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16
• Thermodynamics is the science of
energy transfer as heat.
• The higher the T
the higher the
internal energy
• So, use changes
in T (∆T) to
monitor changes
in energy (∆E).
Heat energy is associated
with molecular motions.
Energy transfers as heat until
thermal equilibrium is established.
∆T measures energy transferred.
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System and Surroundings
17
Directionality of Energy Transfer
18
• Energy transfer as heat is always from a
hotter object to a cooler one.
• SYSTEM
– The object under study
• EXOthermic: energy transfers from
SYSTEM to SURROUNDINGS.
• SURROUNDINGS
– Everything outside the
system
T(system) goes down
T(surr) goes up
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3
Chapter 6 — Thermochemistry
Directionality of Energy Transfer
19
20
Energy and Temperature Summary
• Energy transfer at heat is always from a
hotter object to a cooler one.
• ENDOthermic: heat transfers from
1) Temperature determines the
direction of thermal energy
transfer.
2) The higher the temperature of a
given object, the greater the
thermal energy (molecular
motion) of its atoms, ions, or
molecules.
SURROUNDINGS to the SYSTEM.
T(system) goes up
T (surr) goes down
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21
Energy & Chemistry
Energy and Temperature Summary
All of thermodynamics depends
on the law of
3) Heating and cooling are
processes by which energy is
transferred as heat from an
object at higher temperature to
one at lower temperature.
• Heat is not a substance but a
‘process quantity.’ That is
heating is a process that changes
the energy of a system.
CONSERVATION OF ENERGY.
• The total energy is unchanged
in a chemical reaction.
• If potential energy (PE) of
products is less than reactants,
the difference must be released
as kinetic energy (KE).
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PE of system dropped. KE increased. Therefore,
you often feel a T increase.
23
24
Exothermic Reaction – reaction give off energy.
CH4 + 2O2  CO2 + 2H2O + Heat
Potential energy
Energy Change in
Chemical Processes
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22
CH4 + 2O2
Heat
CO2 + 2H2O
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4
Chapter 6 — Thermochemistry
UNITS OF ENERGY
25
Endothermic Reaction – requires external energy to
proceed. Very strong bonds have low potential
energy.
N2 + O2 + Heat  2NO
1 calorie = heat required to
raise temp. of 1.00 g of H2O
by 1.0 oC.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food
2NO
Potential energy
26
“calorie”)
But we use the unit called the
JOULE
Heat
1 cal = exactly 4.184
joules
James Joule
1818-1889
N2 + O2
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28
Work
Work vs Energy Flow
•
Work = P × A × Δh = PΔV
– P is pressure.
– A is area.
– Δh is the piston moving a distance.
– ΔV is the change in volume.
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29
Work
•
•
For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w
must have opposite signs:
w = –PΔV
To convert between L∙atm and Joules, use 1 L∙atm = 101.3 J.
30
Practice
Which of the following performs more
work?
a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.
b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.
They perform the same amount of work.
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5
Chapter 6 — Thermochemistry
FIRST LAW OF
THERMODYNAMICS
31
32
energy transfer in
(endothermic), +q
energy transfer out
(exothermic), -q
heat energy transferred
SYSTEM
∆E = q + w
energy
change
work done
by the
system
∆E = q + w
Energy is conserved!
w transfer in
(+w)
w transfer out
(-w)
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Practice Problems
33
34
Ch. 6.2 - HEAT CAPACITY
• Calculate ∆E, and determine whether the process
is endothermic or exothermic for the following
cases.
a)q = 1.62 kJ and w = -874 J
b)A system releases 113 kJ of heat to the
surroundings and does 39 kJ of work on the
surroundings.
c)The system absorbs 77.5 kJ of heat while doing
63.5 kJ of work on the surroundings.
The heat required to raise an
object’s T by 1 ˚C.
Which has the larger heat capacity?
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Specific Heat Capacity
How much energy is transferred
due to T difference?
The heat (q) “lost” or “gained” is
related to
a)
b)
sample mass
change in T and
c)
specific heat capacity (Symbolized by s or Cp)
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35
Specific Heat Capacity
36
Substance
Spec. Heat (J/g•K)
H2O
4.184
Ethylene glycol
2.39
Al
0.897
glass
0.84
Aluminum
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6
Chapter 6 — Thermochemistry
37
Specific Heat Capacity Example
38
Specific Heat Capacity Example
If 25.0 g of Al cool from 310 oC to 37 oC, what amount of
energy (J) has been transferred by the Al?
If 25.0 g of Al cool
from 310oC to 37oC,
what amount of
energy (J) is lost by
the Al?
heat gain/lose = q = (mass)(Sp. Heat)(∆T)
or
q = m x Cp x ∆T
where ∆T = Tfinal - Tinitial
q = (25.0 g) (0.897 J/g•K)(37 - 310)K
q = - 6120 J
Notice that the negative sign on q signals
heat “lost by” or transferred OUT of Al.
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39
40
Specific Heat Capacity Practice
Energy Transfer
In an experiment, it was determined that 59.8 J
was required to raise the temperature of 25.0 g of
ethylene glycol by 1.00 K. Calculate the specific
heat capacity of ethylene glycol from these data.
• Use energy transfer as a
way to find specific heat
capacity, Cp
• 55.0 g Fe at 99.8 ˚C
• Drop into 225 g water at
21.0 ˚C
• Water and metal come to
23.1 ˚C
• What is the specific heat
capacity of the metal?
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Energy Transfer
41
Energy Transfer Important Points
• Iron and the water are the system, the beaker
holding the water and everything else is the
surroundings.
• The iron and water end up at the same temp.
• The energy transferred as heat from the iron to the
water is negative (temp of Fe drops). The qwater
increase is positive because the temp of water
increases.
• The values of qwater and qiron are numerically equal
but opposite in sign.
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42
Because of conservation of energy,
q(Fe) = –q(H2O) (energy out of Fe = energy into H2O)
or q(Fe) + q(H2O) = 0
q(Fe) = (55.0 g)(Cp)(23.1 ˚C – 99.8 ˚C)
q(Fe) = –4219 • Cp
q(H2O) = (225 g)(4.184 J/K•g)(23.1 ˚C – 21.0 ˚C)
q(H2O) = 1977 J
q(Fe) + q(H2O) = –4219 Cp + 1977 = 0
Cp = 0.469 J/K•g
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7
Chapter 6 — Thermochemistry
ENTHALPY
43
Most chemical reactions occur at constant P, so
∆H = Hfinal – Hinitial
Heat transferred at constant P = qp
qp =
∆H
where H
44
ENTHALPY
If Hfinal > Hinitial then ∆H is positive
= enthalpy
Process is ENDOTHERMIC
and so ∆E = ∆H + w (and w is usually small)
∆H = energy transferred as heat at constant P ≈ ∆E
∆H = change in heat content of the system
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
∆H = Hfinal - Hinitial
Enthalpy is an extensive property!!
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© 2009 Brooks/Cole - Cengage
Practice Using Enthalpy
Practice Using Enthalpy
45
 Hydrogen peroxide decomposes to water and
oxygen by the following reaction:
46
– Hydrogen peroxide decomposes to water and oxygen by
the following reaction:
2H2O2(l)  2H2O(l) + O2(g) ∆H = ‐196 kJ
–Calculate the value of energy
transferred when 5.00 g of H2O2(l)
decomposes at constant pressure.
2H2O2(l)  2H2O(l) + O2(g)
∆H = ‐196 kJ
 -196 kJ/2 mol peroxide = -98 kJ/mol of peroxide
5.0 g * (1 mol/34 g H2O2) = 0.147 mol H2O2
= 0.147 mol H2O2 * -98 kJ/mol peroxide
= -14.4 kJ
Heat is a stoichiometric value which works the
same way as last year.
Hints: What is enthalpy of reaction if only 1 mole of peroxide is decomposed?
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USING ENTHALPY
USING ENTHALPY
47
48
Making liquid H2O from H2 +
O2 involves two exothermic
steps.
Consider the formation of water
H2(g) + 1/2 O2(g)  H2O(g) + 241.8 kJ
H2 + O2 gas
H2O vapor
Liquid H2O
Exothermic reaction — energy is a “product” and
∆H = – 241.8 kJ
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© 2009 Brooks/Cole - Cengage
8
Chapter 6 — Thermochemistry
Calorimetry
49
50
Constant Pressure
Calorimetry
• Definition of calorimetry: measure heat
flow in a chemical reaction.
• Two kinds of calorimeters (devices that
measure heat flow):
–Constant pressure (coffee cup calorimeter)
–Constant volume (bomb calorimeter)
m = mass of solution
C = heat capacity of the
calorimeter (J/g-⁰C) (or K)
∆T = the change in temperature
in ⁰C or K
• Heat Capacity – amount of heat needed
to raise an object’s temperature by 1°C
or 1 K.
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Constant Pressure Calorimetry
Practice 1
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51
• When a 9.55 g sample of solid NaOH
dissolves in 100.0g of water in a coffee-cup
calorimeter, the temperature rises from
23.6˚C to 47.4˚C. Calculate the ∆H (in kJ/mol
NaOH) for the solution process
NaOH(s)  Na+(aq) + OH-(aq)
• Assume the specific heat of the solution is
the same as that for water (4.184 J/g-˚C)
• Hint: what is system, what are surroundings?
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Constant Volume CALORIMETRY
Measuring Heats of Reaction
Constant Pressure Calorimetry
Practice 2
52
• A 150.0 g sample of a metal at 75.0°C is
added to 150.0 g H2O at 15.0°C. The
temperature of the water rises to 18.3°C.
Calculate the specific heat capacity of the
metal.
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CALORIMETRY
53
54
Measuring Heats of Reaction
Constant Volume
“Bomb” Calorimeter
• Burn combustible
sample.
• Measure heat evolved
in a reaction.
• Derive ∆E for
reaction.
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9
Chapter 6 — Thermochemistry
55
Calorimetry
Some heat from reaction warms
water
qwater = (sp. ht.)(water mass)(∆T)
56
Measuring Heats of Reaction
CALORIMETRY - Example
Calculate energy of combustion (∆E) of
octane.
C8H18 + 25/2 O2  8 CO2 + 9 H2O
• Burn 1.00 g of octane
Some heat from reaction warms
“bomb”
qbomb = (heat capacity, J/K)(∆T)
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1200. g water
• Heat capacity of bomb = 837 J/K
Total heat evolved = qtotal = qwater + qbomb
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© 2009 Brooks/Cole - Cengage
Measuring Heats of Reaction
CALORIMETRY - Example
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58
‘BOMB’ CALORIMETRY PRACTICE
A 1.00 g sample of sucrose (C12H22O11) is
burned in a bomb calorimeter. The
temperature of 1500. g of water in the
calorimeter rises from 25.00˚C to
27.32˚C. The heat capacity of the bomb
is 837 J/K and the specific heat capacity
of water is 4.18 J/g-K. Calculate the heat
evolved
a)Per gram of sucrose
b)Per mole of sucrose (M = 343 g/mol)
Step 1 Calc. energy transferred from reaction to water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2 Calc. energy transferred from reaction to bomb.
q = (bomb heat capacity)(∆T)
= (837 J/K)(8.20 K) = 6860 J
Step 3 Total energy evolved
41,200 J + 6860 J = 48,060 J
Energy of combustion (∆E) of 1.00 g of octane
= - 48.1 kJ
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© 2009 Brooks/Cole - Cengage
Ch. 6.3 – Hess’s Law
59
Hess’s Law
60
Making H2O from H2 involves two steps.
H2(g) + 1/2 O2(g) H2O(g)
∆rH˚ =-242 kJ
H2O(g)  H2O(liq)
∆rH˚ =-44 kJ
-----------------------------------------------------------------------
H2(g) + 1/2 O2(g)  H2O(liq) ∆rH˚ =-286 kJ
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others,
the net ΔH is the sum of the ΔH’s of
the other rxns.
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60
10
Chapter 6 — Thermochemistry
61
62
Hess’s Law
& Energy Level
Diagrams
Hess’s Law
& Energy Level
Diagrams
Forming H2O can occur
in a single step or in a
two steps. ∆rHtotal is the
same no matter which
path is followed.
Forming CO2 can occur
in a single step or in a
two steps. ∆rHtotal is the
same no matter which
path is followed.
NOTE: ∆rH stands for the
enthalpy change for a
reaction, r
Active Figure 5.16
Active Figure 5.16
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© 2009 Brooks/Cole - Cengage
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64
Simple Hess’s Law problem
Enthalpy and Hess’s Law
• C(graphite) + O2(g) → CO2(g)
∆H° = –393.51 kJ mol–1
• If a reaction is reversed, the sign of ∆H is also
reversed.
Xe(g) + 2F2(g)  XeF4(s) ∆H=-251 kJ
XeF4(s)  Xe(g) + 2F2(g) ∆H = +251 kJ
• ∆H is also proportional to the quantities of the
reactants and product (mol-rxn). If you multiply
the coefficients by an integer, the ∆H value is
multiplied by the same integer.
2Xe(g) + 4F2(g)  2XeF4(s) ∆H =-502 kJ
• C(diamond) + O2(g) → CO2(g)
∆H° = –395.40 kJ mol–1
Calculate the ∆H for the conversion of
graphite to diamond.
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65
Simple Hess’s Law problem
• C(graphite) + O2(g) → CO2(g)
∆H° = –393.51 kJ mol–1
• CO2(g) → C(diamond) + O2(g)
∆H° = +395.40 kJ mol–1
Add rxns together, cancel out substances that
appear on both sides and get:
∆H = 1.9 kJ
C(graphite)  C(diamond)
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66
Problem‐Solving Strategy
• Work backward from the required
reaction, using the reactants and products
to decide how to manipulate the other
given reactions at your disposal.
• Reverse any reactions as needed to give
the required reactants and products.
• Multiply reactions to give the correct
numbers of reactants and products.
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66
11
Chapter 6 — Thermochemistry
Harder Example of Hess’s Law
67
C(s) + O2(g)  CO2(g) ∆H= -393.5kJ/mol
2S(s) + 2O2(g)  2SO2(g) ∆H=2*(-296.8 kJ/mol)
=-593.6 kJ
CO2(g) + 2SO2(g)  CS2(l) + 3O2(g)
∆H = 1103.9 kJ/mol
C(s) + 2S(s)  CS2(l) ∆H = 116.8 kJ
Given:
C(s) + O2(g)  CO2(g) ∆H= -393.5kJ/mol
S(s) + O2(g)  SO2(g) ∆H= -296.8 kJ/mol
CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)
∆H = -1103.9 kJ/mol
Use Hess’s Law to calculate enthalpy change for
the formation of CS2(l) from C(s) and S(s)
C(s) + 2S(s)  CS2(l)
Reaction 1 was used as is
Reaction 2 was multiplied by 2 (SO2 cancels)
Reaction 3 was multiplied by -1 (Get CS2 on
product side)
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69
Most ∆H values are labeled ∆Ho
Measured under standard conditions:
• This equation is valid because
∆H is a STATE FUNCTION
• These depend only on the state
of the system and not how it got
there.
• V, T, P, energy — and your bank
account!
• Unlike V, T, and P, one cannot
measure absolute H. Can only
measure ∆H.
•
For a Compound
– For a gas, pressure is exactly 1 atm.
– For a solution, concentration is exactly 1 M.
– Pure substance (liquid or solid)
• For an Element
– The form [N2(g), K(s)] in which it exists at 1
atm and 25⁰C.
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© 2009 Brooks/Cole - Cengage
71
o,
ΔHf standard molar
enthalpy of formation
NIST (Nat’l Institute for Standards and
Technology) gives values of
72
H2(g) + 1/2 O2(g)  H2O(g)
ΔHfo = standard molar enthalpy of
∆Hfo (H2O, g) = -241.8 kJ/mol
formation
— the enthalpy change when 1 mol of
compound is formed from elements under
standard conditions.
By definition,
∆Hfo = 0 for elements in their
standard states.
See Appendix 4 (A19 – A22)
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70
Standard Enthalpy Values
∆H along one path =
∆H along another path
Ch. 6.4 – Standard Enthalpies
of Formation
68
Multiply equations as needed:
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12
Chapter 6 — Thermochemistry
73
Enthalpy of Formation Practice
74
Given:
C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l)
∆Hr° = -1300. kJ/mol-rxn
• Show the ∆Hf˚ reaction for the
following compounds:
a) C2H2(g)
b) KCl(s)
• Why is the following not a ∆Hf˚?
4Na(s) + O2(g)  2Na2O(s)
C(s) + O2(g)  CO2(g)
∆Hr° = -394. kJ/mol-rxn
H2(g) + ½O2(g)  H2O(l)
∆Hr° = -286. kJ/mol-rxn
Determine the ∆Hf° for C2H2 (acetylene).
First, determine the overall reaction. Manipulate
above reactions to obtain overall reaction.
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Hess’s Law and ΔHf° Practice
Hess’s Law and ΔHf° Practice
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75
Answer:
-1( C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l) )
Using Standard Enthalpy Values
76
Use ∆H˚’s to calculate enthalpy change for
H2O(g) + C(graphite) H2(g) + CO(g)
(product is called “water gas”)
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Using Standard Enthalpy Values
H2O(g) + C(graphite)  H2(g) + CO(g)
From reference books we find
• H2(g) + 1/2 O2(g) H2O(g)
∆fH˚ of H2O vapor = - 242 kJ/mol
• C(s) + 1/2 O2(g)  CO(g)
∆fH˚ of CO = - 111 kJ/mol
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© 2009 Brooks/Cole - Cengage
77
Using Standard Enthalpy Values
78
H2O(g)  H2(g) + 1/2 O2(g) ∆Hro (= –∆Hr˚) = +242 kJ
C(s) + 1/2 O2(g)  CO(g) ∆Hro = -111 kJ
------------------------------------------------------------------------------------------------------
H2O(g) + C(graphite)  H2(g) + CO(g)
∆Hronet = +131 kJ
To convert 1 mol of water to 1 mol each of H2
and CO requires 131 kJ of energy.
The “water gas” reaction is ENDOthermic.
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13
Chapter 6 — Thermochemistry
Using Standard Enthalpy Values
79
In general, when ALL
Calculate ∆H of
reaction?
enthalpies of formation are
known,
Heats of Reactions
(ΔH⁰(r or rxn) = generic reaction)
80
• ∆H⁰combustion or ∆H⁰comb is amount of heat release when 1
mol of a substance is combusted (burned).
Ex: CH4 + 2O2  CO2 + 2H2O(g) ∆H = -890 kJ/mol
• ∆H⁰neutralization is the amount of heat released or
absorbed when an acid is neutralized by a base
Ex: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
∆Hneut =-57 kJ/mol-rxn
∆Hro =  ∆fHfo (products)
-  ∆Hfo (reactants)
Remember that ∆ always = final – initial
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© 2009 Brooks/Cole - Cengage
Heats of (Physical) Process
(ΔH⁰process = generic process)
81
• ∆H⁰solution is the amount of heat released or absorbed
when a solid compound dissolves and becomes
aqueous.
Ex: NaOH(s)  Na+(aq) + OH-(aq) ∆H⁰sol = -43 kJ/mol
• ∆H⁰fusion is the amount of heat needed to change 1
mole of a solid substance to a liquid, i.e. melting.
Ex: H2O(s)  H2O(l) ∆H ⁰fus= 6.0 kJ/mol
• ∆H⁰vaporization is the amount of heat needed to change
1 mole of a liquid to a vapor. Can use opposite too.
Ex: H2O(l)  H2O(g) ∆H ⁰vap= 40.7 kJ/mol
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82
Calculate the heat of combustion of
methanol, i.e., ∆Hocomb for
CH3OH(g) + 3/2 O2(g)  CO2(g) + 2 H2O(g)
∆Hocomb =  ∆Hfo (prod) -  ∆Hfo (react)
© 2009 Brooks/Cole - Cengage
Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g)  CO2(g) + 2 H2O(g)
∆Hocomb =  ∆Hfo (prod) -  ∆Hfo (react)
∆Hocomb = ∆Hfo (CO2) + 2 ∆Hfo (H2O)
∆Hfo (O2)
- {3/2
+
= (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
o
∆H comb = -675.6 kJ per mol of methanol
© 2009 Brooks/Cole - Cengage
Using Standard Enthalpy Values
∆Hfo (CH3OH)}
83
Standard Enthalpy Values
Example
84
Using Enthalpies of Formation calculate the Enthalpy
of Combustion for the reaction below.
C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(l)
∆H° = 3∆H°(CO2) + 4 ∆H°(H2O(l))–[∆H°(C3H8) + 5∆H°f(O2)]
Hocomb = 3 (-393.5) + 4 (-285.8) – [-103.8 + 5(0)]
H˚comb= -2220 kJ
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14
Chapter 6 — Thermochemistry
Using Enthalpy of
Reaction
85
86
Think of reaction like this:
Formation of elements
3C(s) + 4H2(g)
∆Hrxn = ∆H1 + ∆H2 + ∆H3
-∆Hf1°
Regroup and form new substances
5O2 3C + 3O2 4H2+2O2
-∆Hf2°
3∆Hf3°
4∆Hf4°
C3H8 + 5O2  3CO2 + 4H2O
∆rH = -∆Hf1 – ∆Hf2 + 3∆Hf3 + 4∆Hf4
= +3∆Hf3
Enthalpy Practice Problem
+ 4∆Hf4 – (∆Hf1 + ∆Hf2)
© 2009 Brooks/Cole - Cengage
© 2009 Brooks/Cole - Cengage
87
Calculate the standard enthalpy change for the
reaction below using the standard enthalpies of
formation.
SiCl4(l) + 2H2O(l)  SiO2(s) + 4HCl(aq)
∆H°f’s:
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15