Transcript THERMOCHEMISTRY or Thermodynamics

1
Thermochemistry
CHAPTER 6
© 2009 Brooks/Cole - Cengage
Objectives
• 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
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Geothermal power —Wairakei
North Island, New Zealand
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Ch. 6.1 - Energy & Chemistry
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• 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)
Energy & Chemistry
• These reactions are PRODUCT
FAVORED
• They proceed almost completely from
reactants to products, perhaps with
some outside assistance.
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Energy & Chemistry
ENERGY is the capacity to
do work or transfer heat.
HEAT is the form of energy
that flows between 2
objects because of their
difference in temperature.
Other forms of energy —
• light
• electrical
• kinetic and potential
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Potential & Kinetic Energy
Potential energy —
energy a motionless
body has by virtue of
its position.
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Potential Energy
on the Atomic Scale
• 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
• 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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Potential & Kinetic Energy
Kinetic energy
— energy of
motion
• Translation
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Potential & Kinetic Energy
Kinetic energy —
energy of
motion.
rotate
vibrate
translate
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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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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)
PE + KE = Internal energy (E)
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Internal Energy (E)
• The higher the T
the higher the
internal energy
• So, use changes
in T (∆T) to
monitor changes
in energy (∆E).
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Thermodynamics
• Thermodynamics is the science of
energy transfer as heat.
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
• SYSTEM
– The object under study
• SURROUNDINGS
– Everything outside the
system
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Directionality of Energy Transfer
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• Energy transfer as heat is always from a
hotter object to a cooler one.
• EXOthermic: energy transfers from
SYSTEM to SURROUNDINGS.
T(system) goes down
T(surr) goes up
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Directionality of Energy Transfer
• Energy transfer at heat is always from a
hotter object to a cooler one.
• ENDOthermic: heat transfers from
SURROUNDINGS to the SYSTEM.
T(system) goes up
T (surr) goes down
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Energy and Temperature Summary
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.
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Energy and Temperature Summary
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.
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Energy & Chemistry
All of thermodynamics depends
on the law of
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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Energy Change in
Chemical Processes
PE
Reactants
Kinetic
Energy
Products
PE of system dropped. KE increased. Therefore,
you often feel a T increase.
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Exothermic Reaction – reaction give off energy.
Potential energy
CH4 + 2O2  CO2 + 2H2O + Heat
CH4 + 2O2
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Heat
CO2 + 2H2O
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Potential energy
Endothermic Reaction – requires external energy to
proceed. Very strong bonds have low potential
energy.
N2 + O2 + Heat  2NO
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2NO
Heat
N2 + O2
UNITS OF ENERGY
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
“calorie”)
But we use the unit called the
JOULE
1 cal = exactly 4.184
joules
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James Joule
1818-1889
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Work vs Energy Flow
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Work
•
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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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.
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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.
© 2009 Brooks/Cole - Cengage Copyright © Cengage Learning. All rights reserved
FIRST LAW OF
THERMODYNAMICS
heat energy transferred
∆E = q + w
energy
change
work done
by the
system
Energy is conserved!
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energy transfer in
(endothermic), +q
energy transfer out
(exothermic), -q
SYSTEM
∆E = q + w
w transfer in
(+w)
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w transfer out
(-w)
Practice Problems
• Calculate ΔE, 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.
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Ch. 6.2 - HEAT CAPACITY
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)
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
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Specific Heat Capacity
Substance
Spec. Heat (J/g•K)
H2O
4.184
Ethylene glycol
2.39
Al
0.897
glass
0.84
Aluminum
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Specific Heat Capacity Example
If 25.0 g of Al cool
from 310oC to 37oC,
what amount of
energy (J) is lost by
the Al?
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
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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?
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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Specific Heat Capacity Practice
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.
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Energy Transfer
• 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 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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Energy Transfer
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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ENTHALPY
Most chemical reactions occur at constant P, so
Heat transferred at constant P = qp
qp =
∆H
where H
= enthalpy
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
∆H = Hfinal - Hinitial
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ENTHALPY
∆H = Hfinal – Hinitial
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
Enthalpy is an extensive property!!
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Practice Using Enthalpy
 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.
Hints: What is enthalpy of reaction if only 1 mole of
peroxide is decomposed?
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Practice Using Enthalpy
– Hydrogen peroxide decomposes to water and oxygen by
the following reaction:
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.
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USING ENTHALPY
Consider the formation of water
H2(g) + 1/2 O2(g)  H2O(g) + 241.8 kJ
Exothermic reaction — energy is a “product” and
∆H = – 241.8 kJ
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USING ENTHALPY
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Making liquid H2O from H2 +
O2 involves two exothermic
steps.
H2 + O2 gas
H2O vapor
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Liquid H2O
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)
• 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
H  qP
qrxn  qsoln  specificheat of solut ion
 grams of solut ion T
 m CT
m = mass of solution
C = heat capacity of the
calorimeter (J/g-⁰C) (or K)
∆T = the change in temperature
in ⁰C or K
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Constant Pressure Calorimetry
Practice 1
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• 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 ΔH (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 Pressure Calorimetry
Practice 2
• 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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Constant Volume CALORIMETRY
Measuring Heats of Reaction
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CALORIMETRY
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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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Calorimetry
Some heat from reaction warms
water
qwater = (sp. ht.)(water mass)(∆T)
Some heat from reaction warms
“bomb”
qbomb = (heat capacity, J/K)(∆T)
Total heat evolved = qtotal = qwater + qbomb
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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
• Temp rises from 25.00 to 33.20 oC
• Calorimeter contains 1200. g water
• Heat capacity of bomb = 837 J/K
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Measuring Heats of Reaction
CALORIMETRY - Example
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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‘BOMB’ CALORIMETRY PRACTICE
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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)
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Ch. 6.3 – Hess’s Law
Making H2O from H2 involves two steps.
H2(g) + 1/2 O2(g) H2O(g)
ΔrH˚ =-242 kJ
H2O(g)  H2O(liq)
ΔrH˚ =-44 kJ
-----------------------------------------------------------------------
H2(g) + 1/2 O2(g)  H2O(liq) ΔrH˚ =-286 kJ
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others,
the net ∆H is the sum of the ∆H’s of
the other rxns.
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Hess’s Law
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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.
NOTE: ∆rH stands for the
enthalpy change for a
reaction, r
Active Figure 5.16
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Hess’s Law
& Energy Level
Diagrams
Forming CO2 can occur
in a single step or in a
two steps. ∆rHtotal is the
same no matter which
path is followed.
Active Figure 5.16
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Enthalpy and Hess’s Law
• 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
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Simple Hess’s Law problem
• C(graphite) + O2(g) → CO2(g)
ΔH° = –393.51 kJ mol–1
• C(diamond) + O2(g) → CO2(g)
ΔH° = –395.40 kJ mol–1
Calculate the ∆H for the conversion of
graphite to diamond.
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Simple Hess’s Law problem
• C(graphite) + O2(g) → CO2(g)
ΔH° = –393.51 kJ mol–1
• CO2(g) → C(diamond) + O2(g)
ΔH° = +395.40 kJ mol–1
Add rxns together, cancel out substances that
appear on both sides and get:
C(graphite)  C(diamond)
ΔH = 1.9 kJ
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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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Harder Example of Hess’s Law
Given:
C(s) + O2(g)  CO2(g) ΔH= -393.5kJ/mol
S(s) + O2(g)  SO2(g) ΔH= -296.8 kJ/mol
CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)
ΔH = -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)
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Multiply equations as needed:
C(s) + O2(g)  CO2(g) ΔH= -393.5kJ/mol
2S(s) + 2O2(g)  2SO2(g) ΔH=2*(-296.8 kJ/mol)
=-593.6 kJ
CO2(g) + 2SO2(g)  CS2(l) + 3O2(g)
ΔH = 1103.9 kJ/mol
C(s) + 2S(s)  CS2(l) ΔH = 116.8 kJ
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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∆H along one path =
∆H along another path
• 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.
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Standard Enthalpy Values
Most ∆H values are labeled ∆Ho
Measured under standard conditions:
•
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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Ch. 6.4 – Standard Enthalpies
of Formation
NIST (Nat’l Institute for Standards and
Technology) gives values of
∆Hfo = standard molar enthalpy of
formation
— the enthalpy change when 1 mol of
compound is formed from elements under
standard conditions.
See Appendix 4 (A19 – A22)
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∆Hfo, standard molar
enthalpy of formation
H2(g) + 1/2 O2(g)  H2O(g)
∆Hfo (H2O, g) = -241.8 kJ/mol
By definition,
∆Hfo = 0 for elements in their
standard states.
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Enthalpy of Formation Practice
• Show the ΔHf˚ reaction for the
following compounds:
a) C2H2(g)
b) KCl(s)
• Why is the following not a ΔHf˚?
4Na(s) + O2(g)  2Na2O(s)
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Hess’s Law and ΔHf° Practice
Given:
C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l)
ΔHr° = -1300. kJ/mol-rxn
C(s) + O2(g)  CO2(g)
ΔHr° = -394. kJ/mol-rxn
H2(g) + ½O2(g)  H2O(l)
ΔHr° = -286. kJ/mol-rxn
Determine the ΔHf° 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
Answer:
-1( C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l) )
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Using Standard Enthalpy Values
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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Using Standard Enthalpy Values
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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Using Standard Enthalpy Values
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In general, when ALL
Calculate ∆H of
reaction?
enthalpies of formation are
known,
∆Hro =  ∆fHfo (products)
-  ∆Hfo (reactants)
Remember that ∆ always = final – initial
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Heats of Reactions
(∆H⁰(r or rxn) = generic reaction)
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• ∆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
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Heats of (Physical) Process
(∆H⁰process = generic process)
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• ∆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
© 2009 Brooks/Cole - Cengage
Using Standard Enthalpy Values
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
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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)
- {3/2 ∆Hfo (O2) + ∆Hfo (CH3OH)}
= (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
∆Hocomb = -675.6 kJ per mol of methanol
© 2009 Brooks/Cole - Cengage
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Standard Enthalpy Values
Example
84
Using Enthalpies of Formation calculate the Enthalpy
of Combustion for the reaction below.
r H    n f H   products    m f H   reactants 
C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(l)
ΔH° = 3ΔH°(CO2) + 4 ΔH°(H2O(l))–[ΔH°(C3H8) + 5ΔH°f(O2)]
Hocomb = 3 (-393.5) + 4 (-285.8) – [-103.8 + 5(0)]
 H˚comb= -2220 kJ
© 2009 Brooks/Cole - Cengage
85
ΔHrxn = ΔH1 + ΔH2 + ΔH3
© 2009 Brooks/Cole - Cengage
Using Enthalpy of
Reaction
86
Think of reaction like this:
Formation of elements
3C(s) + 4H2(g)
-ΔHf1°
Regroup and form new substances
5O2 3C + 3O2 4H2+2O2
-ΔHf2°
3ΔHf3°
4ΔHf4°
C3H8 + 5O2  3CO2 + 4H2O
ΔrH = -ΔHf1 – ΔHf2 + 3ΔHf3 + 4ΔHf4
= +3ΔHf3
© 2009 Brooks/Cole - Cengage
+ 4ΔHf4 – (ΔHf1 + ΔHf2)
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Enthalpy Practice Problem
Calculate the standard enthalpy change for the
reaction below using the standard enthalpies of
formation.
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)
ΔH°f’s: 0
-825.5
-1675.7 0
(kJ/mol)
© 2009 Brooks/Cole - Cengage