Energy &
Chemical Change
CHAPTER 7
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
1
CHAPTER 7: Energy & Chemical Change
Learning Objectives
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Potential vs Kinetic Energy
Internal Energy, Work, and Heat Calculations
System vs Surroundings
First Law of Thermodynamics
Units of Energy
Endothermic vs Exothermic
Heat of Temperature Changes (Specific Heat and Heat
Capacity)
Calorimetry: Constant Volume vs Constant Pressure
Heat Stoichiometry (Thermochemical Equations)
Hess’s Law and Equation Summation
Heat of Reaction from Heats of Formation
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
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Energy
Chemical Bonds
Chemical bond
• Attractive forces that bind
– Atoms to each other in molecules, or
– Ions to each other in ionic compounds
– Give rise to compound’s potential energy
Chemical energy
– Potential energy stored in chemical bonds
Chemical reactions
• Generally involve both breaking and making
chemical bonds
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
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Energy
Chemical Reactions
Forming Bonds
– Atoms that are attracted to each other are moved
closer together
– Decrease the potential energy of reacting system
– Releases energy
Breaking Bonds
– Atoms that are attracted to each other are forced
apart
– Increase the potential energy of reacting system
– Requires energy
4
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Exothermic Reaction
• Reaction where products have less chemical energy
than reactants
– Some chemical heat energy converted to kinetic
energy
– Reaction releases heat energy to surroundings
– Heat leaves the system; q is negative ( – )
– Heat energy is a product
– Reaction gets warmer, temperature increases
Example.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat
5
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Endothermic Reaction
Reaction where products have more chemical energy than
reactants
– Some kinetic energy converted to chemical energy
– Reaction absorbs heat from surroundings
– Heat added to system; q is positive (+)
– Heat energy is a reactant
– Reaction becomes colder,
temperature decreases
Example: Photosynthesis
6CO2(g) + 6H2O(g) + solar energy 
C6H12O6(s) + 6O2(g)
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
© Michelle Molinari/Alamy
Energy
Bond Strength
• Measure of how much energy is needed to break bond
or how much energy is released when bond is formed.
• Larger amount of energy equals a stronger bond
• Weak bonds require less energy to break than strong
bonds
• Key to understanding reaction energies
Example: If reaction has
– Weak bonds in reactants and
– Stronger bonds in products
– Heat released
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Why Fuels Release Heat
• Methane and oxygen have weaker bonds
• Water and carbon dioxide have stronger bonds
8
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Heat of Reaction
• Amount of heat absorbed or released in chemical
reaction
• Determined by measuring temperature change they
cause in surroundings
Calorimeter
– Instrument used to measure temperature changes
– Container of known heat capacity
– Use results to calculate heat of reaction
Calorimetry
– Science of using calorimeter to determine heats of
reaction
9
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Heats of Reaction
• Calorimeter design not standard
– Depends on
• Type of reaction
• Precision desired
• Usually measure heat of reaction under one of
two sets of conditions
– Constant volume, qV
• Closed, rigid container
– Constant pressure, qP
• Open to atmosphere
10
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
What is Pressure?
Amount of force acting on unit area
force
Pressure 
area
Atmospheric Pressure
– Pressure exerted by Earth’s
atmosphere by virtue of its weight.
– ~14.7 lb/in2
• Container open to atmosphere
– Under constant P conditions
– P ~ 14.7 lb/in2 ~ 1 atm ~ 1 bar
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
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Energy
Comparing qV and qP
• Difference between qV and qP can be significant
• Reactions involving large volume changes,
– Consumption or production of gas
• Consider gas phase reaction in cylinder
immersed in bucket of water
– Reaction vessel is cylinder topped by piston
– Piston can be locked in place with pin
– Cylinder immersed in insulated bucket
containing weighed amount of water
– Calorimeter consists of piston, cylinder,
bucket, and water
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Comparing qV and qP
• Heat capacity of calorimeter = 8.101 kJ/°C
• Reaction run twice, identical amounts of reactants
Run 1: qV - Constant Volume
– Same reaction run once at constant
volume and once at constant pressure
– Pin locked;
– ti = 24.00 C; tf = 28.91 C
qCal = Ct
= 8.101 J/C  (28.91 – 24.00)C = 39.8 kJ
qV = – qCal = –39.8 kJ
13
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Comparing qV and qP
Run 2: qP
– Run at atmospheric pressure
– Pin unlocked
– ti = 27.32 C; tf = 31.54 C
– Heat absorbed by calorimeter is
qCal = Ct
= 8.101 J/C  (31.54  27.32)C
= 34.2 kJ
qP = – qCal = –34.2 kJ
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
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Energy
Comparing qV and qP
• qV = -39.8 kJ
• qP = -34.2 kJ
• System (reacting mixture) expands, pushes against
atmosphere, does work
– Uses up some energy that would otherwise be heat
– Work = (–39.8 kJ) – (–34.2 kJ) = –5.6 kJ
• Expansion work or pressure volume work
– Minus sign means energy leaving system
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Work Convention
Work = –P × V
• P = opposing pressure against which piston pushes
• V = change in volume of gas during expansion
• V = Vfinal – Vinitial
• For Expansion
– Since Vfinal > Vinitial
– V must be positive
– So expansion work is negative
– Work done by system
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
First Law of Thermodynamics
In an isolated system, the change in internal
energy (E) is constant:
E = Ef – Ei = 0
• Can’t measure internal energy of anything
• Can measure changes in energy
E is state function
E = heat + work
E = q + w
E = heat input + work input
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
First Law of Thermodynamics
Energy of system may be transferred as heat or work,
but not lost or gained.
• If we monitor heat transfers (q) of all materials
involved and all work processes, can predict that
their sum will be zero
– Some energy transfers will be positive, gain in
energy
– Some energy transfers will be negative, a loss
in energy
• By monitoring surroundings, we can predict what is
happening to system
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
First Law of Thermodynamics
E = q + w
q is (+)
Heat absorbed by system (IN)
q is (–)
Heat released by system (OUT)
w is (+)
Work done on system (IN)
w is (–)
Work done by system (OUT)
Endothermic reaction
E = +
Exothermic reaction
E = –
19
Energy
E is Independent of Path
q and w
• NOT path
independent
• NOT state
functions
• Depend on
how change
takes place
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Discharge of Car Battery
Energy
Path a
– Short out with wrench
– All energy converted to heat, no work
• E = q
(w = 0)
Path b
– Run motor
– Energy converted to work and little heat
• E = w + q
(w >> q)
• E is same for each path
– Partitioning between two paths differs
21
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Bomb Calorimeter (Constant V)
• Apparatus for measuring
E in reactions at
constant volume
• Vessel in center with
rigid walls
• Heavily insulated vat
– Water bath
– No heat escapes
• E = qv
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
22
Energy
Calorimeter Problem: Ex 4
When 1.000 g of olive oil is completely burned in pure oxygen
in a bomb calorimeter, the temperature of the water bath
increases from 22.000 ˚C to 26.049 ˚C.
a) How many Calories are in olive oil, per gram? The heat
capacity of the calorimeter is 9.032 kJ/˚C.
t = 26.049 ˚ C – 22.000 ˚C = 4.049 ˚C
qabsorbed by calorimeter = Ct = 9.032 kJ/°C × 4.049 ˚C
= 36.57 kJ
qreleased by oil = – qcalorimeter = – 36.57 kJ
q oil
 36.57 kJ
1 kcal
1 Cal
(in cal/g) 


1.000 g
4.184 kJ 1 kcal
–8.740 Cal/g oil
23
Energy
Calorimeter Problem: Ex 4
b) Olive oil is almost pure glyceryl trioleate, C57H104O6. The
equation for its combustion is
C57H104O6(l) + 80O2(g)  57CO2(g) + 52H2O
What is E for the combustion of one mole of glyceryl trioleate
(MM = 885.4 g/mol)? Assume the olive oil burned in part a)
was pure glyceryl trioleate.
885.4 g C 57H104O 6
 36.57 kJ

1.000 g C 57H104O 6
1 mol C 57H104O 6
E = qV = –3.238 × 104 kJ/mol oil
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Enthalpy (H)
• Heat of reaction at constant Pressure (qP)
H = E + PV
• Similar to E, but for systems at constant P
– Now have PV work + heat transfer
• H = state function
• At constant pressure
H = E + PV = (qP + w) + PV
If only work is P–V work, w = – P V
H = (qP + w) – w = qP
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Enthalpy Change (H)
H is a state function
– H = Hfinal – Hinitial
– H = Hproducts – Hreactants
• Significance of sign of H
Endothermic reaction
• System absorbs energy from surroundings
• H positive
Exothermic reaction
• System loses energy to surroundings
• H negative
26
Energy
Coffee Cup Calorimeter
• Simple
• Measures qP
• Open to atmosphere
– Constant P
• Let heat be exchanged between
reaction and water, and measure
change in temperature
– Very little heat lost
• Calculate heat of reaction
– qP = Ct
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Coffee Cup Calorimetry: Ex 5
NaOH and HCl undergo rapid and exothermic reaction when
you mix 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH.
The initial t = 25.5 °C and final t = 32.2 °C. What is H in
kJ/mole of HCl? Assume for these solutions s = 4.184 J g–1°C–
1. Density: 1.00 M HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g
mL–1
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(aq)
qabsorbed by solution = mass  s  t
massHCl = 50.0 mL  1.02 g/mL = 51.0 g
massNaOH = 50.0 mL  1.04 g/mL = 52.0 g
massfinal solution = 51.0 g + 52.0 g = 103.0 g
t = (32.2 – 25.5) °C = 6.7 °C
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Coffee Cup Calorimetry: Ex 5
qcal = 103.0 g  4.184 J g–1 °C–1  6.7 °C = 2890 J
Rounds to qcal = 2.9  103 J = 2.9 kJ
qrxn = –qcalorimeter = –2.9 kJ
1 mol HCl
0.0500 L HCl soln 
1 L HCl soln
= 0.0500 mol HCl
Heat evolved per mol HCl =
-2.9 kJ
= -58 kJ/mol
DH =
0.0500 mol HCl
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Enthalpy Changes in Chemical Reactions
• Focus on systems
• Endothermic
– Reactants + heat  products
• Exothermic
– Reactants  products + heat
• Want convenient way to use enthalpies to calculate
reaction enthalpies
• Need way to tabulate enthalpies of reactions
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
The Standard State
A standard state specifies all the necessary parameters to
describe a system. Generally this includes the pressure,
temperature, and amount and state of the substances
involved.
Standard state in thermochemistry
– Pressure = 1 atmosphere
– Temperature = 25 °C = 298 K
– Amount of substance = 1 mol (for formation reactions
and phase transitions)
– Amount of substance = moles in an equation
(balanced with the smallest whole number
coefficients)
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Thermodynamic Quantities
• E and H are state functions and are also extensive
properties
• E and H are measurable changes but still extensive
properties.
– Often used where n is not standard, or specified
– E ° and H ° are standard changes and intensive
properties
– Units of kJ /mol for formation reactions and phase
changes (e.g. H °f or H °vap)
– Units of kJ for balanced chemical equations
(H °reaction)
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
H in Chemical Reactions
Standard Conditions for H 's
– 25 °C and 1 atm and 1 mole
Standard Heat of Reaction (H ° )
– Enthalpy change for reaction at 1 atm and 25 °C
Example:
N2(g) + 3H2(g)  2 NH3(g)
1.000 mol
3.000 mol
2.000 mol
• When N2 and H2 react to form NH3 at 25 °C and 1
atm 92.38 kJ released
• H= –92.38 kJ
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Thermochemical Equation
• Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g)
H = –92.38 kJ
• Must give physical states of products and reactants
– H different for different states
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) H ° rxn = –890.5 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H ° rxn = –802.3 kJ
– Difference is equal to the energy to vaporize water
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Thermochemical Equation
• Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g) H= –92.38 kJ
• Assumes coefficients is the number of moles
– 92.38 kJ released when 2 moles of NH3 formed
– If 10 mole of NH3 formed
5N2(g) + 15H2(g)  10NH3(g) H= –461.9 kJ
• H° = (5 × –92.38 kJ) = – 461.9 kJ
– Can have fractional coefficients
• Fraction of mole, NOT fraction of molecule
½N2(g) + 3/2H2(g)  NH3(g) H°rxn = –46.19 kJ
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
State Matters!
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH °rxn= –2043 kJ
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )
ΔH °rxn = –2219 kJ
Note: there is difference in energy because
states do not match
If
H2O(l ) → H2O(g) ΔH °vap = 44 kJ/mol
4H2O(l ) → 4H2O(g) ΔH °vap = 176 kJ/mol
Or –2219 kJ + 176 kJ = –2043 kJ
36
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Thermochemical Equations in Reverse
To get energy out when form products, must put energy in to
go back to reactants
Consequence of Law of Conservation of Energy
– If you know H ° for reaction, you also know H ° for the
reverse
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H°reaction = – 802.3 kJ
•
Reverse thermochemical equation
•
Must change sign of H
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H°reaction = 802.3 kJ
37
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Multiple Paths; Same H °
Can often get from reactants to products by several
different paths
Products
Reactants
Intermediate A
Intermediate B
 Should get same H °
 Enthalpy is state function and path independent
 Let’s see if this is true
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Multiple Paths; Same H °: Ex 6
Path a: Single step
C(s) + O2(g)  CO2(g)
H°rxn = –393.5 kJ
Path b: Two step
Step 1: C(s) + ½O2(g)  CO(g)
H °rxn = –110.5 kJ
Step 2: CO(g) + ½O2(g)  CO2(g) H °rxn = –283.0 kJ
Net Rxn: C(s) + O2(g)  CO2(g) H °rxn = –393.5 kJ
• Chemically and thermochemically, identical results
• True for exothermic reaction or for endothermic
reaction
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Multiple Paths; Same H °rxn: Ex 7
Path a: N2(g) + 2O2(g)  2NO2(g)
H °rxn = 68 kJ
Path b:
Step 1:
N2(g) + O2(g)  2NO(g)
Step 2: 2NO(g) + O2(g)  2NO2(g)
H °rxn = 180. kJ
H °rxn = –112 kJ
Net rxn: N2(g) + 2O2(g)  2NO2(g) H °rxn =
68 kJ
Hess’s Law of Heat Summation
– For any reaction that can be written into steps,
value of H °rxn for reactions = sum of H °rxn values
of each individual step
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Enthalpy Diagrams
Graphical description of Hess’ Law:
– Vertical axis = enthalpy scale
– Horizontal line =various states of reactions
– Higher up = larger enthalpy
– Lower down = smaller enthalpy
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
41
Energy
Enthalpy Diagrams
• Use to measure Hrxn
– Arrow down Hrxn =
negative
– Arrow up Hrxn = positive
• Calculate cycle
– One step process = sum
of two step process
Example: H2O2(l )  H2O(l ) + ½O2(g)
–286 kJ = –188 kJ + Hrxn
Hrxn = –286 kJ – (–188 kJ )
Hrxn = –98 kJ
42
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Hess’s Law
Hess’s Law of Heat Summation
• Going from reactants to products
• Enthalpy change is same whether reaction takes
place in one step or many
• Chief Use
– Calculation of H °rxn for reaction that can’t be
measured directly
– Thermochemical equations for individual steps
of reaction sequence may be combined to obtain
thermochemical equation of overall reaction
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Rules for Manipulating
Thermochemical Equations
1.
When equation is reversed, sign of H°rxn must also
be reversed.
1.
If all coefficients of equation are multiplied or divided
by same factor, value of H°rxn must likewise be
multiplied or divided by that factor
1.
Formulas canceled from both sides of equation must
be for substance in same physical states
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Strategy for Adding Reactions Together:
1. Choose most complex compound in equation for one-step path
2. Choose equation in multi-step path that contains that compound
3. Write equation down so that compound
 is on appropriate side of equation
 has appropriate coefficient for our reaction
4. Repeat steps 1 – 3 for next most complex compound, etc.
5. Choose equation that allows you to

cancel intermediates

multiply by appropriate coefficient
6. Add reactions together and cancel like terms
7. Add energies together, modifying enthalpy values in same way
equation modified

If reversed equation, change sign on enthalpy

If doubled equation, double energy
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
45
Energy
Calculate H°rxn: Ex 8
C (s, graphite)  C (s, diamond)
Given C (s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ
–1[ C (s, dia) + O2(g)  CO2(g) H°rxn = –396 kJ ]
• To get desired equation, must reverse second
equation and add resulting equations
C(s, gr) + O2(g)  CO2(g)
H°rxn = –394 kJ
CO2(g)  C(s, dia) + O2(g)
H°rxn = –(–396 kJ)
C(s, gr) + O2(g) + CO2(g)  C(s, dia) + O2(g) + CO2(g)
H° = –394 kJ + 396 kJ = + 2 kJ
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
•
•
•
•
Tabulating H° values
Need to Tabulate H° values
Major problem is vast number of reactions
Define standard reaction and tabulate these
Use Hess’s Law to calculate H° for any other reaction
Standard Enthalpy of Formation, Hf°
– Amount of heat absorbed or evolved when one mole
of substance is formed at 1 atm (1 bar) and 25 °C
(298 K) from elements in their standard states
– Standard Heat of Formation
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Energy
Standard State
• Most stable form and physical state of element at 1 atm (1
bar) and 25 °C (298 K)
Element
O
C
Standard
state
O2(g)
C (s, gr)
H
Al
Ne
H2(g)
Al(s)
Ne(g)
Note: All Hf° of elements
in their standard states =
0
Forming element from itself.
 See Appendix C in back of textbook and Table 7.2
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Jesperson, Brady, Hyslop. Chemistry:
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Energy
Uses of Standard Enthalpy (Heat) of
Formation, Hf°
1. From definition of Hf°, can write balanced
equations directly
Hf° of C2H5OH(l )
2C(s, gr) + 3H2(g) + ½O2(g)  C2H5OH(l )
Hf° = –277.03 kJ/mol
Hf° of Fe2O3(s)
2Fe(s) + 3/2O2(g)  Fe2O3(s) Hf° = –822.2 kJ/mol
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Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
Using Hf°
Energy
2. Way to apply Hess’s Law wighout needing to manipulate
thermochemical equations
Sum of all H°f of
all of the products
H°reaction =
–
Sum of all H°f of all
of the reactants
Consider the reaction:
aA + bB  cC + dD
H°reaction = c × H°f(C) + d × H°f(D)
– {a×H°f(A) + b×H°f(B)}
• H°rxn has units of kJ because
• Coefficients  heats of formation have units of mol  kJ/mol
(
) (
)
o
DH rxn
= åéëDH fo products ´ moles of product ùû éDH o reactants ´ moles of reactant ù
åë f
û
(
H°rxn has units of kJ
) (
)
H°f has units of kJ/mol
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
50
Calculate H°rxn Using Hf°: Ex 9
Energy
Calculate H°rxn using Hf° data for the reaction
SO3(g)  SO2(g) + ½O2(g)
1. Multiply each Hf° (in kJ/mol) by the number of moles in the equation
2. Add the Hf° (in kJ/mol) multiplied by the number of moles in the
equation of each product
3. Subtract the Hf° (in kJ/mol) multiplied by the number of moles in the
equation of each reactant
(
) (
)
) (
o
DH rxn
= åéëDH fo products ´ moles of product ùû åéëDH fo reactants ´ moles of reactant ùû
(
H°rxn has units of kJ

H rxn
 H f (SO 2 ( g )) 
)
H°f has units of kJ/mol
1



H
(O
(
g
)
)


H
f
2
f (SO 3 ( g ))
2
DHrxn = -297 kJ/mol + 1 2 (0 kJ/mol) - ( - 396 kJ/mol)
H°rxn = 99 kJ
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
51
Other Calculations
Energy
• Don’t always want to know H°rxn
• Can use Hess’s Law and H°rxn to calculate Hf°
for compound where not known
Example 10: Given the following data, what is the value
of
Hf°(C2H3O2–, aq)?
Na+(aq) + C2H3O2–(aq) + 3H2O(l )  NaC2H3O2·3H2O(s)
H°rxnkJ/mol
= –19.7 kJ/mol
Na aq)
H
= –239.7
f
NaC2H3O2•3H2O(s)
H
f
= 710.4 kJ/mol
H2O(l)
H
f
= 285.9 kJ/mol
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
52
Energy
Other Calculations: Ex 10 Continued
H°rxn = Hf° (NaC2H3O2·3H2O, s) – Hf° (Na+, aq) – Hf°
(C2H3O2–, aq) – 3Hf° (H2O, l )
Rearranging
Hf°(C2H3O2–, aq) = Hf°(NaC2H3O2·3H2O, s) –
Hf°(Na+, aq) – H°rxn – 3Hf° (H2O, l)
Hf°(C2H3O2–, aq) =
–710.4 kJ/mol – (–239.7kJ/mol) – (–19.7 kJ/mol) – 3(–
285.9 kJ/mol)
= +406.7 kJ/mol
53
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E