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1 Chapter 5 — Thermo
Jeffrey Mack
California State University, Sacramento
Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions
Questions that need to be addressed:
• How do we measure and calculate the energy
changes that are associated with physical changes
and chemical reactions?
• What is the relationship between energy changes,
heat, and work?
• How can we determine whether a chemical reaction is
product-favored or reactant-favored at equilibrium?
• How can we determine whether a chemical reaction
or physical change will occur spontaneously, that is,
without outside intervention?
Energy & Chemistry
1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food
“calorie”)
SI units for energy: joule (J)
1 cal = exactly 4.184 J
James Joule 1818-1889
Units of Energy
Some Basic Principles
• Energy as the capacity to do work or transfer heat (q).
• Heat (q) is NOT temperature, temperature is a
measure of kinetic energy.
• Energy is divided into two basic categories: – Kinetic energy (the energy associated with motion)
– Potential energy (energy that results from an object’s position).
• The law of conservation of energy requires that
energy can neither be created nor destroyed.
• However, energy can be converted from one type into
another.
Energy & Chemistry
Energy can be divided into two forms:
Kinetic: Energy of motion:
Thermal
Mechanical
Electrical
Potential: Stored energy:
Gravitational
Electrostatic
Chemical & Nuclear
Energy & Chemistry
Kinetic energy — energy of motion
• Translation
Potential & Kinetic Energy
2 Chapter 5 — Thermo
Potential energy — energy a motionless body has by virtue of its composition and position.
Potential & Kinetic Energy
• Burning peanuts supply sufficient energy to boil a cup of water.
• Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)
Energy & Chemistry
• Energy transfer as heat will occur spontaneously from an object at a higher temperature to an object at a lower
temperature.
• Transfer of energy as heat continues until both objects are at
the same temperature and thermal equilibrium is achieved.
• At thermal equilibrium, the object with a temperature increase has gained thermal energy, the object with a
temperature decrease has lost thermal energy.
Thermal Equilibrium
• SYSTEM – The object under study
• SURROUNDINGS – Everything outside the
system
• Energy flows between the two
System & Surroundings
• When energy leaves the system and goes into the surroundings, the process is said to be EXOTHERMIC.
• In the case of thermal energy, the temperature of the system decreases. (qsystem < 0)
• Tsystem = (Tfinal – Tinitial) < 0
Directionality of Energy Transfer
• When energy enters the system and from the surroundings, the process is said to be ENDOTHERMIC.
• In the case of thermal energy, the temperature of the system increases. (qsystem > 0)
• Tsystem > 0
Directionality of Energy Transfer
3 Chapter 5 — Thermo
• Heat flows between the system and
surroundings.
• U is defined as the internal energy of the
system.
• q = the heat absorbed or lost
= - ±system final initialÄU U U = q
Heat & Changes in Internal Energy
If heat enters the system:
U > 0 therefore q is positive (+)
If heat leaves the system:
U < 0 therefore q is negative (–)
The sign of q is a “convention”, it designates the direction of heat flow between the system and
surroundings.
= - ±system final initialÄU U U = q
Heat & Changes in Internal Energy
Surroundings
System
heat in
qsystem > 0 (+)
Usystem > 0
heat out
qsystem < 0 (–)
Usystem < 0
increases decreases
q = the heat absorbed or lost by the system.
Usys
tem
Usys
tem
= - ±system final initialÄU U U = q
Heat & Changes in Internal Energy
= ´Dq C T
The amount of heat (q) transfer related to an object
and temperature is given by:
When heat is absorbed or lost by a body, the
temperature must change as long as the phase
(s, g or l) remains constant.
q = heat lost or gained (J)
C = Heat Capacity of an object
J
C or K
T = Tfinal Tinitial is the temperature change (°C or K)
Heat Capacity
q = m S ∆T
The amount of heat (q) transfer per unit mass of a substance is related to the mass and temperature by:
When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l)
remains constant.
q = heat lost or gained (J) m = mass of substance (g)
S= the Specific Heat Capacity of a compound J
g C or K
T = Tfinal Tinitial is the temperature change (°C or K)
Heat & Specific Heat Capacity
Recall that T = Tf –Tin
50.0 °C = 50.0 °C + 273.2 = 323.2 K
25.0 °C = (25.0 °C + 273.2) = 298.2 K
= 25.0 °C = 25.0 K
The are the same!
T T
Does it matter if we calculate a temperature
change in Kelvin or degrees C?
Tf =
–Tin =
let Tin = 25.0 °C and Tf = 50.0 °C
4 Chapter 5 — Thermo
Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial
temperature of the block was 17.0 C
The specific heat capacity of the metal is:
J2.72
g C
Data Information: Mass, initial temp, heat capacity
of metal, heat (q) absorbed.
Solve q = mST for Tfinal,
plug in data.
Final temperature of the metal is determined.
Find Tfinal of the metal after it absorbs the energy
Strategy Map:
Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial
temperature of the block was 17.0 C
The specific heat capacity of the metal is:
J2.72
g C
rearranging:
Tf =
+255 cal 4.184 J
1 cal
25.0 g J
2.72 g C
+ 17.0 °C = 32.7
°C Tf > Tin
as
expected
q = m S ∆T q = m S (TFinal – Tinitial)
´final initial
qT = +T
m C
Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at
initially at 21.0 °C.
• At thermal equilibrium, the water and iron are both at 23.1 °C
• What is the specific heat capacity of the metal?
Solution:
Heat is transferred from the hot metal to the colder
water.
Energy is conserved so:
+ =
= -
Fe water
Fe water
q q 0
q q
Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at
initially at 21.0 °C.
• At thermal equilibrium, the water and iron are both at 23.1 °C
• What is the specific heat capacity of the metal?
Solution:
Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at
initially at 21.0 °C.
• At thermal equilibrium, the water and iron are both at 23.1 °C
• What is the specific heat capacity of the metal?
mFe Sfe ∆Tfe = -mw Sw ∆Tw
Solution:
Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at
initially at 21.0 °C.
• At thermal equilibrium, the water and iron are both at 23.1 °C
• What is the specific heat capacity of the metal?
- ´ ´D=
´D
water water waterFe
Fe Fe
m C TC
m T
5 Chapter 5 — Thermo
Solution:
Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at
initially at 21.0 °C.
• At thermal equilibrium, the water and iron are both at 23.1 °C
• What is the specific heat capacity of the metal?
Fe
J 1 K225 g 4.184 (23.1 C 21.0 C)
Jg K 1 CC 0.469
1 K g K55.0 g (23.1 C 99.8 C)
1 C
• When matter absorbs heat, its temperature will rise until
it undergoes a Phase Change.
• The matter will continue to absorb energy, however
during the phase change its temperature remains
constant: Phase changes are “Isothermal” processes.
Energy & Changes of State
• Some changes of state (phase
changes) are endothermic:
• When you perspire, water on your
skin evaporates.
• This requires energy.
• Heat from your body is absorbed by
the water as it goes from the liquid
state to the vapor state, as a result
you cool down.
+ energy
+ ®2 2H O(l) Heat H O(g)
Energy Transfer & Changes of State
• Some changes of state (phase changes) are exothermic:
• When it is muggy outside, water condenses on your skin.
• This releases energy.
• Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state.
• As a result you feel hot.
+ energy
® +2 2H O(g) H O(l) Heat
Energy Transfer & Changes of State
Heating/Cooling Curve for Water
Note the isotherms
at each phase
transition.
• The energy associated with a change of state is given by the Enthalpy or Heat of the phase change.
• Since there is no temperature change associated with the
process, the units are most often in J/g or J/mol.
Sublimation: subH > 0 (endothermic)
Vaporization: vapH > 0 (endothermic)
Melting or Fusion: fusH > 0 (endothermic)
Deposition: depH < 0 (exothermic)
Condensation: conH < 0 (exothermic)
Freezing: freH < 0 (exothermic)
Where H refers to the “Heat” of a phase change
Heat & Changes of State
6 Chapter 5 — Thermo
q(heating or cooling) = m C T
q(phase change) = (phase change)H n
Heating & Cooling:
Heat absorbed or lost in a Phase change:
(n = moles or grams)
Energy Change Calculations Problem:
What quantity of heat is required to melt 500. g
of ice at 0 °C then heat the resulting water to
steam at 100 °C?
+333 J/g +2260 J/g
melt the ice form liquid water at 0 °C heat the water to 100 °C boil
water
fusH
Swater
VapH
×
Constants Needed:
JHeat of fusion of ice = 333
g
JSpecific heat of water = 4.18
g K
JHeat of vaporization = 2260
g
Equations:
qphase change =
mH
qheat = mST
Problem:
What quantity of heat is required to melt 500. g
of ice at 0 °C then heat the resulting water to
steam at 100 °C?
Ice H2O(s)
(0 °C)
fusH
water H2O(l)
(0 °C)
water H2O(l)
(100
°C)
Cwater vapH
steam H2O(g)
(100
°C)
Problem:
What quantity of heat is required to melt 500. g
of ice at 0 °C then heat the resulting water to
steam at 100 °C?
= ´D1 ice fusq n H + = ´ ´D2 water water q m C T + = ´D3 water vap q n Hqtotal =
melt ice heat water boil water
Problem:
What quantity of heat is required to melt 500. g
of ice at 0 °C then heat the resulting water to
steam at 100 °C?
Ice H2O(s)
(0 °C)
fusH
water H2O(l)
(0 °C)
water H2O(l)
(100
°C)
Cwater vapH
steam H2O(g)
(100
°C)
total
6
Jq 500. g 333
g
J 500. g 4.18 (100.0 C 0.0 C)
g C
J 500. g 2260 1.51 10 J
g
Problem:
What quantity of heat is required to melt 500. g
of ice at 0 °C then heat the resulting water to
steam at 100 °C?
Ice H2O(s)
(0 °C)
fusH
water H2O(l)
(0 °C)
water H2O(l)
(100
°C)
Cwater vapH
steam H2O(g)
(100
°C)
7 Chapter 5 — Thermo
The total internal energy of an isolated system is constant.
The “system” is that which we are interested in.
The “surroundings” are
everything in contact with the
system.
Together: System + Surroundings = Universe
First Law of Thermodynamics
∆Usystem = Ufinal – Uinitial
Any energy lost by the
system is transferred to the
surroundings and vice versa.
Any change in energy is
related to the final and initial
states of the system.
The same holds for the surrounding!
Energy
Energy
First Law of Thermodynamics
Energy cannot be created or destroyed.
Energy of the universes (system + surroundings) is constant.
Any energy transferred from a system must be transferred to
the surroundings (and vice versa).
From the first law of thermodynamics: When a system undergoes
a physical or chemical change, the change in internal energy is
given by the heat added to or absorbed by the system plus the
work done on or by the system:
U q w change in
system
energy =
heat lost or gained
by the system +
work done by or on the system
Relating U to Heat and Work energy transfer out
(exothermic), -q
energy transfer in
(endothermic), +q
SYSTEM
∆U = q + w
w transfer in
(+w) w transfer out
(-w)
U initial
U Final
ene
rgy
q in
work in
Uf > Ui
Usystem > 0 (+)
q out
work out
Uf < Ui
Usystem < 0 (–)
U Final
U initial
Energy in Energy out
Heat
Work
Usystem = 0
work and heat can
balance!
Energy is the capacity to do work. Work equals a force applied through a distance.
When you do work, you expend energy.
When you push down on a bike pump, you do work.
=force
Parea
=2
force
x
change in volume V = x3
3
2
FP V = x = F x
x
P × V = work
Therefore work is equal to a change of volume at constant pressure.
8 Chapter 5 — Thermo
Enthalpy, “H” is the heat transferred between the system and surroundings under conditions of constant pressure.
H = qp the subscript “p” indicates constant
pressure = +H U PV
( )D = D +
= D + D
H U PV
U P Vif no “PV” work is done by the system,
V = 0
0
H = Up
the change in
enthalpy is the
change in internal
energy at constant
pressure
Enthalpy
For a “system” the overall change in Enthalpy is path independent.
No matter which path is taken
(AB) the results are the
same:
Final – Initial
Since individual Enthalpies
cannot be directly measured,
we only deal with enthalpy
changes
(H = Hf – Hi)
A
B
Enthalpy is a “State Function”
• Since Enthalpies are state functions, one
must specify the conditions at which they are
measured.
• H(T,P): Enthalpy is a function of temperature
and pressure.
• “H°” indicates that the Enthalpy is taken at Standard State conditions.
• Standard State Conditions are defined as:
• 1 atm = 760 mm Hg or 760 torr & 298.15 K or
25 °C
Enthalpy Conditions
Reactants Products
Ene
rgy
H (Products)
H (Reactants) H (Products)
H (Reactants)
rH > 0 rH < 0
Endothermic Exothermic
H = Hfinal Hinitial
Enthalpy of reaction = rH = Hproducts Hreactants
Enthalpies & Chemical Reactions:
rH
Just like a regular chemical equation, with an energy term.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rHo = 802 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ
What does this imply? CONVERSION FACTORS!!!
From the equation:
+
4
802 kJ of Energy Released
1 mol CH (g) consumed2
+802 kJ of Energy Released
2 mol H O (g) produced
energy out… Exothermic
Energy is a product just like CO2 or H2O!
Thermochemical Equations Problem: How many kJ of energy are released when 128.5 g of methane, CH4(g) is combusted?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
4128.5g CH´ 41 mol CH
16.04 g´ =
+802 kJ
1mol rxn6.43103 kJ
g mols J
molar
mass
Reaction
enthalpy
rHo = 802 kJ
´4
1 mol rxn
1 mol CH
9 Chapter 5 — Thermo
When the reaction is reversed , the sign of H reverses:
CO2(g) + 2H2O(g) CH4(g) + 2O2(g) rHo = +802 kJ
Endothermic
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rHo = 802 kJ
Exothermic
Thermochemical Equations
[ ] [ ]
H scales with the reaction:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rHo = 802 kJ
1
2´
1
2´
®4 2 2 2
1 1CH (g) + O (g) CO (g) + H O(g)
2 2rH
o = 401 kJ
Yes you can write the reaction with fractions, so long as you are writing it on a mole basis…
Thermochemical Equations
Change in enthalpy depends on state:
H2O(g) H2O(l)
This means that water’s liquid state lies
44 kJ/mol lower than the gas state
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rH° = 802 kJ
rH° = 890 kJ
H = 44 kJ/mol
= 88kJ
From your text:
[ ] 2 = 88kJ The difference in the rxn H is due to the change in state!!
Thermochemical Equations
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
CO2(g) + 2H2O(l)
rHo = 802 kJ
rHo = 890 kJ
2 44 kJ
The difference is the
88 kJ released when 2
mols of water go from
gas to liquid.
Comparing the reaction with water as a gas or
liquid:
Enth
alp
y (
H)
products
products
reactants
either pathway gives the same
results!
• A constant pressure
calorimeter can be used
to measure the amount of
energy transferred as
heat under constant
pressure conditions, that
is, the enthalpy change
for a chemical reaction.
Constant Pressure Calorimetry, Measuring H
• The constant pressure
calorimeter used in general
chemistry laboratories is
often a “coffee-cup
calorimeter.” This
inexpensive device
consists of two nested
Styrofoam coffee cups with
a loose-fitting lid and a
temperature-measuring
device such as a
thermometer or
thermocouple.
Constant Pressure Calorimetry, Measuring H
10 Chapter 5 — Thermo
• Because the “coffee-cup
calorimeter.” is an
isolated system, “What happens in the coffee cup,
stays in the coffee cup!” • No mass loss to the
surroundings.
• No heat loss to the
surroundings.
Constant Pressure Calorimetry, Measuring H
Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffee-
cup calorimeter.
• This resulted in a decrease in temperature from 18.6 °C to 16.2 °C.
• Calculate the enthalpy change for dissolving NH4NO3(s) in water in
kJ/mol.
• Assume the solution has a specific heat capacity of 4.18 J/g ? K.
Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffee-
cup calorimeter.
• This resulted in a decrease in temperature from 18.6 °C to 16.2 °C.
• Calculate the enthalpy change for dissolving NH4NO3(s) in water in
kJ/mol.
• Assume the solution has a specific heat capacity of 4.18 J/g ? K.
Data Information: Mass of
reactant, C, water & temperature change.
Cals. Moles of
NH4NO3
Step 1:
Eq. gives mole ratios
(stoichiometry)
Determine qsolution =
mCT
Step 2:
qsolution
qsolution + qrxn = 0 Step 3:
qrxn = q(NH4NO3)
rH =qrxn/mol NH4NO3 Step 4:
Enthalpy per mole of
reactant
Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffee-
cup calorimeter.
• This resulted in a decrease in temperature from 18.6 °C to 16.2 °C.
• Calculate the enthalpy change for dissolving NH4NO3(s) in water in
kJ/mol.
• Assume the solution has a specific heat capacity of 4.18 J/g ? K.
4 34 3 4 3
solution solution
solution
3
solution
1 mol NH NO5.44 g NH NO 0.0680 moles NH NO
80.04 g
q m C T
Jq 154.4 g 4.18 (16.2 C 18.6 C)
g C
q 1.55 10 J
Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffee-
cup calorimeter.
• This resulted in a decrease in temperature from 18.6 °C to 16.2 °C.
• Calculate the enthalpy change for dissolving NH4NO3(s) in water in
kJ/mol.
• Assume the solution has a specific heat capacity of 4.18 J/g ? K.
= - ´
+ =
= - = + ´
+ ´D = = ´ = +
3
solution
rxn solution
3
rxn solution
3
rxnr 3
4 3
q 4.55 10 J
q q 0
q q 1.55 10 J
q 1.55 10 J 1kJ kJH 22.8
moles of reaction 0.0680 moles NH NO 10 J mol
The sign is positive indicating an endothermic process.
Under conditions of
constant volume, any
heat transferred is equal
to a change of internal
energy rU.
qV = rU
Constant Volume Calorimetry, Measuring U
11 Chapter 5 — Thermo
Heats of combustion
(rUocombustion ) are measured
using a device called a Bomb
Calorimeter.
A combustible sample is reacted with excess O2
rxn cal
cal water bomb
q C T
C C C
= - ´D
= +
The heat capacity of the bomb is constant.
The heat of reaction is found by:
Constant Volume Calorimetry, Measuring U
Octane, the primary component of gasoline combusts by the reaction:
C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9H2O(l)
A 1.00 g sample of octane is burned in a bomb calorimeter that
contains 1.20 kg of water surrounding the bomb.
The temperature of the water rises to 33.20 °C from 25.00 °C
when the octane is reacted.
If the heat capacity of the bomb is 837 J/°C, calculate the heat of
reaction per mole of octane.
–qrxn = qwater + qbomb
Since the temperature of the water rose, the reaction must have been exothermic:
Therefore one can write:
Calculating Heat in an Exothermic Reaction
–qrxn = mwaterCwaterTwater + qbombTwater
- =RXNq4.184J
g C´ ´ 31.20 10 g 33.20 25.00 C
J837
C
33.20 25.00 C qwater
qbomb
qRXN = – 4803 J or – 48.0 kJ
Heat transferred per mole qV: 3 48.0 kJ kJ
5.48 10mol mol
1.00g114.2g
-= - ´
´
Calculating Heat in an Exothermic Reaction
The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.
Reactants Products rH = ?
unknown!
Intermediate Reaction
rH1 + rH2 = rH
The sum of the H’s in one direction must equal the sum in the other direction.
What if the enthalpy changes through
another path are know?
Why?
Because enthalpy is a state function… Path independent!
so we can write…
Hess’s Law
Forming CO2 can
occur in a single
step or in a two
steps.
∆rHtotal is the same
no matter which
path is followed.
Hess’s law & Energy Level Diagrams
Notice that the path from reactants to products in the
desired reaction goes through an “intermediate
compound” in the given reactions.
This means that the path for hydrogen and nitrogen to
produce ammonia goes through hydrazine (N2H4).
Therefore, the path to the enthalpy of the reaction
must be a sum of the two given reactions!
Hess’s Law Problem:
Example: Determine the rH for the reaction:
3H2(g) + N2(g) 2NH3(g) rHo = ???
Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ
(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ
12 Chapter 5 — Thermo
adding equations (1) & (2) yields:
2 H2(g) + N2(g) + N2H4(g) + H2(g) N2H4(g) + 2NH3(g)
Hess’s Law Problem:
Example: Determine the rH for the reaction:
3H2(g) + N2(g) 2NH3(g) rHo = ???
Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ
(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ
Hess’s Law Problem:
Example: Determine the rH for the reaction:
3H2(g) + N2(g) 2NH3(g) rHo = ???
Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ
(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ
adding equations (1) & (2) yields:
2 H2(g) + N2(g) + N2H4(g) + H2(g) N2H4(g) + 2NH3(g)
Look what happens…
/ /
3H2(g) + N2(g) 2NH3(g)
Hess’s Law Problem:
Example: Determine the rH for the reaction:
3H2(g) + N2(g) 2NH3(g) rHo = ???
Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ
(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ
adding equations (1) & (2) yields:
2 H2(g) + N2(g) + N2H4(g) + H2(g) N2H4(g) + 2NH3(g)
Look what happens…
/ /
3H2(g) + N2(g) 2NH3(g)
rH = ∆rH°1 + ∆rH°2 = +95.4 kJ + (–187.6 kJ) = –92.2 kJ
therefore… • Reaction (1) is endothermic by 95.4 kJ
• Reaction (2) is exothermic by 187.6 kJ
• Since the exothermicity has a greater magnitude
than the endothermicity, the overall process is
exothermic (rH < 0).
intermediates
H2(g) +
N2H4(g)
reactants
3H2(g) +
N2(g)
products
2N3 (g)
(Step 1)
Ho = 95.4kJ (Step 2)
Ho = 187.6kJ
(Overall)
Ho = 92.2kJ
When 1 mole of compound is formed from its elements, the enthalpy change for the reaction is called the enthalpy of
formation, fHo (kJ/mol).
These enthalpies are always reported at Standard conditions:
1 atm and 25 °C (298 K).
The standard enthalpies of formation of the most stable form of any element is zero:
fH (element) = 0
fH° O2(g) = 0 fH° O(g) ≠ 0 elemental form NOT the elemental form
Standard Enthalpies of Formation
H2(g) + ½ O2(g) H2O(g) ∆rH˚ = -242 kJ
H2(g) + ½ O2(g) H2O(liq) ∆rH˚ = -286 kJ
Same reaction, different phases, different enthalpies.
Since Enthalpy is a state function, enthalpy values
depend on the reaction conditions in terms of the
phases of reactants and products.
Enthalpy Values
13 Chapter 5 — Thermo
The formation of water is given by the reaction:
H2(g) + ½ O2(g) H2O(l)
Each element and the compound are represented by
the physical state they take on at 25.0 °C and 1 atm pressure. (Standard State conditions)
A chemical reaction that describes the formation of
one mole of a compound from its elements at
standard state conditions is known as a “formation
reaction”.
Formation Reactions Question:
What is the formation reaction for potassium
permanganate?
KMnO4
• salts are solids at standard state conditions.
• metals are solids at standard state conditions.
• oxygen is a gas at standard state conditions.
• balance for one mole of the product
(s) K + Mn + O2 (s) (s) (g) 2
standard state conditions = 25oC and 1 atm
compound elements
• All components of a reaction can be related back
to their original elements.
• Each compound has an enthalpy of formation
associated with it.
• Reactants require energy to return to component
elements.
• Products release energy when formed form
component elements.
• Since enthalpy is a state function, the sum of the
above must relate somehow to the overall
enthalpy of a reaction.
Enthalpy Changes for a Reaction: Using Standard Enthalpy Values
C3H8(g) + 5O2(g) 3CO2(g) + 4 H2O(l)
• In order to make CO2(g) and H2O(l) one must break
the propane up into its elements.
• This takes energy.
• The elements carbon, hydrogen and oxygen then
combine to make the new compounds, CO2 and
H2O.
• This process releases energy.
3C(s) + 8H2(g) + 5O2(g)
Consider the Combustion of Propane
( ) ( )° °D ° = D - Då år f fH n H products m H reactants
The sum of the fH° for the products multiplied by the respective coefficients, subtracted by the sum of
the fH° for the reactants multiplied by the
respective coefficients, yields the fH° for the reaction.
n and m are the stoichiometric balancing coefficients.
In other words: Energy gained – Energy spent = Net Energy
Enthalpy Changes for a Reaction: Using Standard Enthalpy Values
fH° [CaO(s)] = – 635.5 kJ/mol
CaO(s) + CO2(g) CaCO3(s)
fH° [CO2(g)] = – 393.5 kJ/mol
fH° [CaCO3(s)] = –1207 kJ/mol
Problem:
Calculate the rH° for CaCO3(s) given the following fH°
14 Chapter 5 — Thermo
fH° [CaO(s)] = – 635.5 kJ/mol
CaO(s) + CO2(g) CaCO3(s)
fH° [CO2(g)] = – 393.5 kJ/mol
fH° [CaCO3(s)] = –1207 kJ/mol
( )
( )
°
°
D ° = D
D
å
å
r f
f
H n H products
m H reactants
Problem:
Calculate the rH° for CaCO3(s) given the following fH°
Problem:
Calculate the rH° for CaCO3(s) given the following fH°
fH° [CaO(s)] = – 635.5 kJ/mol
CaO(s) + CO2(g) CaCO3(s)
fH° [CO2(g)] = – 393.5 kJ/mol
fH° [CaCO3(s)] = –1207 kJ/mol
( )
( )
°
°
D ° = D
D
å
å
r f
f
H n H products
m H reactants
rH° = –1207 kJ/mol
1mol CaCO3
– (–635.5 kJ/mol
1mol CaO
–393.5 kJ/mol)
1mol CO2
rH° = – 178 kJ/mol
Problem:
The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol.
Calculate the fH° for naphthalene given the following enthalpies of formation:
fH° [CO2(g)] = –393.5 kJ/mol
fH° [H2O(l)] = –285.7 kJ/mol
C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l)
Step 1: Write the balanced equation for the reaction…
Problem:
The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol.
Calculate the fH° for naphthalene given the following enthalpies of formation:
fH° [CO2(g)] = –393.5 kJ/mol
fH° [H2O(l)] = –285.7 kJ/mol
10 fH° [CO2(g)] + 4 fH° [H2O(l)]
– {fH° [C8H10(l)] + 12 fH° [O2(g)]}
combustionH° =
( ) ( )° °D ° = D - Då år f fH n H products m H reactants
Next recall that:
C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l)
From the problem, all quantities are know but f 10 8H C H
Problem:
The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol.
Calculate the fH° for naphthalene given the following enthalpies of formation:
fH° [CO2(g)] = –393.5 kJ/mol
fH° [H2O(l)] = –285.7 kJ/mol
fH° [C10H8(l)] =
10 fH° [CO2(g)]
+ 4 fH° [H2O(l)]
–
{combH°
+ 12 fH° [O2(g)]}
elements = 0
15 Chapter 5 — Thermo
Problem:
The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol.
Calculate the fH° for naphthalene given the following enthalpies of formation:
fH° [CO2(g)] = –393.5 kJ/mol
fH° [H2O(l)] = –285.7 kJ/mol
fH° [C10H8(l)] =
10 fH° [CO2(g)]
+ 4 fH° [H2O(l)]
–
{combH°
+ 12 fH° [O2(g)]}
elements = 0 H° f [C10H8(l)] =
10 (–393.5 kJ/mol + 4 (–285.7 kJ/mol) – (– 5156 kJ/mol)
= + 79 kJ/mol
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