Download Chapter 5

Chapter 5 — Thermo
1
Energy & Chemistry
Questions that need to be addressed:
Chapter 5
Principles of Chemical
Reactivity:
Energy and Chemical
Reactions
Jeffrey Mack
California State University,
Sacramento
Units of Energy
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
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:
1 kcal = 1 Calorie (a food
“calorie”)
– Kinetic energy (the energy associated with motion)
– Potential energy (energy that results from an object’s position).
SI units for energy: joule (J)
1 cal = exactly 4.184 J
James Joule
1818-1889
Energy & Chemistry
Energy can be divided into two forms:
Kinetic: Energy of motion:
Thermal
Mechanical
Electrical
Potential: Stored energy:
Gravitational
Electrostatic
Chemical & Nuclear
• 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?
• 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.
Potential & Kinetic Energy
Kinetic energy — energy of motion
• Translation
Chapter 5 — Thermo
Potential & Kinetic Energy
2
Energy & Chemistry
Potential energy — energy a motionless body
has by virtue of its composition and position.
• Burning peanuts
supply sufficient
energy to boil a
cup of water.
Thermal Equilibrium
• Burning sugar
(sugar reacts with
KClO3, a strong
oxidizing agent)
System & Surroundings
• SYSTEM
– The object under study
• SURROUNDINGS
– Everything outside the
system
• 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.
Directionality of Energy Transfer
• 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
• Energy flows between
the two
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
Chapter 5 — Thermo
3
Heat & Changes in Internal Energy
Heat & Changes in Internal Energy
• Heat flows between the system and
surroundings.
• U is defined as the internal energy of the
system.
• q = the heat absorbed or lost
ÄUsystem = Ufinal - Uinitial = ± q
If heat enters the system:
U > 0 therefore q is positive (+)
If heat leaves the system:
ÄUsystem = Ufinal - Uinitial = ± q
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.
Heat Capacity
Heat & Changes in Internal Energy
ÄUsystem = Ufinal - Uinitial = ± q
When heat is absorbed or lost by a body, the
temperature must change as long as the phase
(s, g or l) remains constant.
Usystem
heat in
heat out
System
Usystem
Surroundings
qsystem < 0
(–)
Usystem < 0
decreases
qsystem > 0
(+)
Usystem > 0
increases
q = the heat absorbed or lost by the system.
Heat & Specific Heat Capacity
The amount of heat (q) transfer related to an object
and temperature is given by:
q = C ´ DT
J


 C or K 


q = heat lost or gained (J)
C = Heat Capacity of an object
T = Tfinal  Tinitial is the temperature change (°C or K)
Does it matter if we calculate a temperature
change in Kelvin or degrees C?
 When heat is absorbed or lost by a body, the
temperature must change as long as the phase (s, g or l)
remains constant.
 The amount of heat (q) transfer per unit mass of a
substance is related to the mass and temperature by:
let Tin = 25.0 °C and Tf = 50.0 °C
Recall that T = Tf –Tin
Tf =
50.0 °C
=
50.0 °C + 273.2
=
323.2 K
q = m S ∆T
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)
–Tin =  25.0 °C
T = 25.0 °C
=  (25.0 °C + 273.2)
T =
The are the same!
=  298.2 K
25.0 K
Chapter 5 — Thermo
Determine the final temperature of a 25.0 g block of
metal that absorbs 255 cal of energy. The initial
J
temperature of the block was 17.0 C
2.72
The specific heat capacity of the metal is:
g  C
Strategy Map:
4
Determine the final temperature of a 25.0 g block of
metal that absorbs 255 cal of energy. The initial
J
temperature of the block was 17.0 C
2.72
The specific heat capacity of the metal is:
g  C
Find Tfinal of the metal after
it absorbs the energy
Data Information: Mass,
initial temp, heat capacity
of metal, heat (q) absorbed.
Solve q = mST for Tfinal,
plug in data.
Final temperature of the
metal is determined.
q = m S ∆T
rearranging:
q = m S (TFinal – Tinitial)
Tfinal =
q
+Tinitial
m´C
4.184 J
1 cal
+ 17.0 °C = 32.7
J
°C
25.0 g  2.72
Tf > Tin
g  C
as
expected
+255 cal 
Tf =
Example Problem:
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?
• 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:
qFe + qwater = 0
qFe = -qwater
Example Problem:
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?
• 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:
Solution:
CFe =
mFe Sfe ∆T fe = -mw Sw ∆Tw
-mwater ´ Cwater ´ DTwater
mFe ´ DTFe
Chapter 5 — Thermo
5
Energy & Changes of State
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:
J
1K
 (23.1 C  21.0 C) 
J
gK
1 C
 0.469
1K
gK
55.0 g  (23.1 C  99.8 C) 
1 C
225 g  4.184
CFe 
Energy Transfer & 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
H2O(l) + Heat ® H2O(g)
Heating/Cooling Curve for Water
Note the isotherms
at each phase
transition.
• 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 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
H2O(g) ® H2O(l) + Heat
Heat & Changes of State
• 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:
Vaporization:
Melting or Fusion:
Deposition:
Condensation:
Freezing:
subH > 0
vapH > 0
fusH > 0
depH < 0
conH < 0
freH < 0
(endothermic)
(endothermic)
(endothermic)
(exothermic)
(exothermic)
(exothermic)
Where H refers to the “Heat” of a phase change
Chapter 5 — Thermo
6
Energy Change Calculations
Heating & Cooling:
q(heating or cooling) = m  C  T
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
Heat absorbed or lost in a Phase change:
q(phase change) = (phase change)H  n
+2260 J/g
melt the ice  form liquid water at 0 °C  heat the water to 100 °C  boil
water
(n = moles or grams)

fusH

Swater

VapH
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?
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?
Constants Needed:
Equations:
J
Heat of fusion of ice = 333
g
qphase change =
mH
Ice
H2O(s) fusH
(0 °C)
Specific heat of water = 4.18
Heat of vaporization = 2260
J
g×K
melt ice
water
H2O(l)
(100
°C)
vapH
steam
H2O(g)
(100
°C)
qheat = mST
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) fusH
(0 °C)
water
H2O(l) Cwater
(0 °C)
water
H2O(l) Cwater
(0 °C)
water
H2O(l)
(100
°C)
heat water
vapH
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?
steam
H2O(g)
(100
°C)
boil water
qtotal = q1 = nice ´ D fusH + q2 = mwater ´ Cwater ´ DT + q3 = nwater ´ D vapH
Ice
H2O(s) fusH
(0 °C)
water
H2O(l) Cwater
(0 °C)
qtotal  500. g  333
water
H2O(l)
(100
°C)
vapH
J
g
J
 (100.0 C  0.0 C)
g C
J
 500. g  2260  1.51 10 6 J
g
 500. g  4.18
steam
H2O(g)
(100
°C)
Chapter 5 — Thermo
7
First Law of Thermodynamics
First Law of Thermodynamics
The total internal energy of an isolated system is constant.
Any energy lost by the
system is transferred to the
surroundings and vice versa.
The “system” is that which we are
interested in.
The “surroundings” are
everything in contact with the
system.
Together:
System + Surroundings = Universe
Any change in energy is
related to the final and initial
states of the system.
Energy
Energy
∆Usystem = Ufinal – Uinitial
The same holds for the surrounding!
Relating U to Heat and Work
energy transfer in
(endothermic), +q
 Energy cannot be created or destroyed.
energy transfer out
(exothermic), -q
 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:
SYSTEM
∆U = q + w
U  q  w
change in
system
energy
heat
work done by or
= lost or gained +
on the system
by the system
Energy in
U Final
Uf > U i
Work
Uf < U i
w transfer out
(-w)
Energy is the capacity to do work.
Work equals a force applied through a distance.
When you do work, you expend energy.
Energy out
U initial
work out
When you push down on a bike
pump, you do work.
P=
energy
work in
work and
heat can
balance!
w transfer in
(+w)
force
force
=
area
x2
change in volume V = x3
q in
Heat
P  V =
q out
U Final
U initial
Usystem >
0 (+)
Usystem = 0
Usystem <
0 (–)
F
 x3 = F  x
x2
P × V = work
Therefore work is equal to a change
of volume at constant pressure.
Chapter 5 — Thermo
8
Enthalpy
Enthalpy is a “State Function”
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
DH = D (U + PV )
if no “PV” work is
done by the system,
V = 0
= DU + PDV
0
H = Up
the change in
enthalpy is the
change in internal
energy at constant
pressure
No matter which path is taken
(AB) the results are the
same:
Final – Initial
Since individual Enthalpies
cannot be directly measured,
we only deal with enthalpy
changes
(H = Hf – Hi)
For a “system” the overall change in Enthalpy is path
independent.
• 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.
Thermochemical Equations
Just like a regular chemical equation, with an energy term.
rHo = 802 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 802 kJ
Reactants
Enthalpy of reaction = rH = Hproducts  Hreactants
H (Reactants)
rH > 0
rH < 0
H (Reactants)
H (Products)
Exothermic
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) rHo = 802 kJ
g
energy out…
Exothermic
What does this imply? CONVERSION FACTORS!!!
+802 kJ of Energy Released
2 mol H2O (g) produced
H (Products)
Endothermic
From the equation:
+802 kJ of Energy Released
1 mol CH4 (g) consumed
Products
H = Hfinal  Hinitial
Energy
• “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
Energy is a product just like CO2 or H2O!
A
Enthalpies & Chemical Reactions:
rH
Enthalpy Conditions
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
B
128.5g CH4´
 mols  J


molar Reaction
mass enthalpy
1 mol CH4 1 mol rxn +802 kJ
´
´
= 6.43103 kJ
16.04 g 1 mol CH4 1mol rxn
Chapter 5 — Thermo
9
Thermochemical Equations
When the reaction is reversed , the sign of H reverses:
Thermochemical Equations
H scales with the reaction:
1
1
´ [ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)] [rHo =  802 kJ] ´
2
2
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
rHo = 802 kJ
Exothermic
rHo = +802 kJ
Endothermic
Thermochemical Equations
1
1
CH4 (g) + O2 (g) ® CO2 (g) + H2O(g)
2
2
Yes you can write the reaction with fractions, so long as
you are writing it on a mole basis…
Comparing the reaction with water as a gas or
liquid:
CH4(g) + 2O2(g)
Change in enthalpy depends on state:
reactants
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) rH° = 890 kJ
=  88kJ
H2O(g)  H2O(l)
This means that water’s liquid state lies
44 kJ/mol lower than the gas state
The difference in the rxn H
 88kJ is due to the change in state!!
[ H = 44 kJ/mol ]  2 =
Constant Pressure Calorimetry,
Measuring H
• 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.
Enthalpy (H)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rH° = 802 kJ
From your text:
rHo =  401 kJ
r
Ho
= 802 kJ either pathway
gives the same
results!
rHo = 890 kJ
products
CO2(g) + 2H2O(g)
The difference is the
88 kJ released when 2
mols of water go from
gas to liquid.
2  44 kJ
products
CO2(g) + 2H2O(l)
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.
Chapter 5 — Thermo
10
Constant Pressure Calorimetry,
Measuring H
• 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.
Problem:
• 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup 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:
Problem:
• 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup 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.
• 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup 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.
Step 1:
Cals. Moles of
NH4NO3
qsolution
Step 3:
Eq. gives mole ratios
(stoichiometry)
qsolution + qrxn = 0
qsolution
qrxn = q(NH4NO3)
Step 2:
Determine qsolution =
mCT
1 mol NH4NO3
 0.0680 moles NH4NO3
80.04 g
 msolution  C  T
5.44 g NH4NO3 
Step 4:
rH =qrxn/mol NH4NO3
qsolution  154.4 g  4.18
J
 (16.2 C  18.6 C)
g  C
qsolution  1.55  103 J
Enthalpy per mole of
reactant
Problem:
• 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffeecup 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.
qsolution = -4.55 ´ 103 J
qrxn + qsolution = 0
qrxn = -qsolution = +1.55 ´ 103 J
DrH =
qrxn
+1.55 ´ 103 J
1kJ
kJ
=
´
= +22.8
moles of reaction 0.0680 moles NH4NO3 103 J
mol
The sign is positive indicating an endothermic process.
Constant Volume Calorimetry,
Measuring U
Under conditions of
constant volume, any
heat transferred is equal
to a change of internal
energy rU.
qV = rU
Chapter 5 — Thermo
11
Constant Volume Calorimetry,
Measuring U
Heats of combustion
(rUocombustion ) are measured
using a device called a Bomb
Calorimeter.
Calculating Heat in an Exothermic
Reaction
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.
A combustible sample is
reacted with excess O2
The temperature of the water rises to 33.20 °C from 25.00 °C
when the octane is reacted.
The heat capacity of the
bomb is constant.
If the heat capacity of the bomb is 837 J/°C, calculate the heat of
reaction per mole of octane.
The heat of reaction is found
by:
qrxn = -Ccal ´ DT
Since the temperature of the water rose, the reaction must
have been exothermic:
Therefore one can write:
Ccal = Cwater + Cbomb
–qrxn = qwater + qbomb
Calculating Heat in an Exothermic
Reaction
–qrxn = mwaterCwaterTwater + qbombTwater
-qRXN =
Hess’s Law
The overall enthalpy change for a reaction is equal to the sum
of the enthalpy changes for the individual steps in the reaction.
4.184J
´1.20 ´ 103 g   33.20  25.00  C
g C
qwater
Intermediate
Reaction
J
  33.20  25.00  C
837
C
qbomb
rH1 + rH2 =
Reactants
qRXN = – 4803 J or – 48.0 kJ
Heat transferred per mole qV:
- 48.0 kJ
kJ
= -5.48 ´ 103
mol
mol
114.2g
1.00g ´
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.
so we can write…
rH = ?
unknown!
rH
Products What if the enthalpy
changes through
another path are know?
The sum of the H’s in one direction must equal the sum
Why?
in the other direction.
Because enthalpy is a state function… Path independent!
Hess’s Law Problem:
Example:
Given:
(1)
(2)
Determine the rH for the reaction:
3H2(g) + N2(g)  2NH3(g)
rHo = ???
2 H2(g) + N2(g)  N2H4(g)
∆rH°1 = +95.4 kJ
N2H4(g) + H2(g)  2NH3(g)
∆rH°2 = –187.6 kJ
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!
Chapter 5 — Thermo
12
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: Look what happens…
adding equations (1) & (2) yields:
2 H2(g) + N2(g)+ N2H4(g) + H2(g)  N2H4(g) + 2NH3(g)
2 H2(g) + N2(g)+ N2H4(g) + H2(g)  N2H4(g) + 2NH3(g)
/
/
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
intermediates
H2(g) +
N2H4(g)
(Step 1)
Ho = 95.4kJ
(Step 2)
Ho = 187.6kJ
reactants
3H2(g) +
N2(g)
(Overall)
Ho = 92.2kJ
adding equations (1) & (2) yields: Look what happens…
2 H2(g) + N2(g)+ N2H4(g) + H2(g)  N2H4(g) + 2NH3(g)
/
therefore…
/
3H2(g) + N2(g)  2NH3(g)
rH = ∆rH°+
+95.4 kJ + (–187.6 kJ) = –92.2 kJ
1 ∆rH°2 =
Standard Enthalpies of Formation
products
2N3 (g)
• 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).
Enthalpy Values
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).
Since Enthalpy is a state function, enthalpy values
depend on the reaction conditions in terms of the
phases of reactants and products.
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:
H2(g) + ½ O2(g)  H2O(g)
∆rH˚ = -242 kJ
H2(g) + ½ O2(g)  H2O(liq)
∆rH˚ = -286 kJ
fH (element) = 0
fH° O2(g) = 0
elemental form
fH° O(g) ≠ 0
NOT the elemental form
Same reaction, different phases, different enthalpies.
Chapter 5 — Thermo
Formation Reactions
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”.
13
Question:
What is the formation reaction for potassium
permanganate?
elements
K (s)
+ Mn (s)
compound
+ 2 O2(g)
KMnO4 (s)
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)
•
•
•
•
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
standard state conditions = 25oC and 1 atm
Enthalpy Changes for a Reaction:
Using Standard Enthalpy Values
• 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
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.
Consider the Combustion of
Propane
3C(s) + 8H2(g) + 5O2(g)
C3H8(g) + 5O2(g) 
• 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.
Problem:
Calculate the rH° for CaCO3(s) given the following fH°
CaO(s) + CO2(g)  CaCO3(s)
fH° [CaO(s)] = – 635.5 kJ/mol
fH° [CO2(g)] = – 393.5 kJ/mol
fH° [CaCO3(s)] = –1207 kJ/mol
In other words: Energy gained– Energy spent = Net Energy
DrH° = å nD f H° (products ) - å mD f H° (reactants )
n and m are the stoichiometric balancing coefficients.
3CO2(g) + 4 H2O(l)
Chapter 5 — Thermo
14
Problem:
Problem:
Calculate the rH° for CaCO3(s) given the following fH°
Calculate the rH° for CaCO3(s) given the following fH°
CaO(s) + CO2(g)  CaCO3(s)
CaO(s) + CO2(g)  CaCO3(s)
fH° [CaO(s)] = – 635.5 kJ/mol
DrH° = å nD f H° (products )
fH° [CO2(g)] = – 393.5 kJ/mol
fH° [CaCO3(s)] = –1207 kJ/mol
å mD H (reactants )
°
f
fH° [CaO(s)] = – 635.5 kJ/mol
DrH° = å nD f H° (products )
fH° [CO2(g)] = – 393.5 kJ/mol
å mD H (reactants )
°
f
fH° [CaCO3(s)] = –1207 kJ/mol
1mol CaCO3
rH° =
1mol CaO
1mol CO2
–1207 kJ/mol – (–635.5 kJ/mol–393.5 kJ/mol)
rH° = – 178 kJ/mol
Problem:
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
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° [H2O(l)] = –285.7 kJ/mol
Step 1: Write the balanced equation for the reaction…
C10H8(l) + 12 O2(g)  10CO2(g) + 4H2O(l)
Problem:
Next recall that:
DrH° = å nD f H° (products ) - å mD f H° (reactants )
C10H8(l) + 12 O2(g)  10CO2(g) + 4H2O(l)
combustionH° =
10  fH° [CO2(g)]+ 4  fH° [H2O(l)]
– {fH° [C8H10(l)]+ 12  fH° [O2(g)]}
From the problem, all quantities are know but
 f H C10H8
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°
+ 4  fH°
=
[CO2(g)]
[H2O(l)]
–
+ 12  fH°
{combH° [O2(g)]}
elements = 0
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°
+ 4  fH°
=
[CO2(g)]
[H2O(l)]
–
+ 12  fH°
{combH° [O2(g)]}
H° f [C10H8(l)] =
elements = 0
10  (–393.5 kJ/mol
+ 4  (–285.7 kJ/mol)– (– 5156 kJ/mol)
= + 79 kJ/mol
15