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Chapter 17 - Thermochemistry
-is the study of energy relationships
in chemical and physical reactions

almost nothing happens in chemistry
unless there is an energy advantage; all
reactions are an attempt to arrive at a
lower system energy.
a) 1st law of Thermodynamics: Law of
conservation of energy.
 Energy is never lost or gained , it only
changes form.
This law states that energy can change form
but is never really lost in any closed system.
b) 2nd law of Thermodynamics:
 Heat Energy always travels spontaneously
from a warmer body (body with higher
temperature) to a colder body.
-droping a hot rock in water
-frostbite
 OBJECTIVES:
• Explain the relationship between
energy and heat.
• Distinguish between heat capacity
and specific heat.
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temperature is simply an indirect
measure of the average kinetic energy
of the particles in a sample of matter.
The scale of measurement is arbitrary:

i)
Celsius scale sets the zero
value at the freezing point of water at
101.325 kPa and 100 at the boiling point
of water at the same pressure.

ii) Kelvin scale sets the zero
value at absolute zero.
 The degrees are the same size as
celsius degrees.

Energy and Heat
 Thermochemistry
- concerned with
heat changes that occur during
chemical reactions
 Energy - capacity for doing work or
supplying heat
• weightless, odorless, tasteless
• if within the chemical substancescalled chemical potential energy
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- represented by “q”, is energy
that transfers from one object to
another, because of a temperature
difference between them.
• only changes can be detected!
• flows from warmer  cooler object
 Heat
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Exothermic and Endothermic
Processes
 Essentially
all chemical reactions,
and changes in physical state,
involve either:
• release of heat, (exothermic) or
• absorption of heat (endothermic)
 In
studying heat changes, think of
defining these two parts:
• the system - the part of the
universe on which you focus your
attention
• the surroundings - includes
everything else in the universe
 Together,
the system and it’s
surroundings constitute the
universe
 Thermochemistry is concerned with
the flow of heat from the system to
it’s surroundings, and vice-versa.
 The
Law of Conservation of Energy
states that in any chemical or
physical process, energy is neither
created nor destroyed.
• All the energy is accounted for as
work, stored energy, or heat.
flowing into a system from it’s
surroundings:
• defined as positive
• q has a positive value
• called endothermic
–system gains heat as the
surroundings cool down
 heat
flowing out of a system into it’s
surroundings:
• defined as negative
• q has a negative value
• called exothermic
–system loses heat as the
surroundings heat up
 heat
 Every
reaction has an energy
change associated with it
 Exothermic reactions release energy,
usually in the form of heat.
 Endothermic reactions absorb
energy
 Energy is stored in bonds between
atoms
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Heat Capacity and Specific Heat
 A calorie
is defined as the quantity of
heat needed to raise the temperature
of 1 g of pure water 1 oC.
• Used except when referring to food
• a Calorie, written with a capital C,
always refers to the energy in food
• 1 Calorie = 1 kilocalorie = 1000 cal.
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Heat Capacity and Specific Heat
 The
calorie is also related to the
joule, the SI unit of heat and energy
• named after James Prescott Joule
• 4.184 J = 1 cal
 Heat Capacity - the amount of heat
needed to increase the temperature
of an object exactly 1 oC
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Heat Capacity and Specific Heat
 Specific
Heat Capacity - the
amount of heat it takes to raise the
temperature of 1 gram of the
substance by 1 oC often called
simply “Specific Heat”

mass; measured in g or kg. The greater
the mass, the more heat required to
change the temperature.
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water, C = 4.18 J/(g oC),
 Thus, for water:
• it takes a long time to heat up, and
• it takes a long time to cool off!
 Water is used as a coolant!
 For
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Heat Capacity and Specific Heat
 To
calculate, use the formula:
 q = m C T
 heat abbreviated as “q”
 m = mass
 T = change in temperature (Tf - Ti)
 C = Specific Heat
 Units are either J/(g oC) or kJ/(kg oC)
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Table of Specific Heats
• Ex: Determine the heat required to
raise the temperature of 100g of water
from 298.0 K to 373.0 K .
Q = m c ΔT
Q = 100g (4.184 J/g K)( 373.0 K –
298.0 K)
Q = 418.4 J/K (75 K)
Q = 31350 J
Q = 31.4 kJ
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Calorimetry
 Calorimetry
- the accurate and
precise measurement of heat
change for chemical and physical
processes.
 The device used to measure the
absorption or release of heat in
chemical or physical processes is
called a Calorimeter
Calorimetry
 Foam
cups are excellent heat
insulators, and are commonly used
as simple calorimeters
A Cheap
Calorimeter
 For
systems at constant pressure,
the heat content is the same as a
property called Enthalpy (H) of the
system
Calorimetry
in enthalpy = H
 q = H These terms will be used
interchangeably in this textbook
 Thus, q = H = m x C x T
 H is negative for an exothermic
reaction
 H is positive for an endothermic
reaction
 Changes
Calorimetry
 Calorimetry
experiments can be
performed at a constant volume
using a device called a “bomb
calorimeter” - a closed system
Energy
C + O2  CO2+ 395 kJ
C + O2
395kJ
C O2
Reactants

Products
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In terms of bonds
C
O
O
O
C
O
Breaking this bond will require energy.
O
C
O C O
O
Making these bonds gives you energy.
In this case making the bonds gives you
more energy than breaking them.
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Exothermic
 The
products are lower in energy
than the reactants
 Releases energy
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Energy
CaCO
 CaO
CaCO
CaO
+ CO+2 CO2
3 + 176
3 kJ
CaO + CO2
176 kJ
CaCO3
Reactants

Products
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Endothermic
 The
products are higher in energy
than the reactants
 Absorbs energy
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Chemistry Happens in
MOLES
An equation that includes energy is
called a thermochemical equation
 CH4 + 2O2  CO2 + 2H2O + 802.2 kJ
 1 mole of CH4 releases 802.2 kJ of
energy.
 When you make 802.2 kJ you also
make 2 moles of water

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What is the molar enthalpy of CO2 (g) in the
reaction for the burning of butane below?
2 C4H10 +13 O2  8 CO2 +10 H2O
∆H=-5315 kJ
Answer:
Molar enthalpy is the enthalpy change in
equation divided by the balance of CO2
 Molar enthalpy, H substance = 5315 kJ ÷ 8
mol

= 664 kJ / mol.

Thermochemical Equations
 A heat
of reaction is the heat
change for the equation, exactly
as written
• The physical state of reactants
and products must also be given.
• Standard conditions for the
reaction is 101.3 kPa (1 atm.)
and 25 oC
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CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ

If 10. 3 grams of CH4 are burned
completely, how much heat will be
produced?
10. 3 g CH4
1 mol CH4
16.05 g CH4
802.2 kJ
1 mol CH4
= 514 kJ
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CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ
 How
many liters of O2 at STP would
be required to produce 23 kJ of
heat?
 How many grams of water would be
produced with 506 kJ of heat?
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
How much heat will be released if 65 grams of butane
is burned in a lighter according the equation:
2 C4H10 +13 O2  8 CO2 +10 H2O
∆H=-5315 kJ
 1molC4 H10  5315kJ 

65 gC4 H10 

 58.14 g  2molC4 H10 
= 2976.4 kJ
= 3.0 MJ
Calculate the heat released when 120 grams
of Iron (III) oxide is formed by the following
equation
2 Fe2O3 (s) → 4 Fe(s)+3 O2 (g)
∆H=1625 kJ

 1molFe2O3  1625kJ 
120 gFe2O3 


 159.70 g  2mol 
= 610.5 kJ
= 610 kJ

Q = n ∆H (substance)
Where n = # of moles
What mass of carbon dioxide must form
to create 1200 kJ of heat when the
following reaction occurs?
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
∆H=- 2808kJ
Answer: 110 grams

3) What mass of oxygen is needed to
completely react and release 550 kJ of
heat in the following reaction?
 4Fe (s)+3O2 (g) → 2 Fe2O3 (s)

∆H=- 1625 kJ
Answer: 32 grams
Summary, so far...
Enthalpy
 The
heat content a substance has at a
given temperature and pressure
 Can’t be measured directly because
there is no set starting point
 The reactants start with a heat content
 The products end up with a heat
content
 So we can measure how much
enthalpy changes
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Enthalpy
 Symbol
is H
 Change in enthalpy is H (delta H)
 If heat is released, the heat content of
the products is lower
H is negative (exothermic)
 If heat is absorbed, the heat content
of the products is higher
H is positive (endothermic)
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Energy
Change is down
H is <0
Reactants

Products
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Energy
Change is up
H is > 0
Reactants

Products
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Heat of Reaction
The heat that is released or absorbed in a
chemical reaction
 Equivalent to H
 C + O2(g)  CO2(g) + 393.5 kJ
 C + O2(g)  CO2(g)
H = -393.5 kJ
 In thermochemical equation, it is important
to indicate the physical state
 H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ
 H2(g) + 1/2O2 (g) H2O(l) H = -285.8 kJ

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Heat of Combustion
 The
heat from the reaction that
completely burns 1 mole of a
substance
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 OBJECTIVES:
• Classify, by type, the heat changes
that occur during melting, freezing,
boiling, and condensing.
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 OBJECTIVES:
• Calculate heat changes that occur
during melting, freezing, boiling,
and condensing.
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Heats of Fusion and
Solidification
 Molar
Heat of Fusion (Hfus) - the
heat absorbed by one mole of a
substance in melting from a solid to
a liquid
 Molar Heat of Solidification (Hsolid)
- heat lost when one mole of liquid
solidifies
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Heats of Fusion and
Solidification
 Heat
absorbed by a melting solid is
equal to heat lost when a liquid
solidifies
• Thus, Hfus = -Hsolid
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Heats of Vaporization and
Condensation
 When
liquids absorb heat at their
boiling points, they become vapors.
 Molar Heat of Vaporization (Hvap) the amount of heat necessary to
vaporize one mole of a given liquid.
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Heats of Vaporization and
Condensation
 Condensation
is the opposite of
vaporization.
 Molar Heat of Condensation (Hcond)
- amount of heat released when one
mole of vapor condenses
 Hvap = - Hcond
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Heats of Vaporization and
Condensation
The large values for Hvap and Hcond
are the reason hot vapors such as
steam is very dangerous
• You can receive a scalding burn
from steam when the heat of
condensation is released!
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Heats of Vaporization and
Condensation
 H20(g)
 H20(l)
Hcond = - 40.7kJ/mol
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Heat of Solution
 Heat
changes can also occur when
a solute dissolves in a solvent.
 Molar Heat of Solution (Hsoln) heat change caused by dissolution
of one mole of substance
 Sodium hydroxide provides a good
example of an exothermic molar
heat of solution:
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Heat of Solution
NaOH(s)
H2O(l)

Na1+(aq) + OH1-(aq)
Hsoln = - 445.1 kJ/mol
 The heat is released as the ions
separate and interact with water,
releasing 445.1 kJ of heat as Hsoln
thus becoming so hot it steams
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 OBJECTIVES:
• Apply Hess’s law of heat
summation to find heat changes
for chemical and physical
processes.
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 OBJECTIVES:
• Calculate heat changes using
standard heats of formation.
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Hess’s Law
 If
you add two or more
thermochemical equations to give a
final equation, then you can also
add the heats of reaction to give
the final heat of reaction.
Called Hess’s law of heat summation
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Why Does It Work?
If you turn an equation around, you change
the sign:
 If H2(g) + 1/2 O2(g) H2O(g) H=-285.5 kJ
 then,
H2O(g) H2(g) + 1/2 O2(g) H =+285.5 kJ
 also,
 If you multiply the equation by a number,
you multiply the heat by that number:
 2 H2O(g) H2(g) + O2(g) H =+571.0 kJ

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You make the products, so you need
their heats of formation
 You “unmake” the products so you have
to subtract their heats.
 How do you get good at this?

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Standard Heats of Formation
The H for a reaction that produces 1
mol of a compound from its elements at
standard conditions
 Standard conditions: 25°C and 1 atm.
0
 Symbol is H
f

The
standard heat of formation of an
element = 0
This
includes the diatomics
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What good are they?
Table 11.6, page 316 has standard
heats of formation
 The heat of a reaction can be calculated
by:
• subtracting the heats of formation of
the reactants from the products

Ho = (H 0f Products) - (H 0f Reactants)
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Examples

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
0
H f CH4 (g) = - 74.86 kJ/mol
0
H f O2(g) = 0 kJ/mol
0
H f CO2(g) = - 393.5 kJ/mol
0
H f H2O(g) = - 241.8 kJ/mol

H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]
H= - 802.4 kJ
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Examples
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