Transcript Chapter 6

Chapter 5
Energy
Thermodynamics
(rev. 0910)
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Definition

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Thermodynamicsis the study of energy transformations.
Chemical Reactions
Chemical reactions involve not just the
conversion of reactants into products,
but also involve an energy change in
the form of heat—heat released as the
result of a reaction, or heat absorbed as
a reaction proceeds.
 Energy changes accompany all chemical
reactions and are due to rearranging of
chemical bonding.
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Making Bonds
Addition of energy is always a
requirement for the breaking of bonds
but the breaking of bonds in and of
itself does not release energy.
 Energy release occurs when new bonds
are formed.
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Bond Energy
If more energy is released when new
bonds form than was required to break
existing bonds, then the difference will
result in an overall release of energy.
 If, on the other hand, more energy is
required to break existing bonds than is
released when new bonds form, the
difference will result overall in energy
being absorbed.
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Overall Reaction Energy
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Whether or not an overall reaction
releases or requires energy depends
upon the final balance between the
breaking and forming of chemical
bonds.
Energy is...
the ability to do work or produce heat.
 conserved.
 made of heat and work.
 a state function. (Energy is a property that is

determined by specifying the condition or “state” (e.g.,
temperature, pressure, etc.) of a system or substance.)
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independent of the path, or how you get
from point A to B.
Energy
While the total internal energy of a
system (E) cannot be determined,
changes in internal energy (E) can be
determined.
 The change in internal energy will be
the amount of energy exchanged
between a system and its surroundings
during a physical or chemical change.

Δ E = E final - E initial
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Definitions

Work is a force acting over a distance.

Heat is energy transferred between
objects because of temperature
difference. (Heat is not a property of a system or
substance and is not a state function. Heat is a
process—the transfer of energy from a warm to a
cold object.)
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System vs Surroundings

The Universe is divided into two
halves.
• system and the surroundings.
In a chemistry setting, a system
includes all substances undergoing a
physical or chemical change.
 The surroundings would include
everything else that is not part of the
system.
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Heat
Most commonly, energy is exchanged
between a system and its surroundings
in the form of heat.
 Heat will be transferred between objects
at different temperatures.
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Thermochemistry is the study of thermal
energy changes.
Exo vs Endo
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Exothermic reactions release energy to
the surroundings.

Endothermic reactions absorb energy
from the surroundings.
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
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CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
Potential energy
2NO
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Heat
N2 + O2
Three Parts
Every energy measurement has three
parts:
1. A unit ( Joules of calories).
2. A number.
3. a sign to tell direction.
negative - exothermic
positive- endothermic
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Surroundings
System
Energy
DE <0
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Surroundings
System
Energy
DE >0
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Same rules for heat and work
Heat given off is “negative”.
 Heat absorbed is “positive”.

Work done by the system on the
surroundings is negative.
 Work done on the system by the
surroundings is positive.
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First Law of Thermodynamics
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The energy of the universe is constant.
It is also called the:
Law of conservation of energy.
q = heat
w = work
In a chemical system, the energy exchanged
between a system and its surroundings can be
accounted for by heat (q) and work (w).
DE = q + w
Take the system’s point of view to decide signs.
Conservation of Energy
Energy exchanged between a system
and its surroundings can be considered
to off set one another.
 The same amount of energy leaving a
system will enter the surroundings (or
vice versa), so the total amount of
energy remains constant.
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Metric Units
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The SI (Metric System) unit for all forms
of energy is the joule (J).
Heat and Work
DE = q + w
 - q is exothermic
-q = -∆H
 +q is endothermic
 -w is done “by” the system
 +w is done “on” the system
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Note:
 ∆H stands for enthalpy which is the
heat of reaction
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Practice Problem
A gas absorbs 28.5 J of heat and then
performs 15.2 J of work. The change in
internal
 energy of the gas is:
(a) 13.3 J
(b) - 13.3 J
(c) 43.7 J
(d) - 43.7 J
(e) none of the above
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Answer
(b) E = q + w
 28.5 J - 15.2 J = + 13.3 J
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Practice Problem
Which of the following statements correctly
describes the signs of q and w for the
following exothermic process at 1 atmosphere
pressure and 370 Kelvin?
H2O(g) → H2O(l)
(a) q and w are both negative
(b) q is positive and w is negative
(c) q is negative and w is positive
(d) q and w are both positive
(e) q and w are both zero
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Answer
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(c). An exothermic indicates q is
negative and the gas is condensing to a
liquid so it is exerting less pressure on
its surroundings indicating w is
positive.
What is work?
Work is a force acting over a distance.
w= F x Dd
P = F/ area
d = V/area
w= (P x area) x D (V/area)= PDV
 Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
 Use units of liter•atm or L•atm
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Pressure and Volume Work
Work refers to a force that moves an
object over a distance.
 Only pressure/volume work (i.e., the
expansion/contraction of a gas) is of
significance in chemical systems and
only when there is an increase or
decrease in the amount of gas present.
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Work needs a sign
If the volume of a gas increases, the
system has done work on the
surroundings.
 work is negative
 w = - PDV
 Expanding work is negative.
 Contracting, surroundings do work on
the system W is positive.
 1 L•atm = 101.3 J
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Example
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When, in a chemical reaction, there are
more moles of product gas compared to
reactant gas, the system can be thought
of as performing work on its
surroundings (making w < 0) because it
is “pushing back,” or moving back the
atmosphere to make room for the
expanding gas.
When the reverse is true, w > 0.
Compressing and Expanding Gases
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Compressing gas
– Work on the system is positive
– Work is going into the system
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Expanding gas
– Work on the surroundings is negative
– Work is leaving the system
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Clarification Info
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
If the reaction is performed in a rigid
container, there may be a change in pressure,
but if there is no change in volume, the
atmosphere outside the container didn’t
“move” and without movement, no work is
done by or on the system.

If there is no change in volume (V= 0), then
no work is done by or on the system (w= 0)
and the change in internal energy will be
entirely be due to the heat involved ( ΔE = q).
Examples
What amount of work is done when 15 L
of gas is expanded to 25 L at 2.4 atm
pressure?
 If 2.36 J of heat are absorbed by the gas
above. what is the change in energy?
 How much heat would it take to change
the gas without changing the internal
energy of the gas?
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Enthalpy
The symbol for Enthalpy is H
 H = E + PV (that’s the definition)
 at constant pressure.
 DH = DE + PDV
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the heat at constant pressure qp can be
calculated from
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DE = qp + w = qp - PDV

qp = DE + P DV = DH
DH = DE + PDV

Using DH = DE + PDV
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the heat at constant pressure qp can be
calculated from:
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
DE = qp + w

DE = qp – PDV (now rearrange)
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qp = DE + P DV = DH
(if w = - PDV then…)
Examples of Enthapy Changes

KOH(s) → K(aq) + OH-1 (aq)
ΔHsolution = - 57.8 kJ mol1
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C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔHcombustion = -2221kJ/mol
H2O(s) → H2O(l)
ΔHfusion = 6.0 kJ/mol
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Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s)
ΔHreaction - 852 kJ/mol
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Ca(s) + O2(g) H2(g) → Ca(OH)2(s)
ΔHformation - 986 kJ/mol1
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3 Methods
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1.
2.
3.
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There are a variety of methods for
calculating overall enthalpy changes
that you should be familiar with.
The three most common are the:
the use of Heats of Formation
Hess’s Law
the use of Bond Energies
Heat of Reaction
To compare heats of reaction for
different reactions, it is necessary to
know the temperatures at which heats
of reaction are measured and the
physical states of the reactants and
products.
 Look in the Appendix of the textbook to
find Standard Enthapy tables.
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Standard Enthalpy of Formation
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Measurements have been made and tables
constructed of Standard Enthalpies of Formation
with reactants in their “standard states”.
Use the symbol DHºf
Standard state is the most stable physical state of
reactants at:
1 atmosphere pressure
specified temperature—usually 25 °C
1 M solutions
For solids which exist in more than one allotropic
form, a specific allotrope must be specified.
ΔH°formation
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It is important to recognize that the
ΔH°formation (abbreviated as ΔH°f) is really
just the heat of reaction for a chemical
change involving the formation of a
compound from its elements in
their standard states.
Standard Enthalpies of Formation

The standard heat of formation is the
amount of heat needed to form 1 mole of
a compound from its elements in their
standard states.
See the table in the Appendix
 Remember: For an element the value is 0
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Equation Practice

You need to be able to write the
equation correctly before solving the
problem.
Try…
 What is the equation for the formation
of NO2 ? Try writing the equation.
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Practice Answer
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
½N2 (g) + O2 (g)  NO2 (g)

You must make one mole to meet the
definition.
Since we can manipulate the equations
We can use heats of formation to figure
out the heat of reaction.
 Lets do it with this equation.
 C2H5OH +3O2(g)  2CO2 + 3H2O
 which leads us to this rule.
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( DH of products) - ( DH of reactants) = DH o
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Hess’s Law
Definition:
 When a reaction may be expressed as
the algebraic sum of other reactions, the
enthalpy change of the reaction is the
algebraic sum of the enthalpy changes
for the combined reactions.
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Hess’ Law
Enthalpy is a state function.
 It is independent of the path.
 We can add equations to come up with
the desired final product, and add the
DH values.
 Two rules to remember:
 If the reaction is reversed the sign of DH
is changed
 If the reaction is multiplied, so is DH
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Enthalpy
As enthalpy is an extensive property, the
magnitude of an enthalpy change for a
chemical reaction depends upon the
quantity of material that reacts.
 This means:
if the amount of reacting material in
an exothermic reaction is doubled,
twice the quantity of heat energy
will be released.
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For the oxidation of sulfur dioxide gas

SO2(g) + ½O2(g) → SO3(g)
ΔH° = - 99 kJ/mol
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Doubling the reaction results in:
2SO2(g) + O2(g) → 2SO3(g)
ΔH° = - 198 kJ/mol
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Notice that if you double the reaction, you must
double the ΔH° value.
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Sign Change of ΔH
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
2SO2(g) + O2(g) → 2SO3(g)
ΔH° = - 198 kJ/mol
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If the reaction is written as an endothermic
reaction:
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2SO3(g) → 2SO2(g) + O2(g)
ΔH° = + 198 kJ/mol
Tips for Hess’s Law Problems
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It is always a good idea to begin by looking
for species that appear as reactants and
products in the overall reaction.
This will provide a clue as to whether a
reaction needs to be reversed or not.
Second, consider the coefficients of species
that appear in the overall reaction.
This will help determine whether a reaction
needs to be multiplied before the overall
summation.
Example
C(s) + O2(g) → CO2(g)
ΔH°f = - 394 kJ/mol
2H2(g) + O2(g) → 2H2O(l)
ΔH°f = - 572 kJ/mol
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol
----------------------------------------------------C(s) + 2H2(g) → CH4(g)
ΔH°f = - 75 kJ/mol
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Hess’s Law Example
Given
O 2 (g) + H 2 (g)  2OH(g) DHº= +77.9kJ
O 2 (g)  2O(g) DHº= +495 kJ
H 2 (g)  2H(g) DHº= +435.9kJ
Calculate DHº for this reaction
O(g) + H(g)  OH(g)
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Example
Given
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C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l)
2
DHº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
DHº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
2
DHº= -286 kJ
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calculate DHº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
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Problems to Try
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Try 12-3 Practice Problems
H (kJ)
O2 NO2
-112 kJ
180 kJ
N2 2O2
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NO2
68 kJ
Calorimetry
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Calorimetry is the study of the heat released
or absorbed during physical and chemical
reactions.
For a certain object, the amount of heat
energy lost or gained is proportional to
the temperature change. The initial
temperature and the final temperature in the
calorimeter
are measured and the temperature difference
is used to calculate the heat of reaction.
Equipment: Calorimeter
 There
are two kinds of
calorimeters:
– constant pressure
– bomb
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Calorimetry
 Constant
pressure calorimeter
(called a coffee cup calorimeter)
A coffee cup calorimeter measures DH.
 The calorimeter can be an insulated
cup, full of water.
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Definitions:
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
Heat capacity is the amount of energy
required to raise the temperature of an object
1 kelvin or 1 °C.

Specific heat capacity is the heat capacity of
1 gram of a substance.

Molar heat capacity is the heat capacity of
1 mole of a substance.
Heat Capacity
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Heat capacity is an extensive property, meaning
it depends on the amount present—a large
amount of a substance would require more
heat to raise the temperature 1 K than a small
amount of the same substance.
Specific Heat Capacity
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Definition:
The specific heat capacity of each substance is
an intensive property which relates the heat
capacity to the mass of the substance.
Specific Heat Capacity “c”

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
This is the main equation for calorimetry
calculations

mass will be in grams
Units for “c” are J/g K or J/g °C
The specific heat of water is 1 cal/g ºC
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q = mass x c x DT
Or written as
q = mcDT
Molar Heat Capacity

Molar Heat Capacity is the heat capacity of 1
mole of a substance.

molar heat capacity = c/moles

heat = molar heat x moles x DT

Remember that heat is shown as ∆H
Make the units work and you’ve done the
problem right.
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Sign of q
• If a process results in the sample losing heat energy,
the loss in heat is designated as q is negative.
The temperature of the surroundings will increase
during this exothermic process.
• If the sample gains heat during the process, then q is
positive. The temperature of the
surroundings will decrease during an endothermic
process.
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The amount of heat that an object gains or loses is
directly proportional to the change in
temperature.
Examples
The specific heat of graphite is 0.71
J/gºC.
 Calculate the energy needed to raise the
temperature of 75 kg of graphite from
294 K to 348 K.
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Extra Problem
A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The
final temperature of both the water and
the copper is 21.8ºC.
 What is the specific heat of copper?
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Calorimetry
Constant volume calorimeter is called a
bomb calorimeter.
 Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
 The heat capacity of the calorimeter is
known and tested.
 Since DV = 0, PDV = 0, DE = q
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Bomb Calorimeter
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
thermometer

stirrer

full of water

ignition wire

Steel bomb

sample
Properties
intensive properties are not related to
the amount of substance.
 Examples: density, specific heat,
temperature.

Extensive property - does depend on
the amount of substance.
 Examples: Heat capacity, mass, heat
from a reaction.
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Collegeboard. (2007-2008). Professional Development workshop materials:
Special focus thermochemistry.
http://apcentral.collegeboard.com/apc/public/repository/58863_Chemistry_pp.ii-88.pdf