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88ChemicalChemical
thermodynamicsthermodynamics
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8.1 Introduction to chemical8.1 Introduction to chemicalthermodynamicsthermodynamics
Chemical thermodynamics allows us
to predict both the direction and the
extent of spontaneous chemical andphysical change under particular
conditions using a property called the
Gibbs free energy, G.
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8.2 Gibbs free energy8.2 Gibbs free energy
Chemical reactions and physical
changes almost always either absorb
or release energy as heat.
Energy may be distributed throughout
a chemical system in a large number
of different ways, some of which have
significantly higher probabilities than
others
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8.2 Gibbs free energy8.2 Gibbs free energy
G =H TS
Henthalpy of the system
Function related to the heat absorbed orevolved by a chemical system
Sentropy of the system
Measure of number of ways energy isdistributed throughout a chemical system
Ttemperature in kelvin
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8.2 Gibbs free energy8.2 Gibbs free energy
G =H TS
G allows us to determine whether a
particular chemical reaction or physicalchange is spontaneous
IfG < 0, the process is spontaneous
If
G > 0, the process isnonspontaneous
IfG = 0, the system is at equilibrium
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System refers to the particular
chemical species being studied
Surroundings are everything else
8.2 Gibbs free energy8.2 Gibbs free energy
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Universe refers to the system and the
surroundings
Boundary defined as region acrosswhich heat flows
8.2 Gibbs free energy8.2 Gibbs free energy
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8.2 Gibbs free energy8.2 Gibbs free energy
Open systems
Can gain or lose mass and energy
across their boundaries Closed systems
Can absorb or release energy, but not
mass, across the boundary
Isolated systems
Can not exchange matter or energy with
their surroundings
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8.2 Gibbs free energy8.2 Gibbs free energy
SI unit of energy
Joule
1 J = 1 kg m2
s-2
1 kJ = 103 J
Thermodynamic equations require the
temperature in kelvin
Temperature difference (T) of 1 K is
numerically equal to Tof 1 C
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8.3 Enthalpy8.3 Enthalpy
Internal energy (U)
The sum of energies for all of the
individual particles in a sample of matter
Interested in the CHANGE in internal
energy that accompanies a process
U = Ufinal Uinitial
U = Uproducts Ureactants
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8.3 Enthalpy8.3 Enthalpy
Internal energy
Chemical system can exchange energy
with its surroundings in two ways
Either absorbing heat from or emitting
heat to the surroundings
Doing work on the surroundings ofhaving the surroundings do work on it
Work may be defined simply as motion
against an opposing force
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8.3 Enthalpy8.3 Enthalpy
Internal energy
Type of work most
often encountered inchemical systems is
the compression or
expansion of gas
Often called pressure-volume orpVwork
w= pV
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8.3 Enthalpy8.3 Enthalpy
First law of thermodynamic
U = q + w
q - heat w- work
Energy can be transferred between
systems as either heat or work
It can never be created or destroyed
Law of conservation of energy
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8.3 Enthalpy8.3 Enthalpy
State functions
Value is dependent only on the current
state of the system
X = Xfinal Xinitial
Independence from the method or
mechanism by which a change occurs is
the important feature of all statefunctions
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8.3 Enthalpy8.3 Enthalpy
State functions
q and ware NOT state functions
The values ofqand ware
dependent on
the path
of change
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8.3 Enthalpy8.3 Enthalpy
Heat Capacity
Heat and temperature are not the same
thing
Heat is a transfer of energy due to a
temperature difference
q =CT
q - heat (J) C- heat capacity (J K-1)
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8.3 Enthalpy8.3 Enthalpy
Heat Capacity
Heat capacity depends on the size of the
sample
A property with a value that depends on
the size of the sample is an extensive
property
A property with a value independent ofthe size of the sample is an intensive
property
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8.3 Enthalpy8.3 Enthalpy
Heat Capacity
Divide heat capacity (extensive property)
by the mass of the sample to form
specific heat capacity (intensive
property)
c - specific heat capacity (J g-1 K-1)
Divide by amount instead of mass to
form molar heat capacity (J mol-1 K-1)
m
Cc !
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8.3 Enthalpy8.3 Enthalpy
Heat Capacity
q = c T
If a gold ring with a mass of 5.50 g changesin temperature from 25.0 to 28.0 C, how
much heat has it absorbed?
m = 5.50 g c= 0.129 J g1 K1 T= 3 K
q = cmT
= (0.129 J g-1 K-1) (5.50 g) (3 K)
= 2.1 J
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The determination of heat
Calorimeter
Apparatus designed to
minimise heat lossbetween the system
and surroundings
Bomb calorimeter
System remains atconstant volume
U=q + w
U=qv
8.3 Enthalpy8.3 Enthalpy
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Enthalpy: the heat of reaction at
constant pressure
U
= q + wU = qp pV
Inconvenient as need
to know V
Define a new thermodynamic
property called enthalpy (H)
H = qp
8.3 Enthalpy8.3 Enthalpy
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8.3 Enthalpy8.3 Enthalpy
Enthalpy: the heat of reaction at
constant pressure
H = U + pV
H = U + pV
Substituting U = qp pVgives
H = qp pV+ pV
H = qp
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8.3 Enthalpy8.3 Enthalpy
Enthalpy: the heat of reaction at
constant pressure
The heat of reaction at constant pressureis equal to H
The heat of reaction at constant volume
is equal to U
H> 0 reaction is endothermic H< 0 reaction is exothermic
The difference between Hand U for a
reaction is pV
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8.3 Enthalpy8.3 Enthalpy
Standard enthalpy change
Standard states
Pressure of 105 Pa
Concentration of 1 M
Standard enthalpy of reaction
Value ofHfor a reaction occurring understandard conditions (H(kJ or kJ mol1)
Involves the actual numbers of MOLES
specified by the coefficients of the equation
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8.3 Enthalpy8.3 Enthalpy
Standard enthalpy change
N2(g) + 3H2(g) 2NH3(g)H=92.38 kJ
The above is a thermochemical equation Always gives the physical states of the
reactants and products
Its value ofH is only true when
coefficients of reactants and productsare numerically equal to the number of
moles of the corresponding substances
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8.3 Enthalpy8.3 Enthalpy
Hesss law
Method for combining known
thermochemical equations in a way that
allows us to calculate Hfor another
reaction
One step
C(s) + O2(g)
CO2(g)
H
= 393.5 kJ Two step
Step 1: C(s) + O2(g) CO(g) H= 110.5 kJ
Step 2: CO(g) + O2(g) CO2(g)H= 283.0 kJ
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Hesss law2Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g) H
= +26.7 kJ
3CO(g) + O2(g) 3CO2(g) H = -849.0 kJ
2Fe(s) + O2(g)
Fe2O3(s) H
= -822.3 kJ Rules for manipulating thermochemical
equations:
1. When an equation is reversed the sign of
H must also be reversed.Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) H
= -26.7 kJ
2Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g) H = +26.7 kJ
8.3 Enthalpy8.3 Enthalpy
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Rules for manipulating thermochemical
equations:
2. Formulae can be cancelled from both
sides of an equation only if the substanceis an identical physical state.
Hesss law2Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g) H
= +26.7 kJ
3CO(g) + O2(g) 3CO2(g) H = -849.0 kJ
2Fe(s) + O2(g) Fe
2O
3(s) H = -822.3 kJ
8.3 Enthalpy8.3 Enthalpy
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Rules for manipulating thermochemical
equations:
3. If all the coefficients of an equation are
multiplied or divided by the same factor,the value ofH must likewise be
multiplied or divided by that factor.
2Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g) H = +26.7 kJ
3CO(g) + O2(g) 3CO2(g) H = -849.0 kJ
2Fe(s) + O2(g) Fe2O3(s) H = -822.3 kJ
CO(g) + O2(g) CO2(g) H = -283.0 kJ
3CO(g) + 3/2O2(g) 3CO2(g) H = -849.0 kJ
Hesss law
8.3 Enthalpy8.3 Enthalpy
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8.3 Enthalpy8.3 Enthalpy
Standard enthalpy of combustion
cH
Enthalpy change at temperature Twhen1 mole of a substance is completely
burned in pure oxygen gas
Combustion reactions are always
exothermiccH
always negative
kJ mol1
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8.3 Enthalpy8.3 Enthalpy
Standard enthalpy of formation (fH)
Enthalpy change when 1 mole of
substance is formed at 105 Pa and the
specified temperature from its elements in
their standard states
An element is in its standard state when it
is in its most stable form and physical stateat 105 Pa and the specified temperature
fH for the elements in their standard
states are 0
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8.3 Enthalpy8.3 Enthalpy
Standard enthalpy of formation
aA + bB cC + dD
Hesss law equation
Use either enthalpies of combustion or enthalpies of
formation
f f f f D A BCH c H d H a H b H
U U U U U ( ! ( ( ( (-
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8.3 Enthalpy8.3 Enthalpy
Bond enthalpies
A bond enthalpy is the enthalpy change on
breaking 1 mole of a particular chemical
bond to give electrically neutral fragments
Atomisation enthalpy (atH) is the enthalpy
change that occurs on rupturing all the
chemical bonds in 1 mole of gaseousmolecules
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Bond enthalpies and Hesss law
8.3 Enthalpy8.3 Enthalpy
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8.3 Enthalpy8.3 Enthalpy
Lattice enthalpies and Hesss law
Lattice enthalpies for ionic solids calculable
using Hesss law and thermodynamic data
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Entropy and probability
Spontaneous processes tend to proceed from states of
low probability to states of higher probability.
Spontaneous processes tend to disperse energy
8.4 Entropy8.4 Entropy
No energy is transferred
One unit of energy is
transferred
Two units of energy
are transferred
Three units of energy
are transferred
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Entropy and entropy change
Entropy (S) describes the number of
equivalent ways that energy can be
distributed in the system
Entropy is a state function
S = Sproducts Sreactants
Any event that is accompanied by an
increase in the entropy of the system has
a tendency to occur spontaneously
8.4 Entropy8.4 Entropy
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Entropy and entropy change
8.4 Entropy8.4 Entropy
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Factors that affect entropy
Often possible to predict whetherSis positive
or negative for a particular change
Volume
For gases the entropy increases with increasing
volume
8.4 Entropy8.4 Entropy
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Factors that affect entropy
Temperature
The higher the temperature the higher the entropy
8.4 Entropy8.4 Entropy
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Factors that affect entropy
Physical state
One of the major factors that affects the entropy of a
system is its physical state
8.4 Entropy8.4 Entropy
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Factors that affect entropy
Number of particles
When all other things are equal, reactions that
increase the number of particles in the system tendto have a positive entropy change
8.4 Entropy8.4 Entropy
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8.4 Entropy8.4 Entropy
The second law of thermodynamics
Whenever a spontaneous event takes
place in our universe, the total entropy of
the universe increases (Stotal>0)
Stotal= Ssystem + Ssurroundings
qsurroundings = -qsystem
qsystem =H
suroundingssurroundings
qS
T( !
systemsurr undin s
HS
T
(( !
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Absolute entropy and the third law of
thermodynamics
At absolute zero, the entropy of a perfectly
ordered pure crystalline substance is 0
S= 0 at T= 0 K
Point at which entropy equal to 0 is known,hence by experimental measurement and
calculation entropy can be determined for
a substance at temperatures above 0 K
8.4 Entropy8.4 Entropy
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8.4 Entropy8.4 Entropy
Absolute entropy and the third law of
thermodynamics
Standard entropy (S) entropy of
1 mole of a substance determined under
standard conditions at temperature
of 298 K
Standard entropy of reactionaA + bB cC + dD
f f f f D A BCS c S d S a S b S U U U U U ( ! ( ( ( (-
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The sign ofG
G = H TS
G < 0 reaction
spontaneous
8.5 Gibbs free energy and8.5 Gibbs free energy andreaction spontaneityreaction spontaneity
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8.5 Gibbs free energy and8.5 Gibbs free energy andreaction spontaneityreaction spontaneity
Standard Gibbs free energy change
When G is determined at 105 Pa, this is
called the standard free energy change
(G)
There are several ways for determining
G for a reaction
G
=
H
T
S
aA + bB cC + dD
f f f f D A BCG c G d G a G b GU U U U U ( ! ( ( ( (-
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Gibbs free energy and work
Maximum conversion of chemical energy to work
occurs if a reaction is carried out under conditions
that are said to be thermodynamically reversible
8.5 Gibbs free energy and8.5 Gibbs free energy andreaction spontaneityreaction spontaneity
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Gibbs free energy and work
The maximum amount of energy produced
by a reaction that can be
THEORETICALLY harnessed as work is
equal to G
G = 0 the system is in a state of
equilibrium
Gproducts =Greactants
When G = 0 the amount of work
available is 0
8.5 Gibbs free energy and8.5 Gibbs free energy andreaction spontaneityreaction spontaneity
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8.5 Gibbs free energy and8.5 Gibbs free energy andreaction spontaneityreaction spontaneity
Gibbs free energy and work
For phase changes such as
H2
O(l) H2
O(s)
equilibrium can be established only at
one particular temperature at
atmospheric pressure
G = 0 =H TSH = TS
HS
T
(( !
HT
S
(!
(
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8.5 Gibbs free energy and8.5 Gibbs free energy andreaction spontaneityreaction spontaneity
Gibbs free
energy and
work
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