Post on 03-Jan-2016
ChemicalThermodynamics
Chapter 17 Chemical Thermodynamics
ChemicalThermodynamics
Spontaneous Processes
Entropy
Second Law of Thermodynamics
Third Law of Thermodynamics
Gibbs Free Energy
Predicting Spontaneity
Standard Enthalpies of Formation
Gibbs Free Energies of Formation
Free Energy Changes
Contents
ChemicalThermodynamics
• Thermodynamics is
the study of energy
relationships that
involve heat,
mechanical work, and
other aspects of
energy and heat
transfer.
ChemicalThermodynamics
First Law of Thermodynamics• You will recall from Chapter 6 that energy
cannot be created nor destroyed.
• Therefore, the total energy of the universe is a constant.
• Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.
ChemicalThermodynamics
First law of thermodynamics, the law of
conservation of energy, dictates the relationship
between heat (q), work (w), and changes in internal
energy ( ΔU ).
ΔU = q + w
Notice: assigning the correct signs to the quantities of heat and work.
ChemicalThermodynamics
Spontaneous Processes• Spontaneous processes
are those that can
proceed without any
outside intervention.
• The gas in vessel B will
spontaneously effuse into
vessel A, but once the
gas is in both vessels, it
will not spontaneously
separate from each other.
ChemicalThermodynamics
Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
ChemicalThermodynamics
Spontaneous Processes• Processes that are spontaneous at one
temperature may be nonspontaneous at other temperatures.
• Above 0C it is spontaneous for ice to melt.• Below 0C the reverse process is spontaneous.
ChemicalThermodynamics
Goal of chemical thermodynamics:
predicting which changes will be
spontaneous.
ChemicalThermodynamics
Reversible Processes
In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.
ChemicalThermodynamics
Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
• Spontaneous processes are irreversible.
ChemicalThermodynamics
Entropy
• Entropy (S) is a term coined by Rudolph Clausius in
the 19th century.
• Clausius was convinced of the significance of the
ratio of heat delivered and the temperature at which
it is delivered, qT
ChemicalThermodynamics
Entropy
• Entropy can be thought of as a measure of the
randomness of a system.
• It is related to the various modes of motion in
molecules.
• The entropy of a system in a given state is a
measure of the number of different microscopic
states that correspond to a given macroscopic
state.
ChemicalThermodynamics
The more the particles and their positions, the more disordered the system is.
ChemicalThermodynamics
Entropy on the Molecular Scale
• The number of microstates and, therefore, the entropy tends to increase with increases inTemperature.Volume.The number of independently moving
molecules.
ChemicalThermodynamics
Entropy
• Like total energy, U, and enthalpy, H,
entropy is a state function.
• Therefore,
S = Sfinal Sinitial
ChemicalThermodynamics
Entropy
• For a process occurring at constant temperature (an
isothermal process), the change in entropy is equal
to the heat that would be transferred if the process
were reversible divided by the temperature:
S =qrev
T
ChemicalThermodynamics
Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
• Therefore,
S(g) > S(l) > S(s)
ChemicalThermodynamics
Second Law of Thermodynamics
The second law of thermodynamics states
that the entropy of the universe increases for
spontaneous processes, and the entropy of
the universe does not change for reversible
processes.
ChemicalThermodynamics
S and Isolated Systems
• For an equilibrium process in an isolated system, S = 0
• For a spontaneous process in an isolated system, S > 0
ChemicalThermodynamics
Second Law of Thermodynamics
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
ChemicalThermodynamics
Second Law of Thermodynamics
These last truths mean that as a result
of all spontaneous processes the
entropy of the universe increases.
ChemicalThermodynamics
Entropy Changes
• In general, entropy increases whenGases are formed from
liquids and solids.Liquids or solutions are
formed from solids.The number of gas
molecules increases.The number of moles
increases.
ChemicalThermodynamics
Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
ChemicalThermodynamics
Entropy and Disorder
• The entropy S of a system is a measure of the disorder or randomness in the system.
• Disorder can be defined as the number of equivalent ways of distributing the conserved matter and energy through the system.
• Example: a substance has higher entropy in the gaseous state than in the solid state.
ChemicalThermodynamics
Entropy Increasing Processes
• The entropy is expected to increase for
processes in which
1. liquids or solutions are formed from solids.
2. gases are formed from either liquids or solids.
3. the number of molecules of gas increases in
going from reactants to products.
4. The number of degrees of freedom increases.
ChemicalThermodynamics
Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
ChemicalThermodynamics
Third Law of Thermodynamics
• The entropy of all pure crystalline substances approach zero as the temperature approaches absolute zero – since all disorder has been removed.
as T 0 K, S0 0
• This defines the absolute entropy scale S0.
ChemicalThermodynamics
Standard Entropies
• The standard entropy of a substance S0 is the
entropy change required to heat 1 mole of the
substance from 0 K to the temperature of 298 K.
• Standard molar entropies are used to calculate
S for reactions, just as Hf values are used to
calculate H for reactions.
• Note that S0 for an element in its standard state
is not zero – unlike the case for Hf
ChemicalThermodynamics
Standard Entropies
• These are molar entropy
values of substances in
their standard states.
• Standard entropies tend
to increase with
increasing molar mass.
ChemicalThermodynamics
Standard Entropies
Larger and more complex molecules have greater entropies.
ChemicalThermodynamics
Entropy Changes
Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:
Srxn° = npS°(products) - nrS°(reactants)
where n and m are the coefficients in the balanced chemical equation.
ChemicalThermodynamics
Sample problem to calculate Srxn°:
Use standard entropies to calculate Srxn°for the
reaction: 2SO2(g)+O2 (g)= 2SO3(g)
Solution:
Equation: 2SO2(g)+O2 (g)= 2SO3(g)
S°, J/K/mol 248.12 205.03 256.72
Srxn° = npS°(products) - nrS°(reactants)
=2× S°(SO3) - 2× S°(SO2) - × S°(O2)
=2 ×256.72-2 ×248.13-205.03
=-187.83 J/K/mol
ChemicalThermodynamics
Entropy Exercise
Predict whether the entropy change in each of the
following reactions is positive or negative:
a) CaCO3(s) CaO(s) + CO2(g)
b) N2(g) + 3H2(g) 2NH3(g)
c) H2O(l) H2O(g)
d) Ag+(aq) + Cl- (aq) AgCl(s)
e) 4Fe(s) + 3O2(g) 2Fe2O3(s)
ChemicalThermodynamics
Entropy Changes in Surroundings
• Heat that flows into or out of the system changes the entropy of the surroundings.
• For an isothermal process:
Ssurr =qsys
T
• At constant pressure, qsys is simply H for the system.
ChemicalThermodynamics
Entropy Change in the Universe
• The universe is composed of the system and the surroundings.
• Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
ChemicalThermodynamics
Entropy Change in the Universe
• This becomes:
Suniverse = Ssystem +
Multiplying both sides by T,
TSuniverse = Hsystem TSsystem
Hsystem
T
ChemicalThermodynamics
Gibbs Free Energy
TSuniverse is defined as the Gibbs free
energy, G.
• When Suniverse is positive, G is negative.
• Therefore, when G is negative, a process
is spontaneous.
ChemicalThermodynamics
Criterion for Spontaneity
• According to the 2nd Law, a reaction is
spontaneous at constant pressure and
temperature if and only if:
Hsystem TSsystem<0
• Or …
H - T S < 0
ChemicalThermodynamics
Gibbs Free Energy
These two factors are combined in the
Gibbs free energy, defined: G = H –TS or G = H - T S
A reaction is spontaneous (under conditions of
constant T and P) when G < 0
• Spontaneous reactions are favored by: H < 0 (exothermic)
S > 0 (increasing entropy)
ChemicalThermodynamics
Predicting Spontaneity
1. If a reaction has G < 0 it is
spontaneous (in the forward direction).
2. If a reaction has G > 0 its reverse is
spontaneous.
3. If a reaction has G =0 then it is already
at equilibrium.
ChemicalThermodynamics
Gibbs Free Energies of Formation
• The Gibbs free energy of formation Gf for a
substance is defined in the same way as the
enthalpy of formation (Hf)..
Gf for a substance is the Gibbs free energy
change when one mole of the substance is formed
under standard conditions from its elements in
their standard states.
ChemicalThermodynamics
Standard Free Energy Changes
Analogous to standard enthalpies of formation are standard free energies of formation, G.
f
Grxn = npGf(products) nrGf(reactants)
where np and nr are the stoichiometric coefficients.
ChemicalThermodynamics
Grxn from Gf Values
• Just as S of reaction can be calculated from standard entropy values S0, and H of reaction can be calculated from Hf values, so Grxn of reaction can be
calculated from Gf values.
Grxn = npGf(products) nrGf(reactants)
ChemicalThermodynamics
The difference between Grxn and Grxn
Grxn(or G ): is the free energy change
that accompanies a change from reactants
in their standard states to products in their
standard states.
Grxn(or G ): is the free energy change that
accompanies a change from reactants in
nonstandard states to products in
nonstandard states.
ChemicalThermodynamics
Example for calculation of G°from H°and S°:
Use standard heat of formation and standard entropies to calculate G°for the reaction at 25 and 1atm partial ℃pressure of each gas:
3H2(g)+N2 (g)= 2NH3(g)
Solution:
(1) Calculation of H° and S°:
Equation: 3H2(g)+N2 (g)= 2NH3(g)
Hf°, kJ/mol 0 0 -46.11
S°, J/K/mol 248.12 205.03 256.72
H°=npHf°, (products) - nrHf°, (reactants)
=2 ×(-46.11) - 3×0 - 2×0
=-92.22 kJ/mol
ChemicalThermodynamics
Srxn° = npS°(products) - nrS°(reactants)
=2× S°(NH3) - 3× S°(H2) - 2× S°(N2)
=2 ×192.3 - 3 ×130.68 - 2 × 191.50
=-198.9 J/K/mol
(2)Calculation of G°:
G°=H°-TS°= -92.22 –(273.2+25) ×(-198.9 ×10-3)
=-32.91kJ/mol
ChemicalThermodynamics
Example for G0 of Reaction from Gf0
• Calculate the standard Gibbs free energy of
reaction for the combustion of methane CH4.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
Gf (CH4 ,g ) = -50.8 kJ/mol
Gf (CO2 ,g) = -393.4 kJ/mol
Gf (H2O ,l) = -237.13 kJ/mol
G0 = 2×Gf (H2O ,l) + Gf (CO2 ,g) - Gf (CH4 ,g )
G0 = -816.9 kJ
ChemicalThermodynamics
S ° and H ° for Reactions
where np is the stoichiometric coefficient of product,
nr is the stoichiometric coefficient of reactant.
Srxn° = npS°(products) - nrS°(reactants)
Hrxn° = npHf°(product) - nrHf°( reactant)
ChemicalThermodynamics
Free Energy Changes
At temperatures other than 25°C,
G° = H TS
How does G change with temperature?
ChemicalThermodynamics
Free Energy and Temperature
• There are two parts to the free energy
equation:
H— the enthalpy term
TS — the entropy term
• The temperature dependence of free energy,
then comes from the entropy term.
ChemicalThermodynamics
Example for G0 from H0 and S0
• Consider N2(g) + 3H2(g) 2NH3(g). Assume that
H0 and S0 do not change much with
temperature. Calculate G0 for the reaction at
500 K.
H0 = -92.38 kJ/mol
S0 = -198.3 J/K-mol
G0 = H0 - T S0 = (-92.38) - (500) ×(-0.1983)
G0 = +6.77 kJ
ChemicalThermodynamics
• S0: depend markedly on temperature.
S0: however, because increasing temperature
increase the entropy of all substance, S0 often do
not change greatly with temperature at ordinary
temperature.
H0: are also often quite constant as temperature
changes because of the same formation and
cleavage in a certain reaction.
ChemicalThermodynamics
Predicting Spontaneity
• Two factors determine the spontaneity
of a chemical or physical change:Enthalpy change H
Entropy change S
H < 0 (exothermic) favors the process.
S > 0 (more randomness) favors the
process.
ChemicalThermodynamics
H, S and Spontaneity
• There are four possible combinations of
positive and negative H and S:
1. H < 0 and S > 0 G < 0 : spontaneous at any
temperature
2. H > 0 and S < 0 G > 0 : not spontaneous at
any temperature
3. H > 0 and S > 0 favored at high T
4. H < 0 and S < 0 favored at low T
ChemicalThermodynamics
Temperature Dependence
• Since G=0 at equilibrium, a process will
reach equilibrium when H = T S or at
the temperature T = H/S.
• If a process is non spontaneous at T<
H/S, it will become spontaneous for T >
H/S and vice versa.
ChemicalThermodynamics
Free Energy and Temperature
ChemicalThermodynamics
Energy and EquilibriumE
nerg
y
reactants productsequilibrium
Q: reaction quotient; K: equilibrium constant
Equilibrium: A system’s macroscopic properties do
not change spontaneously. (Vforward =Vreverse)
Q < Kc means the reaction will go spontaneously in the forward direction.Q > Kc means the reaction will go spontaneously in the reverse direction.
ChemicalThermodynamics
Gibbs Free Energy
1. If G is negative, the forward reaction is spontaneous.
2. If G is 0, the system is at equilibrium.
3. If G is positive, the reaction is spontaneous in the reverse direction.
ChemicalThermodynamics
Free Energy and Equilibrium
Under any conditions, standard or
nonstandard, the free energy change can
be found this way:
G = G + RT lnQ
(Under standard conditions, all concentrations are 1
M, so Q = 1 and lnQ = 0; the last term drops out.)
ChemicalThermodynamics
Free Energy and Equilibrium
• At equilibrium, Q = K, and G = 0.
• The equation becomes
0 = G + RT lnK
• Rearranging, this becomes
G = RT lnK
or,
K = eG/RT
What's Free about Free Energy
• The change in the Gibbs free energy for a
process is the maximum amount of useful
work that can be done by the system at
constant temperature and pressure.
maxwG
CaCO3 Example
• At what temperature will CaCO3(s) just begin
to decompose to CaO(s) and CO2(g) under
standard conditions?
CaCO3(s) CaO(s) + CO2(g)
H0 = +178.3 kJ
S0 = +159.0 J/K = 0.159 kJ/K
G0 = 0 at equilibrium.
H0 - T S0 = 0 T = H0 / S0 = 1121 K
∴When T>1121K, the reaction is spontaneous.
K from G Example
Consider N2(g) + 3H2(g) 2NH3(g). Calculate
the equilibrium constant at 500 K.
H0 = -92.38 kJ/mol
S0 = -198.3 J/K-mol
G0 = H0 - T S0 = (-92.38) – (500)(-0.1983)
G0 = +6.77 kJ
K = exp(- G0 / RT) = 0.196 (K = eG/RT)