Download Chapter 20 Thermodynamics

Chapter 20
Thermodynamics: Entropy, Free Energy and
Thermodynamics:
Entropy, Free Energy,
and the Direction
of Chemical
Reactions
the Direction
of Chemical
Reactions
Thermodyamics considers the energetics of a
reaction but says nothing about the rate of a
chemical reaction (kinetics).
Potential Energy
20.1 The Second Law of Thermodynamics:
Predicting Spontaneous Change
20.2 Calculating the Change in Entropy of a Reaction
20.3 Entropy, Free Energy, and Work
20.4 Free Energy, Equilibrium, and Reaction Direction
Activation Energy
Time
Spontaneity refers to any chemical process that
appears to proceed “naturally” to a final state
without outside intervention.
In the 1870’s, it was thought that !H determined the
“spontaneity of a chemical reaction”. This notion
was proved wrong as both exothermic and
endothermic reactions are spontaneous.
T > 0˚C
Examples of spontaneous reactions:
CH4 (g) + 2O2 (g)
T < 0˚C
Freezing and melting of water
Our goal in this Chapter
is to identify those
factors which determine
whether a chemical
reaction will proceed
spontaneously.
A
H2O (l)
H2O (s) !H0 = - 6.01 kJ
H2O (s)
H2O (l) !H0 = 6.01 kJ
NH4NO3 (s)
H 2O
!Hrnx does not
determine whether
a reaction is
spontaneous!
NH4+(aq) + NO3- (aq) !H0 = 25 kJ
The rusting of a nail
Our goal is to identify those factors that will
allow us to predict whether a chemical reaction
will proceed spontaneously (no outside
intervention).
Suppose a chemical reaction:
CO2 (g) + 2H2O (l) !H0 = -890.4 kJ
Review: The internal energy, E, of a system is a
bulk property of matter that describes all energy
within a defined system.
B
Can we develop a parameter or a function that can
predict whether A ==>B will occur “spontaneously”?
The answer is yes and the details are explained in
thermodynamic theory. First, let’s review thermodynamics
from Chem 7 (internal energy, formalism, enthalpy)
It’s the sum of energies
that exists in atoms/
molecules including:
•
•
•
•
•
•
•
Translational kinetic energy.
Molecular rotation.
Bond vibrations.
Intermolecular attractions.
Electrons.
Nuclear
Electrochemical
• Translational kinetic energy.
• Rotational energy
• Vibrational energy
• Electrostatic energy
Molecules translate, vibrate, rotate in quantized
states. The more atoms there are, the more ways
there are a molecule can vibrate or “wiggle” and
store energy (degrees of freedom).
Molecular Level: Internal energy is distributed over all
quantized translational, vibronic and rotational energy
states in molecules and atoms.
Quantized energy
levels are of different
energy. Rotational
gaps are on the order
of the energy of
microwave radiation.
Vibrational spacing is
higher in energy (IR).
Energy
Spacing
Trans
Roto
Vibro
It is the distribution of quanta in these energy levels that is
essential to understanding the entropy and the “dispersal of
energy” that occurs during a spontaneous chemical reaction.
Macroscopic: The internal energy of a system
can be changed only by exchanging heat (q) or
work (w) with its surroundings.
Change in
Internal
energy
Heat Work
!ESys = q + w
Surroundings
+q = heat
added
+w = work
done on
the
system
System with
internal energy E
Surroundings
It’s a state
function!
“energy is neither created nor destroyed just
transformed between the two boundries.”
!Esystem + !Esurroundings = 0
-q = heat lost
from the
system
-w = work
done by the
system
We can not measure the “absolute” internal energy
only changes in internal energy between two
states.
It’s a state
function!
!ESys = q + w = Efinal - Einitial
The 1st Law of Thermodynamics says that total energy
of a system + surroundings remains constant—i.e.
!E is the measure of
a change relative to
an initial state:
State 1: P1, V1, T1 to
State 2: P2, V2, T2
!Esystem = -!Esurroundings
The 1st Law of Thermodynamics does not
tell us anything about the direction of
change in the universe!
A state function is a special type of mathematical
function that has ONE UNIQUE VALUE between an
initial and a final state. That value does not depend on
the path taken between the two states.
h
Final State
All are state functions!
!T = Tfinal - Tinitial
Recall our “tallest mountain analogy” to measuring changes in quantities.
Elevation is relative to another point that we define as sea level.
!E = Efinal - Einitial
1. Mount Everest is king if measured from sea level
(8,848 meters)
!P = Pfinal - Pinitial
2. Mauna Kea is king when measured from the depths of the
Pacific Ocean floor. It rises 10,203 meters.
PE = mgh
!H = Hfinal - Hinitial
!V = Vfinal - Vinitial
0
Initial State
The potential energy of hiker 1 and hiker 2
is the same even though they took very
different paths to get to the mountain top.
This is the power of all the state functions
in thermodynamics---we only need to now
the start and the finish bulk parameters.
The relevance and power of a state function is that
there is one unique value between any two states
(which represents a change or a reaction) and this
change does not depend on the path between the two
states.
Internal energy is hard to measure the lab, so chemists
define a new lab-friendly function at constant pressure
and give it the name: enthalpy, H.
!E = qP - P!V
mgh2
Final State
restrict system to constant
pressure (Pinital = Pfinal)
Solve for qp and call it enthalpy
Arbitray State
Initial State
!h
!H = qP = !E + P!V
h=1
Two paths up the mountain, one hard--the other easy. Both height
and potential energy are state functions in this scenario, but work
required, or calories burned, time, or distance traveled to the top
are not state functions!
An Exothermic reaction releases heat to the
surroundings, !H < 0.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + heat
!H = <0
H2O (l) + heat
heat written as a
product (given off)
An Endothermic reaction absorbs heat from the
surroundings into the system, !H > 0.
heat + 2HgO (s)
heat + H2O (s)
units of energy (Joule)
Enthalpy is the “heat gained or lost by a system”
under conditions of constant pressure”.
What determines whether a reaction will occur
spontaneously? It was thought that !H determined
the “spontaneity of a chemical reaction”. This
notion was proved wrong as both exothermic and
endothermic reactions are spontaneous .
Examples of spontaneous reactions:
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) !H0 = -890.4 kJ
H2O (l)
H2O (s) !H0 = - 6.01 kJ
H2O (s)
H2O (l) !H0 = 6.01 kJ
2Hg (l) + O2 (g)
H2O (l)
!H = >0
heat written as a
reactant (consumed)
NH4NO3 (s)
Entropy can be described (incorrectly) as a
“positionally messy room” or “disordered
cards” or“disorder”. This is not correct in the
thermodynamic sense, but it illuminates the
statistical nature of entropy.
H 2O
!Hrnx does not
determine whether
a reaction is
spontaneous!
NH4+(aq) + NO3- (aq) !H0 = 25 kJ
Boltzmann founded a thermodynamic state function
called entropy, S, that is a measure of the “dispersion or
spread of energy” that occurs when all spontaneous
reactions occur.
# of microstates
How many ways
can you have your
room or a deck of
cards organized
vs disorganized?
S = k ln W
Entropy
Boltzman’s constant
1.38 " 10#23 J/K.
!S = k ln Wf - k ln Wi
Change in
Entropy
Boltzmann’s Tomb In Vienna, Austria
When Wf > Wi then !S > 0
When Wf < Wi then !S < 0
There is a thermodynamic state function called
entropy, S, that is a measure of the energy
dispersal that occurs when a change of state
occurs.
1. Macroscopic or bulk
defintion (it needs no
molecules). Represents
heat flow.
!S = -qsys reversible
T
Boltzman’s statistical interpretation of entropy
(dispersal or spread of energy) is connected to a
macroscopic definition of heat flow per unit
temperature.
2. Microscopic or
Molecular Description
Macroscopic and Bulk
defintion (it needs no
molecules)
!S = k (ln Wf - ln Wi)
!S = -qsys reversible
T
Microscopic or
Molecular Description
!S = k (ln Wf - ln Wi)
Change in
Entropy
Change in
Entropy
This value quantifies the “level of spreading of energy”.
A system can be described by bulk or macroscopic
properties that we can observe and measure, and the
microscopic or molecular that we can’t see but we
can model statistically.
Consider 4 labeled molecules A,B,C,D
Microstate is particular distribution
that corresponds to some macrostate.
Multiplicity is the number of microstates
that give a specific macrostate.
Macrostate is the observed state of the
system that represents on a molecular
level that microstate with the highest
probability or number.
5-observable
macrostates
The second law of thermodynamics says that all
processes that are spontaneous produce an
increase in the entropy of the universe.
!Suniv = !Ssys + !Ssurr > 0
1. Criteria for
Spontaneous change!
Microstate
Macrostate
Left
Side
Bulk Property
Right Multiplicity Probability
Side
A,B,C,D
-
1
1/16
A,B,C
A,B,D
A,C,D
B,C,D
D
C
B
A
4
4/16
6
6/16
1
16
1/16
A,B
A,C
A,D
B,C
B,D
C,D
C,D
B,D
B,C
A,D
A,C
A,B
-
A,B,C,D
While it appears that entropy focuses on the most
statistically favored position distribution entropy is
more concerned with the fact that energy levels
become more closely spaced and more occupied.
Reactions tend towards dispersing energy!
Note: Entropy is not energy! It is an index or measure
of the dispersal of energy that occurs in all spontaneous
processes passing from State 1 to State 2.
!Suniv = !Ssys + !Ssurr < 0
2. No spontaneous change
!Suniv = !Ssys + !Ssurr = 0
3. Equilibrium condition
There are many
different ways we can
Energy
Spacing
Trans
Roto
Vibro
Matter being dispersed into a larger number of
statistical microstates disperses energy into a larger
number of thermal energy levels increasing entropy.
Gas expands
spontaneously
into larger volume
Example: two block of different temperatures are
brought together.
evacuated
0.5 atm
Quantum mechancis
dictates closer energy
level spacing as V
increases.
Energy level
Same amount of
energy is dispersed or
spread among more
energy levels.
Energy level
Be careful entropy is often illustrated incorrectly
as a positionally a messy room or disordered
cards and described as “disorder”. This is not
correct in the thermodynamic sense!
Entropy refers
to the degree
to which
energy is
dispersed in a
process. The
higher the
entropy the
more energy is
dispersed in
that system.
1
100˚C
2
10˚C
HOT
COLD
100˚C
10˚C
COMBINED
There are many events that result in a higher
number of microstates or higher entropy.
1. Increasing temperature
!S increases as temperature rises as more energy states are filled
2. Solid to Liquid or Liquid to Gas phase changes
!S increases as phase changes to a more dispersed phase.
WRONG: Neither
case represents
thermodynamic
entropy because
rooms and cards do
not exchange heat
with its
surroundings! This
is a fundamental
requirement.
Qualitative Meaning
of Entropy:
Thermal energy flows
from the higher
occupied energy
levels in the warmer
object into the
unoccupied levels of
the cooler one until
equal numbers of
states are occupied.
3. Dissolution of a solid or liquid
Dissolving a solid or liquid into a solvent increases entropy.
4. Dissolution of a gas decreases entropy.
A gas becomes less dispersed when it dissolves in a liquid or solid.
5. Atomic size or molecular complexity
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g., bond
flexibility) relate directly to entropy.
Phase Change
!S > 0
Phase Change
Entropy increases when a substance is heated as
thermal energy is distributed over more energy
states. Phase changes lead to large increases.
Low T
High T
Entropy(S)
1.0 atm
Increasing entropy or the dispersal or spread of
energy occurs in any spontaneous process. It
provides humans with an “arrow of time” as the
reverse situation never happens.
!S > 0
Dissolution
!S > 0
Temperature
Increasing T
Temp increases populates more
energy levels increasing entropy
Phase transitions: solid
to liquid to gas have very
large increases in
entropy!
When a solid crystalline salt dissociates into a
greater number of ions entropy increases as the
number of microstates that energy can be dispersed
is larger.
pure solid
When two miscible liquids combine the most
probable microstate (the observed macrostate) is a
homogeneous mixture.
!Ssys > 0
The dispersal of matter
leads to the dispersal of
energy among more
energy states in a
mixture just as it does
when gases expand.
!Ssys > 0
MIX
pure liquid
Benzene (C6H6)
Toluene
(C6H5CH3)
Impossible
Result
solution
Solution
Entropy decreases significantly as a gas is
dissolved in a liquid, or as a gas condenses to
liquid, or as a liquid freezes to a solid.
Other trends in absolute entropies (see text p.
657-660)
1. Substances down a group (increasing mass) have
higher absolute entropies.
O2 gas
!Ssys < 0
2. As the number of atoms increase in a molecule
absolute entropy tends to increase (more degrees of
freedom => more microstate available)
3. As the number of atoms increase in a molecule
absolute entropy tends to increase (more degrees of
freedom => more microstate available)
O2 dissolved
How does the entropy of a system change for
each of the following processes?
(a) Condensing water vapor to a liquid
How does the entropy of a system change for
each of the following processes?
(a) Condensing water vapor to a liquid
dispersal of energy decreases Entropy decreases (!S < 0)
(b) Forming sucrose crystals from a supersaturated solution
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
dispersal of energy decreases
Entropy decreases (!S < 0)
(c) Heating hydrogen gas from 600C to 800C
dispersal of energy increases
Entropy increases (!S > 0)
(d) Subliming dry ice
dispersal of energy increases
Entropy increases (!S > 0)
Choose the member with the higher entropy in each
of the following pairs, and justify your choice.
Assume constant temperature, except in part (e).
Choose the member with the higher entropy in each
of the following pairs, and justify your choice.
Assume constant temperature, except in part (e).
(a) 1 mol of SO2(g) or 1 mol of SO3(g)
(b) 1 mol of CO2(s) or 1 mol of CO2(g)
(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)
(a) 1 mol of SO2(g) or 1 mol of SO3(g)
(b) 1 mol of CO2(s) or 1 mol of CO2(g)
(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)
(d) 1 mol of KBr(s) or 1 mol of KBr(aq)
(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC
(d) 1 mol of KBr(s) or 1 mol of KBr(aq)
(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC
(f) 1 mol of CF4(g) or 1 mol of CCl4(g)
SOLUTION:
(d) 1 mol of KBr(aq) - solution
(a) 1 mol of SO3(g) - more atoms > solid
(f) 1 mol of CF4(g) or 1 mol of CCl4(g)
(b) 1 mol of CO2(g) - gas > solid
(c) 3 mol of O2(g) - larger # mols
Boltzman’s statistical interpretation of entropy
(dispersal or spread of energy) is connected to a
macroscopic definition of heat flow per unit
temperature.
Macroscopic and Bulk
defintion (it needs no
molecules)
!S = -qsys reversible
T
(e) 23 oC - higher temperature
(f) CCl4 - larger mass
Bulk or macroscopic measurement of entropy is
harder to see. Molecules not required.
!S = -qsys reversible
T
Microscopic or
Molecular Description
Change pressure by
infinitesimal amounts we
get a reversible change
where work is maximized.
The heat q/T is the
entropy or “measure of
energy dispersion”.
!S = k (ln Wf - ln Wi)
Change in
Entropy
This value quantifies the “level of spreading of energy”.
3rd Law of Thermodynamics: The entropy, S of a
perfect crystal is zero at the 0 Kelvin.
The reference point for entropy is called the “zero
point energy” at 0K. The entropy of a perfect solid
crystal is 0 (say KBr or NaCl perfect has 0 entropy).
A Mountain Peak
S>0
Increasing
Temp
Height > 0
T = 298K
S>0
T = 100K
T = 0K
S=0
Perfect crystal
with 1 microstate
T>0K
S>0
A “perfectly” ordered crystal at 0K gives way to a less-structured
liquid with more available microstates for internal energy to be
dispersed.
T = 0K
S=0
Height = 0
Sea-level
A “standard state” is a defined set of experimental
conditions T, P, [x], phase, pH denoted symbolically by a
superscript degree sign above the thermodynamic
functions: !H˚, !G˚, !S˚
Thermodynammics is concerned primarily with
calculating !H˚rxn and !G˚rxn measured
thermodynamic data under standard state
conditions.
2-Ways or Problem Types
For Hess Law Enthalpy
Calculations
Standand State Reaction Conditions
Standard state
!Ho
rxn
o
!G rxn
Non-standard State
∆Hf◦ = 0
1M concentration
!Hrxn
!Grxn
not standard state
conditions!
•
Hess’s Law: The enthalpy change for any reaction is equal
to the sum of the enthalpy changes for any individual step
in the reaction.
Multiple
Thermochemical
Equations To
Rearrange-Use Rules!
!H3 =
CH4 (g) + 2O2
∆H1 = - 607 kJ
∆H3 =
?
CO(g) + 2H2O + 1/2 O2(g)
∆H2 = - 283 kJ
2-Ways or Problem Types
For Hess Law Enthalpy
Calculations
The Standard Enthalpy of Formation, !Hf°, is the enthalpy
change associated with the formation of 1 mole of product
from its naturally occurring elements under standard state
conditions.
Note: 1 mol product
complete conversion
H2(g) + O2(g) #$
H2O(l )
!Hf° = -285.8 kJ/mol
3C(s) + 4H2(g) #$
C3H8(g)
!Hf° = -103.85 kJ/mol
Ag(s) + 1/2Cl2(g) #$
2
1
Multiple
Thermochemical
Equations To
Rearrange-Use Rules!
Use !H˚f
data
1) standard state
2) formation equation
3) Calculation of !H°rxn
!H°rxn = " ni !Hif°(products) - " mi !Hjf°(reactants)
CO2 + 2H2O
•Examples:
1) standard state
2) formation equation
3) Calculation of !H°rxn
Thermochemistry is concerned primarily with
calculating !Hrxn (the heat of reaction).
? kJ
Because enthalpy is a state function, its value only depends on
the final and initial state and not the path. This gives us a
powerful way to predict enthalpy of many chemical reactions.
Use !H˚f
data
!H°rxn = " ni !Hif°(products) - " mi !Hjf°(reactants)
CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2
!H1 = - 607 kJ
CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O !H2 = - 283 kJ
CH4 (g) + 2O2 ==> CO2 + 2H2O
2
1
AgCl(s)
!Hf° = -127.0 kJ/mol
Note: fractional
•Are familar molecular gases as O2, N2, F2, H2 etc as at 25 °C
•Elemental Carbon exists as solid graphite C(graph) at 25 °C.
•Elemental Sulfur exits as S8 as a solid at 25˚C
•Water is H2O(l ) in its standard state (not ice or water vapor).
METHOD 2: The enthalpy of any reaction, !H°rxn can be
obtained by using !H°f and the following equation:
Calculate the heat of decomposition for the following
process using standard enthalpies of formation found
in Appendix. What kind of reaction is this?
Is energy given off or needed to make this reaction proceed?
!H°rxn = " ni !Hif°(products) - " mi !Hjf°(reactants)
CaCO3 (s)
where " means “the sum of”
CaO (s) + CO2 (g)
!H0rxn = ?
ni is the respective stoich coefficient for ith product
mi is the respective stoich coefficient for each ith reactant
!H°rxn = " ni !Hif°(products) - " mi !Hif°(reactants)
!H0rxn =
Example: Suppose
aA + bB
cC + dD
!H°rx = ?
Reaction is decomposition and endothermic. Energy must
be absorbed to proceed.
!H°rx = [c!H°f (C) + d!H°f (D)] – [a!H°f (A) + b!H°f (B)]
We do the same for Standard Molar Entropy, S°rxn: The
standard molar entropy !S°rxn of a chemical is
determined from the standard molar entropies of
reactants and products found in tables in Handbooks.
aA + bB
cC + dD
!S0rxn = ?
#Ssys° = $ni Si°(products) – $niSi°(reactants)
!S0rxn = [ cS˚(C) + dS˚(D) ] - [ aS˚(A) + bS˚(B)]
!Hf0 (CO2) ] - [ !H0f (CaCO3)]
!H0rxn = [ -635.6 + -393.5 ] – [ -1206.9 ] = 179 kJ
!H°rx = !H°f (Products) – !H°f (Reactants)
•
[ !H0f (CaO) +
•
Standard molar entropies of compounds are
tabulated just like "H˚
• standard conditions are
1 atm pressure, 1M
and at 25°C (298K).
• Standard entropies
tend to increase with
increasing molar mass
as solids==>liquids==>
gases.
S° for each component is found in a table. Also note
that S° = 0 for elements (unlike H).
Calculate !Sorxn for the combustion of 1 mol of
propane at 25 oC.
C3H8(g) + 5O2(g)
SOLUTION:
3CO2(g) + 4H2O(l)
Find standard entropy values in an appropriate table.
Calculate !Sorxn for the combustion of 1 mol of
propane at 25 oC.
C3H8(g) + 5O2(g)
SOLUTION:
3CO2(g) + 4H2O(l)
Find standard entropy values in an appropriate table.
#Ssys° = $ni Si°(products) – $niSi°(reactants)
!Sorxn = [(3 mol)(So CO2) + (4 mol)(So H2O)] - [(1 mol)(So C3H8) + (5 mol)(So O2)]
!Sorxn = [(3 mol)(213.7 J/mol.K) + (4 mol)(69.9 J/mol.K)] - [(1 mol)(269.9
J/mol.K) + (5 mol)(205.0 J/mol.K)]
!Sorxn = - 374 J/K
Calculating the standard entropy of reaction, !Sorxn
Calculating the standard entropy of reaction, !Sorxn
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
From a table of standard entropies we find:
S0(CO) = 197.9 J/K•mol
S0(O2)
S0(CO2) = 213.6 J/K•mol
= 205.0 J/K•mol
#Ssys° = $ni Si°(products) – $niSi°(reactants)
!S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
!S0rxn = 427.2 – [395.8 + 205.0] = -173.6 J
The second law of thermodynamics says that all
“spontaneous processes” are accompanied by
an increase in the entropy of the universe,
In a reversible process, the surroundings can add heat
to the system, or it can accept it from the system.
!Suniv > 0
!Suniv = !Ssys + !Ssurr > 0
+
1. Criteria for
Spontaneous change!
+qsurr
-qp
-qsurr
+qp
Entropy is a measure of the dispersal of energy that
occurs as a system passes from State 1 to State 2.
!Suniv = !Ssys + !Ssurr < 0
2. No spontaneous change
!Suniv = !Ssys + !Ssurr = 0
3. Equilibrium condition
Substituting -!Hsys/T for !Ssurr we get an
intermediate expression that is not so useful.
!Suniv = !Ssys + !Ssurr > 0
Calculate from Standard Entropies of
compounds from a table
!Sosys = " miSofi(products) - " niSofi(reactants)
This is
related to
heat lost
by system
_
! Ssurr =
! q sur
T
=
%! q sys
T
=
%! H sys
T
It is very difficult to measure !Suniv, !Ssys and !Ssurr
so we recast the 2nd Law equation into something
more useful called the Gibb’s Free Energy, !G.
!Suniv = !Ssys + !Ssurr
1. substitute !Ssurr = -qsys/T = -!Hsys/T
!Suniv = !Ssys - !Hsys/T
2. multiply by (-T)
-T!Suniv = - T!Ssys + !Hsys
%! Hsys
!Ssurr = -qsys =
T
T
!Suniv = !Ssys + -!Hsys/T > 0
For an endothermic process:
qsys > 0, qsurr < 0, !Ssurr < 0
For an exothermic process:
qsys < 0, qsurr > 0, !Ssurr > 0
3. Define !Guniverse = -T!Suniv
!Guniverse = !Hsys - T!Ssys
!Grxn
= !Hsys - T!Ssys
Watch the
notation both
are used!
The Gibb’s Free Energy, !Guniverse is the function that
indicates whether a reaction will occur spontaneously.
The sign of !G determines the reaction direction, the
magnitude determines the maximum amount of work the
system can do (if negative).
Temperature
in Kelvin
There are 3 Cases for the Sign of !Gsys:
1. !Gsys < 0 a spontaneous process (!Suniv > 0)
2. !Gsys > 0 a non-spontaneous process (!Suniv < 0)
3. !Gsys = 0 a process at equilibrium (!Suniv = 0)
!Guniverse = !Hsys - T !Ssys
Gibb’s Free
Energy Change
Useless
or dispersed
energy
Enthalpy
term
!Guniverse = !Hsys - T !Ssys
}
!Gsys = !Hsys - T !Ssys
!Grxn = !Hrxn - T !Srxn
Don’t be confused by
symbols. These all mean
the same thing and
pertain to the system!
Using Hess’s Law and thermodynamic data, we can
compute !Gorxn under standard state conditions.
1
!G˚sys
Use Table of
!Hof and !Sof
2
What is Free Energy physically?
!G < 0 is the maximum “useful” work that can be produced by a
chemical reaction. !G = workmax
For !G > 0 (non-spontaneous process) !G is energy that must be
added to the system to make the process take place.
Recall: The Standard Gibb’s Free Energy of Formation,
is the free energy change associated with the
formation of 1 mole of compound from its naturally
occurring elements under standard state conditions
for both. The same rules are followed for !Gf°
Note: 1 mol product!
Use Table of !Gof
Ag(s) + 1/2Cl2(g) #$
!Gorxn = !Horxn - T!Sorxn
!Horxn = " mi!Hofi (products) - " ni!Hofi(reactants)
!Sorxn = " miSofi (products) - " niSofi(reactants)
Elements as they exist
in elemental form
!Gf° = -110.0 kJ/mol
AgCl(s)
Note: fractional
coefficients allowed
•Oxygen exists as O2 gas at 25 °C !Hf° and !Gf° = 0
•Carbon exists as solid graphite (C) at 25 °C. !Hf° = 0
!G˚rxn = [c!G˚f (C) + d!G˚f (D) ]
-
[a!G˚f (A) + b!G˚f (B) ]
The rules of !Gfo of any element are the same
as !Hfo
!H0f (O2) = 0
!G0f (O2) = 0
!H0f (O3) = 142 kJ/mol
!G0f (N2) = 0
!H0f (N2) = 0
!G0f (Na) = 0
!H0f (Na(s) = 0
!H0f (C, graphite) = 0
!H0f (C, diamond) = 1.90 kJ/mol
!H0f (S8, rhombic) = 0 kJ/mol
Both !Hfo and !Gfo of many compounds are tabulated in
handbooks. This gives us predictive capability and is
someone useful in the real world.
•Sulfur exits as S8 as a solid at 25˚C
•Water is H2O(l ) in its standard state (not ice or water vapor).
Thermodynamic Tables
Substance
!H°f
!G°f
S°
!H°f and !G°f
have units of kJ/
mol (energy/mol)
S° has units
of J/mol K
Calculating !Gorxn from !Gof values
Use !Gof values to calculate !Gorxn for the following
reaction:
4KClO (s) !
3KClO (s) + KCl(s)
3
4
KClO3(s)
!Gof -296.3 kJ/mol
KClO4(s) +
KCl(s)
-303.2 kJ/mol
-409.2 kJ/mol
Calculating !Gorxn from !Gof values
Use !Gof values to calculate !Gorxn for the following
reaction:
!
4KClO3(s)
3KClO4(s) + KCl(s)
PLAN:
Use the !Gorxn summation equation and look up the values of
products and reactants in a thermodynamic table.
4KClO3(s) !
!Gof -296.3 kJ/mol
3KClO4(s) +
-303.2 kJ/mol
KCl(s)
-409.2 kJ/mol
!Gorxn = " mi!Gofi (products) - " ni!Gofi (reactants)
!Gorxn = [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] - [(4 mol)(-296.3 kJ/mol)]
!Gorxn = -134 kJ (reaction is spontaneous as written)
Using thermodynamic tables of data we can easily
compute !Gorxn under standard state conditions in
two different ways.
✓
!G˚sys
1
2
Use Table of
!Hof and !Sof
done
Use Table of !Gof
The standard free-energy (!G0rxn ) is the free-energy
change for the complete reaction of reactants to
products with both reactants and products in their
standard-state conditions (1 atm, 1M, 25˚C)
!G0 values are a special hypothetical case of 100% conversion.
We are usually more interested in real-world non-standard
reaction conditions and equilibrium values for !G---which is
different than !G°.
aA + bB
!Gorxn = !Horxn - T!Sorxn
!Horxn = " mi!Hofi (products) - " ni!Hofi(reactants)
!So
rxn =
"
m So
i
fi (products)
-"
n So
i
fi(reactants)
!G˚rxn = [c!G˚f (C) + d!G˚f (D) ]
-
[a!G˚f (A) + b!G˚f (B) ]
Calculating !Go from standard enthalpy and entropy values
Potassium chlorate, one of the common oxidizing agents in
explosives, fireworks and match heads, undergoes a solidstate redox reaction when heated. In this reaction, the
oxidation number of Cl in the reactant is higher in one of
the products and lower in the other (a disproportionation
reaction).
+5
+7
-1
!
4KClO3(s)
3KClO4(s) + KCl(s)
Use !Hof and So values to calculate !Gosys (!Gorxn) at 25 oC for this reaction.
PLAN:
Obtain appropriate thermodynamic data from a data table; insert
them into the Gibbs free energy equation and solve.
cC + dD
!G°rxn = ?
◦
◦
∆G◦rxn = ∆Hrxn
− T ∆Srxn
◦
∆Hrxn
= [d∆Hf◦D + c∆Hf◦C ] − [b∆Hf◦B + a∆Hf◦A ]
◦
∆Srxn
= [dSf◦D + cSf◦C ] − [bSf◦B + aSf◦A ]
Calculating !Go from standard enthalpy and entropy values
Potassium chlorate, one of the common oxidizing agents in explosives,
fireworks and match heads, undergoes a solid-state redox reaction
when heated. In this reaction, the oxidation number of Cl in the
reactant is higher in one of the products and lower in the other (a
disproportionation reaction).
+5
4KClO3(s)
!H°rxn =
!
+7
-1
3KClO4(s) + KCl(s)
" ni !Hif°(products) - " mi !Hjf°(reactants)
!H° = [c!H°f (C) + d!H°f (D)] – [a!H°f (A) + b!H°f (B)]
!Horxn = [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)] - [(4 mol)(-397.7 kJ/mol)]
!Horxn = -144 kJ
!Sorxn = " miSofi (products) - " niSofi(reactants)
= [cS°f (C) + dS°f (D)] – [aS°f (A) + bS°f (B)]
!Sorxn =
[(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] - [(4 mol)(143.1 J/mol.K)]
!Sorxn = -36.8 J/K
What is Gibb’s Free Energy physically?
!Gorxn = !Horxn - T!Sorxn
= -144 kJ - (298 K)(-36.8 J/K)(kJ/103J) = -133 kJ
The reaction is spontaneous or the reaction goes to
proceeds to the right.
Temperature plays an important role in determining
the sign of !G--which determines whether a reaction
is spontaneous.
o
o
o
!G
Case !Horxn !Sosys T!So
-
+
2
+
-
3
+
+
4
-
-
= !H
sys
enthalpy factor
Analyze each term!
1
sys
+
-
+
-
!Go
- T!S sys
T and entropy factor
spontaneous at all T
+
not-spontaneous at all T
spontaneous at high T;
nonspontaneous at low T
spontaneous at low T;
nonspontaneous at high T
SOLUTION:
The Given Data:At 298 K,
!Go = -141.6 kJ, !Ho = -198.4 kJ, and !So = -187.9 J/K.
o
o
(a) !G = -141.6 kJ since !G rxn < 0 it is spontaneous at
25˚C (298K) which is the standard state. At higher
T with S < 0 the reaction will be less spontaneous.
Since !Ho is (-) and !So are both (-), !Go will become less
negative and therefore the reaction less spontaneous as
temperature increases. !Gorxn = !Horxn - T!Sorxn
(b)
!Go
rxn =
!Ho
rxn
-
T!So
rxn
For !G > 0 (non-spontaneous process) !G represents energy that
must be added to the system to make the process take place.
Determining the effect of temperature on !Go
An important reaction in the production of sulfuric acid
is the oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g)
2SO3(g)
!Gorxn = -141.6 kJ, !Horxn = -198.4 kJ, and !Sorxn = -187.9 J/K.
Process Description
+
+-
For a spontaneous chemical reaction wherein !G < 0, !G
represents the maximum “useful” work that can be produced by a
chemical reaction.
Use values for 1173K
!Gorxn = -198.4 kJ - [(1173 K)(-187.9 J/mol.K)(kJ/103J)] = 22 kJ
!Gorxn = +22.0 kJ; the reaction will not be spontaneous at 900.oC
(a) Is this reaction spontaneous at 25oC, and predict how
!Go will change with increasing T.
(b) Assuming !Ho and !So are constant with
increasing T, is the reaction spontaneous at 900. oC?
When !Horxn and !Sorxn have the same sign, then the
reaction temperature will determine spontaneity. We
can compute the “cross-over temperature” that
makes the reaction spontaneous.
Cross-over logic: We can set ∆Gosys = 0 to find that temperature
where G changes sign. The approach assumes that ∆H and ∆S are
constant over temperature.
!Gorxn = !Horxn - T!Sorxn
0 = !Horxn - T!Sorxn
T=
!Horxn
!Sorxn
At cross-over T
!Gorxn = 0
Solve for T
Watch your units!
∆S is usually J and ∆H is
usually kJ
!Grxn (not !G˚rxn) is the free energy for nonstandard thermodynamic reaction conditions. It
connects the free energy to the equilibrium
reaction quotient, Q.
Recall our expressions for Q and equilibrium
constants Kc, Kp, Ka, Ksp
It can be shown that:
Concentrations or partial pressures at point before equilibrium
aA(g) + bB(g)
Non-standard
Free Energy
Standard
Free Energy
Reaction
Quotient
Concentrations or partial pressures at equilibrium
[C]c[D]d
Kc = [A]a[B]b
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Recall what Q meant in our equilibrium chapter:
when compared to Kc it told us the direction of a
chemical reaction.
aA(g) + bB(g)
(Pc)c (Pd)d
Q=
(Pa)a (Pb)b
[C]c[D]d
Q = [A]a[B]b
#G = #G˚ + RT ln Q
cC(g) + dD(g)
Kp =
(Pc)c (Pd)d
(Pa)a (Pb)b
"G˚rxn and the Kc are related!
•
•
#Grxn = #G˚rxn + RT ln K
cC(g) + dD(g)
aA + bB
•
R = 8.314 J/K mol
T in Kelvin
[C]c[D]d
Kc = [A]a[B]b
cC + dD
At equilibrium: !G = 0 and Q = Kc
0 = #G˚ + RT ln Kc
•
Reactant-Favored
!Gsys = 0
!Gsys > 0
!Go and Kc at 25 oC are inversely related....big Kc
means product favored and !G˚ < 0 , and vis versa.
!Go (kJ)
K
3 x 10-18
50
2 x 10-9
10
1
2 x 10-2
-1
7 x 10-1
1
1.5
-10
5 x 101
-50
6 x 108
-100
3 x 1017
-200
1 x 1035
Essentially no forward reaction;
reverse reaction goes to
completion
Forward and reverse
reactions proceed to the
same extent
Forward reaction goes to
completion; essentially no
reverse reaction
REVERSE REACTION
9x
100
R = 8.314 J/K mol
T in Kelvin
Taking antiloge or exp of both sides
•
•
K = exp (-"G˚/RT)
R = 8.314 J/K mol
T in Kelvin
Plots of free energy diagrams for !Go > 0 (left) and
!Go < 0 (right).
Significance
FORWARD REACTION
200
0
10-36
•
•
•
Reactant Favored
!Goproduct
!Go(rxn) > 0
!Goreactant
Q>K
Q<K
Q=K
100%
Equilibrium
Reactants Mixture
100%Pro
ducts
Reaction Progress ===>
!Go(rxn)
> 0, K<1
Product Favored
Total free energy !G
!Gsys < 0
Equilibrium
Total free energy !G
Product-Favored
#G˚rxn = - RT ln Kc
Q> K
Q = Kc
Q< K
Solving for #G˚rxn
!Goreactant
Q<K
!Go(rxn)< 0
!Goproduct
Q=K
Pure
Reactants
Equilibrium
Mixture
Q>K
Pure
Products
Reaction Progress ===>
!Go(rxn)<0, K>1
The equilibrium constant, Kp, for the reaction:
N2(g) + 3H2(g) => 2NH3(g) is Kp = 9.4 X 10-5.
The equilibrium constant, Kp, for the reaction:
N2(g) + 3H2(g) => 2NH3(g) is Kp = 9.4 X 10-5.
Calculate the value of G˚ for the reaction under these
reaction conditions (standard assume 298K).
Calculate the value of G˚ for the reaction under these
reaction conditions (standard assume 298K).
Remember K is related to the standard Gibb’s free energy
!G˚rxn when !G = 0. (!Grxn = !G˚rxn + RT ln K)
Remember K is related to the standard Gibb’s free energy
!G˚rxn when !G = 0. (!Grxn = !G˚rxn + RT ln K)
#G˚rxn = - RT ln Kp
#G˚rxn = - (8.314 J/mol K) (298K) (ln (9.4 x 10-5))
- (-22972.58 J/mol) = 22.97 kJ/mol
Calculating !G at non-standard conditions
The oxidation of SO2 to SO3 is too slow at 298 K to
be useful to industry. To overcome this low rate, the
process is conducted at an elevated temperature.
2SO2(g) + O2(g)
2SO3(g)
SOLUTION:
(a) Calculating K !Go298 and !Go973 from at the
two temperatures:
!Go = -RT ln K thus:
At 298 K: !Go/RT = -
(-141.6 kJ/mol)(103 J/kJ)
(a) Calculate Kp at 298 K and at 973 K. (!Go298 = -141.6 kJ/mol
for the reaction as written using !Ho and !So values. At 973 K,
!Go973 = -12.12 kJ/mol for the reaction as written.)
(b) In experiments to determine the effect of temperature on
reaction spontaneity, two sealed containers are filled with 0.500
atm of SO2,0.0100 atm of O2, and 0.100 atm of SO3 and kept at
25 oC and at 700. oC. In which direction, if any, will the reaction
proceed to reach equilibrium at each temperature?
= e57.2 = 7 x 1024
At 973 K: -!Go/RT = -
(-12.12 kJ/mol)(103 J/kJ)
(8.314 J/mol.K)(973 K)
Q=
pSO32
(0.100)2
= 4.00
=
2
(pSO2) (pO2) (0.500)2(0.0100)
Since Q is < K at both temperatures the reaction equilibrium
will shift right and be product favored; for 298 K there will be
a dramatic shift while at 973 K the shift will be slight.
(c) The non-standard !G is calculated using:
!G = !Go + RT ln Q
!G298 = -141.6 kJ/mol + (8.314 J/mol.K)(kJ/103J)(298 K)(ln 4.00)
!G298 = -138.2 kJ/mol
!G973 = -12.12 kJ/mol + (8.314 J/mol.K)(kJ/103J)(973 K)(ln 4.00)
!G298 = -0.9 kJ/mol
= 1.50
= e1.50 = 4.5
(c) Calculate !G for the system in part (b) at each
temperature.
(b)
= 57.2
(8.314 J/mol.K)(298 K)
Summary of Key Points:
1. !Go tells us the maximum amount of work a spontaneous
reaction can do (when negative) !Gosys = !Hosys - T!Sosys
2. Q and !Grxn are linked
Q = Kc, !Grxn = 0 and !Suniverse = 0 ====> Equilibrium.
Q < Kc, !Grxn < 0 and !Suniverse > 0 ====> Product-Favored
Q > Kc, !Grxn > 0 and !Suniverse < 0 ====> Reactant-Favored
3.
!Gorxn gives the position of equilibrium and can be
calculated three ways:
!Gorxn = " mi!Gofi (products) - " ni!Gofi (reactants)
!Gorxn = !Horxn - T!Sorxn
!Gorxn = -RT ln Kc
4.
!Grxn describes the direction in which a non-standard
reaction proceeds to reach equilibrium calculated by:
!Grxn = !Go + RT ln Q
Consider the reaction of copper(I) oxide with carbon with the
following thermodynamic data. .
kJ/mol !H˚f
kJ/mol !G˚f
J/mol K S˚
Cu2O(s) + C(s) ===> 2Cu(s) + CO(g)
-168.6
0
0
-110.5
-146.0
0
0
-137.2
93.1
5.68
33.1
197.5
1. By inspection only of the chemical equation determine if entropy of
the system increases or decreases?
2. Calculate !H˚rxn
3. Calculate !S˚rxn
4. Calculate !G˚rxn, using !G˚rxn = !H˚rxn – T!S˚rxn
5. Caculate !G˚rxn using the thermodynamic !G˚f data, is it the same?
6. Determine if the reaction is (product-favored) spontaneous under
standard state conditions (298K, 1M, 1atm)?
7. What impact would temperature have on the spontaneity of the
reaction assuming constant enthalpy and entropy vs T?
8. Calculate the “cross-over” temperature which must be exceeded to
make this reaction spontaneous.
9. Determine Kp for this reaction.
kJ/mol !H˚f
kJ/mol !G˚f
J/mol K S˚
Cu2O(s) + C(s) ===> 2Cu(s) + CO(g)
-168.6
0
0
-110.5
-146.0
0
0
-137.2
93.1
5.68
33.1
197.5
5. !G°rxn = [1!G°f (CO(g)) + 2!G°f (Cu(s))]
– [1!G°f (Cu2O(s) ) + 1!G°f (C(s))]
= [1(-137.2) + 2(0)] – [1(-146.0) + 1(0)] = 8.8kJ
kJ/mol !H˚f
kJ/mol !G˚f
J/mol K S˚
Cu2O(s) + C(s) ===> 2Cu(s) + CO(g)
-168.6
0
0
-110.5
-146.0
0
0
-137.2
93.1
5.68
33.1
197.5
1. The entropy of the system increases (>0) as we move from
solid reactants to gaseous products.
2. !H°rxn = [1!H°f (CO(g)) + 2!H°f (Cu(s))]
– [1!H°f (Cu2O(s) ) + 1!H°f (C(s))]
= [1(-110.5) + 2(0)] – [1(-168.6) + 1(0)] = 58.1 kJ
3. !S°rxn = [1S°f (CO(g)) + 2S°f (Cu(s))]
– [1S°f (Cu2O(s) ) + 1S°f (C(s))]
= [1(197.5) + 2(33.1)] – [1(93.1) + 1(5.68)] = 164.9 J/K
= .1649 kJ/ K
4. !Gorxn = !Horxn - T!S˚rxn
= (58.1kJ) - (298K) (0.1649 kJ/K) = +8.9 kJ
8. The “cross-over” temperature that must be exceeded is
found by setting !Gorxn = 0.
0 = !Horxn - T!Sorxn
!Horxn
Watch the units of J/K and kJ!
T=
!Sorxn
T = 58.1 kJ
.1649 kJ/K
6. !Gorxn > 0. Therefore the reaction is not spontaneous
under standard state conditions at 298K.
7. !Horxn > 0 and !S˚rxn > 0. Analysis of the Gibb’s
equation: !Gorxn = !Horxn - T!Sorxn will show that
increasing T will increase T!S˚rxn and drive !G˚rxn <0 as
!Horxn remains unchanged (which is assumed).
> 352.K
9. K298 = exp (-#G˚rxn/RT)
This temperature
has to exceeded for
the reaction to be
spontaneous!
- 8.8 kJ 1000J/1kJ
)
8.314 J/mol K (298K)
= 2.9 x 10-2
= exp (