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Thermodynamics
Entropy, Energy and equilibrium
Chapter 19
1
19.1 Thermodynamics
 Thermo: heat
Dynamics: power
 Study of energy flow and its transformations
(heat and energy flow)
 Determines direction of reactions (spontaneous
or nonspontaneous under given conditions)
 State Functions: considers only initial and final
states
 Does not consider pathways or rate
 Organized into three laws:
1st Law (ΔU = qp + w = qp – PΔV )
2nd Law
3rd Law
2
19.1 Basic Definitions
universe
 System
 Surrounding
system surroundings
 Open system
 Closed system
 Isolated system
 State of system: defined by values of composition,
pressure, T, V.
 State Function: defined only by initial and final
condition of the system (Enthalpy, Entropy, Gibbs
Free Energy).
 Energy change signaled by: accomplishment of
work and/or appearance or disappearance of heat.
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19.1 1st Law of Thermodynamics
1. A
gas does 135 J of work while expanding, and at the
same time it absorbs 156 J of heat. What is the change
in internal energy?
(21 J)
2. The internal energy of a fixed quantity of ideal gas
depends only on its temperature. If a sample of an ideal
gas is allowed to expand against a constant pressure at
a constant temperature,
a) What is ΔU for the gas?
b) Does the gas do work?
c) If any heat exchanged with the surroundings?
a) 0
b) w = -P ΔV
c)no, only work
done by gas is energy leaving the system. Internal energy should decrease, so the temp;
but temp is const.; therefore the int energy does not change; gas has to absorb enough
heat from surroundings to compensate for the work. Q = -w; ΔU = -w+w
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19.1 First Law of Thermodynamics
 1st Law: Energy can be neither created nor destroyed,
but it can be converted from one form to another or
transferred from a system to the surroundings or vice
versa. Energy of the universe is constant
 Important concepts from thermochemistry
 Enthalpy
 Hess’s law
 Purpose of 1st Law
 Energy bookkeeping
 How much energy?
 Exothermic or endothermic?
 What type of energy?
 Δu = q + w
 q = heat; w = work system does on the
surroundings (-PΔV)
5
19.2 Spontaneous Processes: Expansion
 Spontaneous processes are
those that can proceed without
any outside intervention. Product
favored at equilibrium.
 Product-favored at equilibrium
 May be fast or slow
 May be influenced by
temperature
 The gas in vessel B will
spontaneously effuse into vessel
A, but once the gas is in both
vessels, it will not spontaneously
separate
6
19.1 Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in the
reverse direction.
Examples: rusting,
neutralization reaction,
dissolution of sugar in
water, heat flow,
expansion of gas,
spontaneous combustion
(CH4 + O2), reaction of
sodium with water
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19.1 Spontaneous Processes
 Processes that are spontaneous at one
temperature may be nonspontaneous at other
temperatures.
 Above 0C it is spontaneous for ice to melt.
 Below 0C the reverse process is spontaneous.
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19.2 Spontaneous vs. Nonspontaneous
3. Determine if the following processes are spontaneous or
not and if they are exothermic or endothermic.
(a) Gases expand into larger volumes at constant
temperature ________________
(b) H2O(s) melts above 0C ____________
(c) H2O(l) freezes below 0C ____________
(d) NH4NO3 dissolves spontaneously in H2O ___________
(e) Steel (iron) rusts in presence of O2 and H2O ________
(f) Wood burns to form CO2 and H2O __________
(g) CH4 gas burns to form CO2 and H2O __________
Evolution of Heat (Exothermicity) : not enough to predict
spontaneity
9
19.2 Spontaneous vs Nonspontaneous
 Nonspontaneous process
 Does not occur unless there is outside assistance
(energy?)
 Reactants-favored at equilibrium
 All processes which are spontaneous in one
direction cannot be spontaneous in the reverse
direction
 Spontaneous processes have a definite direction
 Spontaneous processes are irreversible. Can be
reversed with considerable input of energy.
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19.2 Factors That Favor Spontaneity
Spontaneous Processes driven by
Enthalpy, H (Joules)
 Many, but not all, spontaneous processes
tend to be exothermic.
Entropy, S (Joules/K)
 Measure of the disorder of a system
 Many, but not all, spontaneous processes
tend to increase disorder of the system
Exothermicity favors spontaneity, but does
not guarantee it.
11
19.2 Factors That Favor Spontaneity: Enthalpy
 Example of spontaneous reaction that is not
exothermic:
NH4NO3(s) → NH4+ (aq) + NO3-(aq) ΔH = 25
kJ/mol
 Expansion of gas: energy neutral
 Phase changes: endothermic processes that
occurs spontaneously (ice to water).
 Chemical system: H2(g) + I2(g) ↔ 2HI(g)
Equilibrium can be approached from both sides
(spontaneous both ways) even though the
forward reaction is endothermic and the reverse
is exothermic.
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19.2 Spontaneity: Examples
4. Based on your experience, predict whether the
following processes are spontaneous, are
spontaneous in reverse direction, or are in
equilibrium:
(a) When a piece of metal heated to 150 ºC is
added to water at 40 ºC, the water gets hotter.
(b) Water at room temperature decomposes into
hydrogen and oxygen gases
(c) Benzene vapor, C6H6(g), at a pressure of 1 atm
condenses to liquid benzene at the normal
boiling point of benzene, 80.1 ºC.
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19.2 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.
Example: melting ice at
its melting point
14
19.2 Irreversible Processes
 Irreversible processes cannot be undone by
exactly reversing the change to the system.
Different path has to be used.
 Spontaneous processes are irreversible.
 Example: expansion a gas into vacuum.
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19.2 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,
q
T
16
19.2 Entropy
 Direct measure of the randomness or disorder of
the system.
 Related to probability
describes # of ways the particles in a system
can be arranged in a given state (position
and/or energy levels)
The most likely state – the most random
More possible arrangements, the higher
disorder, higher entropy
Ordered state – low probability of occurring
Disordered state: high probability of occurring
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19.2 Entropy on the Molecular Scale
 Ludwig Boltzmann described the concept of
entropy on the molecular level.
 Temperature is a measure of the average
kinetic energy of the molecules in a sample.
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19.2 Entropy on the Molecular Scale
 Molecules exhibit several types of motion:
 Translational: Movement of the entire molecule from
one place to another.
 Vibrational: Periodic motion of atoms within a molecule.
 Rotational: Rotation of the molecule on about an axis or
rotation about  bonds.
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19.2 Entropy on the Molecular Scale
 Boltzmann envisioned the motions of a sample of
molecules at a particular instant in time.
 This would be akin to taking a snapshot of all the
molecules.
 He referred to this sampling as a microstate of the
thermodynamic system.
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19.2 Entropy on the Molecular Scale
 Each thermodynamic state has a specific number of
microstates, W, associated with it.
 Entropy is
S = k lnW
where k is the Boltzmann constant, 1.38  1023 J/K;
W: number of microstates
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19.2 Entropy on the Molecular Scale
The change in entropy for a process,
then, is
S = k lnWfinal  k lnWinitial
lnWfinal
S = k ln
lnWinitial
• Entropy increases with the number of
microstates in the system.
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19.2 Spontaneous Processes: Dispersal of Matter
 Isothermal (constant temperature) expansion of gas
Two molecules present:
25% probability
After opening stopcock the molecules could be in any
arrangement shown (4 arrangements)
Probability for each arrangement = (1/2)2
S = k (ln(4)
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19.2 Spontaneous Process: Isothermal Gas Expansion
Consider why gases tend to isothermally (constant
temp.) expand into larger volumes.
Gas Container = two bulbed flask
Ordered State
Gas Molecules
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19.2 Spontaneous Process: Isothermal Gas
Expansion
Gas Container
Ordered State
S = k ln (W) = k (ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K
For 3 particles, probability = (1/2)3
For N particles, probability = (1/2)N
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Disordered States
More probable that the gas molecules will disperse
between two halves than remain on one side 26
Disordered States
Driving force for expansion is entropy (probability);27gas
molecules have a tendency to spread out
Disordered States
S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K
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Total Arrangements
Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) =
2.9 x 10-23 J/K
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19.2 Entropy on the Molecular Scale
The number of microstates and,
therefore, the entropy tends to increase
with increases in
Temperature.
Volume.
The number of independently moving
molecules.
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19.2 Entropy and Temperature
(a) A substance at a higher temperature has greater
molecular motion, more disorder, and greater entropy than
(b) the same substance at a lower temperature.
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19.2 Entropy and Physical States
 Entropy increases with
the freedom of motion
of molecules.
 Therefore,
S(g) > S(l) > S(s)
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19.2 Entropy Changes
 In general, entropy
increases when
 Gases are formed from
liquids and solids.
 Liquids or solutions are
formed from solids.
 The number of gas
molecules increases.
 The number of moles
increases.
33
19.2 Entropy in Temperature
34
Standard entropy, S(J/K)
50
What kind of changes are
represented here?
40
30
20
10
0
50
100
150
200
Temperature (K)
250
300
35
19.2 Solutions
Generally, when a
solid is dissolved
in a solvent,
entropy increases.
36
19.2 Patterns of Entropy Change
6. Describe in words the entropy of the system
37
19.2 Entropy
Like total energy,
E, and enthalpy, H,
entropy is a state function.
Therefore,
S = Sfinal  Sinitial
 S > 0 represents increased randomness
or disorder
Note: The magnitude of change in
entropy depends on temperature.
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19.2 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:
qrev
S =
T
Units: Joule/K
39
19.2 Entropy: Example
5. ΔS = q/T
The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point of
mercury is -38.9 ºC, and its molar enthalpy of
fusion is ΔHfusion = 2.331 kJ/mol. What is the
entropy change when 50.0 g of Hg(l) freezes at
the normal freezing point?
(-2.48 J/K)
40
19.2 Entropy: Examples
7. Predict if ΔS increases, decreases or
does not change
(a) Freezing liquid mercury
(b) Condensing H2O(vapor)
(c) Precipitating AgCl
(d) Heating H2(g) from 60.0 ºC to 80 ºC
(e) Subliming iodine crystals
(f) Rusting iron nail
41
19.2 Entropy - Examples
8.
Predict which substance has the higher entropy:
a) NO2(g) or N2O4(g)
b) I2(g) or l2(s)
9.
Predict whether each of the following leads to increase
or decrease in entropy of a system If in doubt, explain
why.
a) The synthesis of ammonia:
N2(g) + 3H2(g) ↔ 2NH3(g)
b) C12H22O11(s)
→
C12H22O11(aq)
c) Evaporation to dryness of a solution of urea,
CO(NH2)2 in vapor.
CO(NH2)2(aq) → CO(NH2)2(s)
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19.3 Second Law of
Thermodynamics
43
19.3 Second Law of Thermodynamics: System
10. Predict the sign of ΔS0 for each of the
following reactions:
a) Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s)
b) MgCO3(s) → MgO(s) + CO2(g)
d) H2(g) + Br2(g) → 2HBr(g)
44
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.
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
For nonspontaneous Process:
ΔS univ= ΔS syst. +
ΔS surr. <0
45
Second Law of Thermodynamics
These last truths mean that as a result of
all spontaneous processes the entropy of
the universe increases.
46
19.3 Second Law: Entropy Changes
EQUILIBRIUM PROCESSES (reversible)
♦ ΔS universe= ΔS syst. + ΔS surr. =0
♦ ΔS syst = ΔS surr
♦ ΔS syst = ΔSº
0
ΔS
final
initial
47
19.3 Entropy Changes in a System
(Reactions)
 Entropy changes in a system
aA + bB → cC + dD
 Standard entropy change ΔSº (25 ºC, 1atm).
 Only changes in entropy can be measured.
 Each element has an entropy value (compare to
enthalpy).
 Absolute value for each substance can be determined.
 For a chemical system:
S° = nS°(products) - mS°(reactants)
where n and m are the coefficients in the balanced
chemical equation.
Standard Molar entropy, S0, is the entropy of one mole of a
substance in its standard state (298 K)
48
19.2 Standard Entropies
These are molar entropy
values of substances in
their standard states.
Standard entropies tend
to increase with
increasing molar mass.
49
19.2 Standard Entropies
Larger and more complex molecules have
greater entropies.
50
19.3 Second Law: Example
11. Using standard molar entropies,
calculate S°rxn for the following reaction
at 25°C:
2SO2(g) + O2(g) → 2SO3(g)
S° =
248.1
205.1
256.6 (J · K-1mol-1)
(Ans.: -187.9 J/K)
51
19.3 Entropy of Reactions (System)
12. Using thermodynamic tables, calculate the
standard entropy changes for the following
reactions at 25 ºC
a) Evaporation of 1.00 mol of liquid ethanol to
ethanol vapor.
b) The oxidation of one mole pf ethanol vapor
(combustion reaction)
c) Are the reactions spontaneous under the
given conditions.
Answers:
a) 122.0 J/K
b) 96.09 J/K
52
19.3 Entropy Changes in the System
In a reaction
More gas molecules produced: entropy
increases
Less gas molecules produced: entropy
decreases
No net change of # of gas molecules
produced: entropy changes, but slightly
Liquid, solid products: hard to estimateneeds calculations
53
19.3 Entropy of Surrounding
(a) When an exothermic reaction occurs in the system
(ΔH < 0), the surroundings gain heat and their
entropy increases (Δ Ssurr > 0).
(b) When an endothermic reaction occurs in the system
(Δ H > 0), the surroundings lose heat and their
54
entropy decreases (Δ Ssurr < 0).
19.3 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. (negative sign needed to make entropy
positive in an exothermic process)
55
19.3 Second Law of Thermodynamics (System)
13. Examples:
 H2O(s) melts above 0C (endothermic). What about
entropy?
 Steel (iron) rusts in presence of O2 and H2O
(exothermic). What about entropy?
 4Fe(s) + 3O2(g)  2Fe2O3(s)
 CH4 gas burns to form CO2 and H2O (exothermic).
What about entropy?
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
 Each process increases entropy of the universe.
56
19.3 Second Law of Thermodynamics
 Consider transferring the same amount of heat to two different
systems, one at 298K and another at 500K.
Q
298K
q rev
S 
T
Q
500K
q
q

298K 500 K
 More entropy (disorder) is created in the system at lower temp.
 At high temperatures the system is already disordered
59
19.3 Second Law of Thermodynamics
 To determine Suniv for a process, both Ssystem and
Ssurroundings need to be known:
 Ssystem
 related to matter dispersal in system
 Ssurroundings
 determined by heat exchange between system and
surroundings and T at which it occurs
 Sign of Ssurr depends on whether process in
system is endothermic (Ssurr< 0) or exothermic
(Ssurr >0)
Magnitude of Ssurr depends on T
 Ssurr = -Hsystem/T
61
19.3 Second Law of Thermodynamics
14. Reaction: N2(g) + 3H2(g) → 2NH3(g)
ΔHº = -92.6kJ
ΔSsys = -199 J/K at 25 ºC
ΔSsurr = -(-92.6 x1000)J/298K = 311 J/K
ΔSuniv = -199 J/K + 311 J/K = 112 J/K
Reaction spontaneous at 25 ºC
62
19.3 Conditions for Spontaneity
ΔSsystem ΔS surr
ΔSuniv Spontaneity
+
+
+
Yes
-
-
-
+
-
?
No, reaction towards
reactants
Yes, if ΔSsys> ΔSsurroun
-
+
?
Yes, if ΔSsurr> ΔSsyst
63
19.3 Entropy and Spontaneity
15. Consider the vaporization of liquid water to
steam at a pressure of 1 atm.
a) Is this process endothermic or exothermic,
explain.
b) In what temperature range is the process
spontaneous?
c) In what temperature range is it a
nonspontaneous process?
d) At what temperature are the two phases at
equilibrium?
64
19.3 Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
Perfect crystal: its internal arrangement is
absolutely regular. Nothing is in motion
(vibrations, rotations and translations)
65
19.3
Entropy
and
Third
Law
Third Law of Thermo
Gives us a starting point, S at 0K is equal to zero.
All others must be >0.
Standard Entropies Sº ( at 298 K and 1 atm) of
substances are listed.
Products - reactants to find Sº (a state function)
More complex molecules higher Sº.
66
19.4 Gibbs Free Energy
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.
67
19.4 Gibbs Free Energy
The universe is composed of the system
and the surroundings.
Therefore,
Suniverse = Ssystem + Ssurroundings
For spontaneous processes
Suniverse > 0
68
19.4 Gibb’s Free Energy
 This becomes:
Suniverse = Ssystem +
Hsystem
T
Multiplying both sides by T,
TSuniverse = Hsystem  TSsystem
TΔS universe is defined as the Gibbs free
energy, G.
When Suniverse is positive, G is negative.
Therefore, when G is negative, a process is
spontaneous.
69
19.4 Gibb's Free Energy at any Conditions
G=H-TS
Never used this way.
At constant temperature
G=Hsys –TSsys
G function eliminates the need to deal with
entropy of the surroundings
If G is negative at constant T and P, the process
is spontaneous.
We deal only with the SYSTEM.
70
19.4 Gibbs Free Energy
 The Gibbs Free Energy is a measure of the maximum
amount of work, at a given temperature and pressure, that
can be done on the surroundings by a system.
 Never really achieved because some of the free energy is
changed to heat during a change, so it can’t be used to do
work.
Wmax = G
 For a spontaneous process:
 Maximum amount of energy released by the system
that can do useful work on the surroundings
 Energy available from spontaneous process that can
be used to drive non-spontaneous process.
 For a nonspontaneous process:
 Minimum amount of work that must be done to force
the process to occur
71
19.4 Gibbs Free Energy
 Summary of Conditions for Spontaneity
 G < 0
 reaction is spontaneous in the forward direction
(Suniv > 0)
 G > 0
 reaction is nonspontaneous in the forward
direction
(Suniv < 0)
 G = 0
 SYSTEM IS AT EQUILIBRIUM
(Suniv = 0)
Remember- Spontaneity tells us nothing about rate.
72
19.4 Gibbs Standard Energy Free Energy
The standard free energies of formation, G°, are
the free energy values for the formation of a
substance under standard conditions.
The standard free energy of formation for any
element in its standard state is zero. (Like
enthalpy, but not entropy)
Compare:
G = Hsys - TSsys (any conditions)
Gº = H0sys - TS0sys (standard conditions)
73
19.4 Convention for Standard States
State of Matter
Standard State
Gas
1 atm pressure
Liquid
Pure liquid
Solid
Pure solid
Elements
Gºf = 0
Solution
1 molar
74
19.4 Free Energy in Reactions
 Gº = standard free energy change.
 Free energy change that will occur if reactants in their
standard state turn to products in their standard state.
 Can’t be measured directly, can be calculated from other
measurements.
The reaction:
aA + bB → cC + dD
or
Gºrxn = ΣnGº (products) - ΣmGº (reactants)
 Gº=Hº-TSº
 Use adapted Hess’s Law with known reactions.
75
19.4 Free Energy Changes
 At temperatures other than 25°C,
G ° = H   TS 
 How does G  change with temperature?
 There are two parts to the free energy equation:
 H — the enthalpy term
 TS  — the entropy term
 The temperature dependence of free energy,
then comes from the entropy term.
76
19.4 Gibbs free Energy: Example
16. For a particular reaction, Hrxn = 53 kJ and
Srxn = 115 J/K. Is this process spontaneous a)
at 25°C, and b) at 250°C? (c) At what
temperature does Grxn = 0?
Look at the equation G=H-TS
Spontaneity can be predicted from the sign of
H and S.
(ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c)
460.9 K or 188C)
77
19.4 Gibbs Free Energy and Spontaneity
Effect of Temperature on the Spontaneity of Reactions
ΔH ΔS -|TΔS|
1
2
3
4
+
-
+
+
-
-
+
ΔG = ΔH-TΔS
Reaction
Characteristics
Example
Always negative
Spontaneous at all
temperatures
2O3(g)
→3O2(g)
+
Always
positive
Nonspontaneous at 3O2(g)
all temperatures
→2O3(g)
+
Negative at low T; Spontaneous at low H2O(l)
positive at high T T; nonspontaneous →H2O(s)+
at high T. Enthalpy
driven
-
-
Positive at low T;
negative at high T
Nonspontaneous at H2O(s) →
low T; becomes
H2O(l)
spontaneous at
high T. Entropy
driven.
78
19.4 Gibbs Free Energy and Spontaneity
Free energy change as a function of
temperature.
79
19.4 Gibbs Free Energy and Temperature
Spontaneous Processes
H2O(s) melts above 0C (endothermic)
 H > 0 (endo process), S > 0
 ENTROPY DRIVEN
Steel (iron) rusts in presence of O2 and H2O at
25 C (exothermic)
 4Fe(s) + 3O2(g)  2Fe2O3(s)
 H < 0 (Exo) , S < 0 (entropy decreases)
 ENTHALPY DRIVEN
G < 0 for each process
T determines sign of G: G = H -TS
80
19.4 Gibbs Free Energy: Example (p. 748)
17. Predict which of the four cases in Table (slide #78)
you expect to apply to the following reactions:
a) C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(g) ΔH = -2540
kJ
b) Cl2(g) → 2Cl(g)
18. Calculate ΔG0 at 298 K for the reaction
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
ΔH0 = 114.4 kJ
a) using Gibbs free Energy equation
b) from standard free energies of formation.
( -76.0 kJ)
81
19.4 Gibbs free Energy: Example
18. For the reaction
SO2(g) + 2H2S(g) →3S(s) + 2H2O(g)
Calculate the temperature at which ΔG0 = 0
Values of ΔH0, kJ/mol
ΔS0 (kJ/(K mol)
SO2
-296.8
0.2481
H2S
-20.6
0.2057
S
0.0
0.0318
H2O
-241.8
+0.1887
Answer: 780K
82
19.4 Temperature and Chemical Reactions –
Problem Set
19. Calculate the temperature at which the reaction:
CaCO3(s) → CaO(s) + CO2(g) becomes spontaneous.
ΔS º = 160.5 J/K; ΔH º = 177.8 kJ
Answer: 835 ºC
20. At its normal boiling point, the enthalpy of vaporization
of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What
should be its approximate normal boiling point
temperature be? ΔS0 of vaporization is 87 J mol-1 K-1
(570 K)
83
19.4 Gibbs Free Energy
21. Consider the reaction
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH0 = -2220 kJ
(a) Without using data from the Thermodynamic Tables,
predict whether ΔG0 for this reaction is more negative, or
less negative than ΔH0.
(b) Use data from the Tables to calculate ΔG0 at 298 K. Is
your prediction correct?
Values of ΔG0 (in kJ/mol): C3H8(g), -23.5; O2(g), 0;
CO2(g) -394.4; H2O(l) -237.2
(a) less negative; (b) -2108 kJ
84
19.5 Gibbs free Energy and
Equilibrium
85
19.5 Free Energy and Equilibrium
 ΔG and ΔGº are not the same.
ΔG = ΔH – T ΔS
ΔGº = ΔH º – T ΔS º
 ΔGº is only at standard conditions (values
obtained from Tables)
ΔGº = Gº(products) - ΔGº (reactants)
 ΔG any conditions (no Tables of values
available)
86
19.5 Free Energy and Equilibrium
Predicament: we start a reaction with all
reactants in standard state (1 atm, 25ºC, 1M
solution). Is the standard state preserved as
the reaction progresses?
It can be shown mathematically that at nonstandard conditions:
ΔG = ΔGº +RTlnQ
[products]
Q, reaction quotient 
[reac tan ts]
87
K<Q
K>Q
K=Q
If ΔG is negative, the forward reaction is spontaneous.
If Δ G is 0, the system is at equilibrium.
If G is positive, the reaction is spontaneous in the reverse
direction.
88
19.5 Free Energy and the Equilibrium Constant
 At equilibrium:
Useful equation to
determine small Keq
0  G   RT ln K eq
G    RT ln K eq
K = e-(ΔGº /RT)
From the above we can conclude:
If G  < 0, then K > 1
Product favored
If G  = 0, then K = 1
Equilibrium
If G  > 0, then K < 1
Reactants favored
Kc used for solutions and molarities, Kp used for gases
89
19.519.5
FreeTemperature
Energy and Dependence
Chemical Equilibrium
of K
Gº= -RTlnK = Hº - TSº
ln(K) = Hº/R(1/T)+ Sº/R
A straight line of lnK vs 1/T
Slope?
Y-intercept?
90
19.5 Reaction Path and ΔGº
ΔGº = -RTln K
Value of ΔGº
Sign of K
Path of reaction
Negative, <0
K>0
Products are
favored
Positive, >0
K<0
Reactants are
favored
zero, =0
K=1
Equilibrium
91
19.5 Keq: Problem Set – 19.6 (p. 755)
22. Determine the value of Keq at 25 ºC for
the reaction:
2NO2(g) ↔ N2O4(g).
Note: calculate ΔG0 from tables
Use ΔG0 = -RTln Kp
( 6.9)
92
19.5 Keq: Problem Set 19.7 (p. 755)
23. Using the solubility product of of silver iodide at
25 ºC (8.5 x 10-17), calculate ΔGº for the
process:
AgI(s) ↔ Ag+(aq) + I-(aq)
Answer:
24. Estimate the value of ΔS0298 for the dissociation
of copper (II) oxide.
CuO(s) ↔ 2Cu2O(s) + O2(g) ΔH0298 = 283 kJ
( 0.203 kJ K-1
93
19.5 Gibbs and Equilibrium: Example
25. Calculate Grxn for the reaction below:
2A(aq) + B(aq)  C(aq) + D(g)
if G°rxn = 9.9 x 103 J/mol and 150 °C
(a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05 M, and
PD = 0.05 atm, and
(b) (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5 M, and
PD = 0.5 atm.
(c) Is the reaction spontaneous under these
conditions?
94
19.5 Gibbs and Equilibrium: Example
26. Calculate G°rxn for the ionization of acetic
acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this
reaction spontaneous under standard
state conditions?
(ans.: 27 kJ)
27. Calculate G° for the neutralization of a
strong acid with a strong base at 25°C. Is this
process spontaneous under these conditions?
For the reaction below, K = 1.0 x 1014.
H+ + OH-  H2O
(ans.: -80 kJ)
95
19.5 Gibbs and Equilibrium: Example
28. Calculate Keq for a reaction at (a) 25°C
and (b) 250°C if H°rxn = 42.0 kJ and
S°rxn = 125 J/K. At which temperature is
this process product favored?
(ans.: a) 0.15; b) 216; 250C)
96
19.5 Summary
First Law of Thermodynamics:
Second Law:
ΔU = Δq + Δw
ΔSuniv = ΔS sys + ΔSsurr
ΔS sys= ΔH/T
ΔG = ΔH – T ΔS
ΔG = ΔGº +RTlnK
ΔGº = -RTlnK
97
19.5 Gibbs Free Energy
 Many biological reactions essential for life are
nonspontaneous
Spontaneous reactions used to “drive” the
nonspontaneous biological reactions
 Example: photosynthesis
6CO2 + 6H2O  C6H12O6 + 6O2
G > 0
What spontaneous reactions drive
photosynthesis?
98
19.5 Entropy and Life Processes
If the 2nd law is valid, how is the
existence of highly-ordered, sophisticated
life forms possible?
growth of a complex life form
represents an increase in order (less
randomness)
 lower entropy
99
19.5 Entropy and Life Processes
Organisms “pay” for
their increased order by
increasing Ssurr.
heat
CO2
Over lifetime,
Suniv > 0.
H2O
100
102