**Unformatted text preview: **CHAPTER 17
Entropy and Free Energy 11 Review: System and
Surroundings
• SYSTEM
– The object under study • SURROUNDINGS
– Everything outside the system Universe =
System +
Surroundings 22 Review:
Review: FIRST
FIRST LAW
LAW
OF
OF
THERMODYNAMICS
THERMODYNAMICS heat energy transferred ∆E = q + w
energy
change work done
by the
system Energy is conserved! 33 44 Review:
Review: ENTHALPY
ENTHALPY
∆H = q
where H = enthalpy
q = Heat transferred at constant P
If If ∆H
∆H > 0 process is > 0 process is ENDOTHERMIC
ENDOTHERMIC
heat transfers from SURROUNDINGS to SYSTEM.
If If ∆H
∆H < 0 process is < 0 process is EXOTHERMIC
EXOTHERMIC
heat transfers from SYSTEM to SURROUNDINGS. Spontaneous processes
• Spontaneous process is a process that occurs without an outside intervention.
• Examples: a basketball rolling down the hill;
transfer of heat from a hot object to a cold object • A spontaneous process leads the system towards equilibrium. “Spontaneous” does not imply anything about
the time it takes for process to occur. 55 66 Spontaneous processes
• Product-favored reactions
proceed spontaneously
from reactants to products.
• Most of the product-favored
reactions are exothermic.
Fe2O3(s) + 2 Al(s) --->
2 Fe(s) + Al2O3(s)
∆H = - 848 kJ Spontaneous Reactions
But not all spontaneous reactions are exothermic;
many spontaneous reactions or processes are
endothermic or have ∆H = 0. NH4NO3(s) + heat NH4NO3(aq)
∆ H= +25.7 kJ/mol 77 Entropy
• In all spontaneous processes, the final state is more DISORDERED (“RANDOM“) than the original state. Spontaneity is related to an increase in
disorder.
• ENTROPY (S) is a property that describes the degree of disorder.
• Greater disorder = higher value of S. 88 Entropy
• The concept of entropy is based on
the idea that spontaneous change
results in dispersal of matter,
dispersal of energy, or of both.
• This is a statistical effect.
• A state in which energy/matter is
more dispersed has a higher
probability of occurring, that’s why
a system spontaneously goes there. 99 10
10 Dispersal of Matter 11
11 Dispersal of
Matter
• Solubility is a result of a
dispersal of matter.
• A solid KMnO4 dissolves
in water because the ions
tend to disperse over the
entire container. 12
12 13
13 Dispersal of Energy
Another key contribution to entropy is
dispersal of energy over as many
different energy states as possible. 14
14 Dispersal of Energy
in exothermic
reactions
Exothermic reactions involve a release of
stored chemical potential energy to the
surroundings.
The stored potential energy starts out in a few
molecules but is finally dispersed over very
many molecules.
The final state—with energy dispersed—is
more probable; and so the reaction is
spontaneous. 15
15 Boltzmann’s Equation for
Entropy 16
16 S = k ln W where S is entropy,
W is the number of microstates (number of different ways of arranging
the energy of a system),
k is a constant (the Boltzmann constant), which is equal to:
k = R/NA (R = universal gas constant, NA = Avogadro’s number).
•A system with relatively few ways to distribute its energy (smaller W)
has relatively less disorder and low entropy.
•A system with many ways to distribute its energy (larger W) has
relatively more disorder and high entropy.
•The largest possible W (and highest entropy) is achieved at equilibrium. 17
17 Spontaneous expansion of a gas
stopcock closed 1 atm evacuated stopcock opened 0.5 atm 0.5 atm 18
18 Expansion of a gas and the increase in number of microstates. 19
19 Expansion of a gas and the increase in number of microstates. Entropy 20
20 ENTROPY is the thermodynamic property that
describes the degree of disorder.
Like energy (E) and enthalpy (H), ENTROPY (S)
is a state function.
“State function” means: change in entropy
( S) for any process depends only on the
initial and final states of the system, but not
on the pathway by which the process occurs.
In a spontaneous process, the entropy
increases (the 2nd law of thermodynamics) 21
21 The entropy of liquid water is greater than the entropy of solid water (ice).
S (H2O liq) > S (H2O solid) Entropy and States of
Matter Entropy of a gas
(vapor) is greater than
entropy of a liquid. 22
22 Standard Molar Entropies 23
23 24
24 Entropy
SSoo (J/K•mol)
(J/K•mol)
H
H22O(liq)
O(liq) 69.95
69.95 H
H22O(gas)
O(gas) 188.8
188.8 S (gases) > S (liquids) > S (solids)
• Units of entropy are J/K•mol
• Entropy depends on temperature.
• So is the standard entropy of a
substance (entropy at 298K) Entropy
Entropy of a substance increases
with temperature. Molecular motions
of heptane, C7H16 Molecular motions of
heptane at different temps. 25
25 Entropy
Entropy and
and Temperature
Temperature S increases
slightly with T
S increases a
large amount
with phase
changes 26
26 Entropy
Increase in molecular
complexity generally
leads to increase in S. 27
27 28
28 Entropy and vibrational motion. NO NO2 N2O4 29
29 Entropy
Entropies of ionic solids depend on
coulombic attractions.
SSoo (J/K•mol)
(J/K•mol) Mg2+ & O2- Na+ & F- MgO
MgO 26.9
26.9 NaF
NaF 51.5
51.5 Entropy
Entropy Change
Change
Entropy increases as a more ordered state changes to a more disordered state. solid
more order crystal + liquid
more order liquid gas
less order ions in solution
less order S > 0
S > 0 30
30 Entropy
Entropy Change
Change
Entropy usually increases when a
pure liquid or solid dissolves in a
solvent. 31
31 32
32 The entropy change accompanying the dissolution of a salt.
pure solid MIX pure liquid
solution 33
33 The small increase in entropy when ethanol dissolves in water. Ethanol Water Solution of
ethanol
and water 34
34 The large decrease in entropy when a gas dissolves in a liquid O2 gas O2 gas in H2O 35
35 Predicting Relative S0 Values of a System
1. Temperature changes
S0 increases as the temperature rises.
2. Physical states and phase changes
S0 increases as a more ordered phase changes to a less ordered phase.
3. Dissolution of a solid or liquid
S0 of a dissolved solid or liquid is usually greater than the S 0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.
4. Dissolution of a gas
A gas becomes more ordered when it dissolves in a liquid or solid.
5. Atomic
size or molecular complexity
In similar substances, increase in molar mass relates directly to entropy.
Increases in complexity (e.g. bond flexibility) increase the entropy. 36
36
Sample Problem 1 Entropy Comparisons PROBLEM: Choose the member with the higher entropy in each of the following
pairs, and justify your choice:
(a) 1mol of O2(g) or 1mol of O3(g)
(b) 1mol of CF4(g) or 1mol of CCl4(g)
(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)
PLAN: In general less ordered systems have higher entropy than ordered
systems.
SOLUTION:
(a) 1mol of O3(g) - more atoms, larger molecular complexity.
(b) CCl4 - larger molar mass
(c) 3mol of O2(g) - larger #mols, more randomness in arranging molecules. 37
37
Sample Problem 2 Entropy Comparisons PROBLEM: Determine the sign of S for each of the following processes:
(a) CO2(s) ---> CO2(g)
(b) KBr(s) ---> KBr(aq)
(c) 3H2(g) + N2(g) ---> 2NH3(g)
PLAN: Entropy is higher when a system is more disordered; entropy
depends on state of matter.
SOLUTION: (a) S is positive because gas is more disordered than solid;
entropy increases.
(b) S is positive because aqueous solution is more disordered
than solid crystal; entropy increases.
(c) S is negative because we went from 4 moles of gas (add
coefficients: 3+1=4) down to 2 moles of gas; some entropy is lost. Entropy changes in chemical
reactions
The standard entropy change during a reaction
(∆Srxno) is the entropy change that occurs when
all reactants and products are in their standard
states.
oo
oo
oo
∆∆SSrxn
=
S
(products) S
(reactants)
= S
(products) S
(reactants)
rxn ∆Srxno = the sum of standard entropies (So) of the
products minus the sum of standard entropies
of reactants.
Take values of So for each involved substance
from reference table (appendix L). 38
38 Calculating ∆S for a
Reaction
o oo
oo
oo
∆∆SSrxn
= S
(products) S
(reactants)
= S
(products) S
(reactants)
rxn Ex. Calculate ∆Srxno for reaction: 2 H2(g) + O2(g) 2 H2O(l)
∆So = 2So(H2O) - [2So(H2) + So(O2)]
∆So = 2 mol (69.9 J/K•mol) [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]
∆So = -326.9 J/K
Note that there is a decrease in S because 3 mol of gas
give 2 mol of liquid. 39
39 40
40
Sample Problem 3
PROBLEM: Calculating the Standard Entropy of Reaction, S0rxn Calculate S0rxn for the combustion of 1mol of propane at 250C.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) PLAN: Use summation equations. Entropy is being lost because the
reaction goes from 6 mols of gas to 3 mols of gas.
SOLUTION: Find standard entropy values in the Appendix 4. S = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]
S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) +
(5 mol)(205.0J/mol*K)]
S = - 374 J/K Entropy Change
Change of entropy during a reversible
process is: ∆S = qrev/T
where qrev = heat absorbed in a reversible
process
T = temperature (in K) at which the
process occurs 41
41 Entropy Change
∆S = qrev/T
What is a reversible process?
• At every step along a reversible
pathway between the two states, the
system remains at equilibrium.
• Spontaneous processes are not
reversible, because they proceed in one
direction and their pathways involve
non-equilibrium conditions.
• Adding heat slowly and in very small
increments approximates a reversible
process. 42
42 Entropy Change
∆S = qrev/T
Examples of reversible processes:
1) melting of ice;
2) boiling of
water.
For H2O (l) H2O(g)
q = ∆H = +40,700 J/mol qq
40,
700 J/mol
40,
700 J/mol = + 109 J/K • mol
S = = S = = = + 109 J/K • mol
TT
373.15 K
373.15 K 43
43 Entropy Changes and
Spontaneity
We mentioned that spontaneity is related to
increase of entropy.
More specifically, In any spontaneous
process, the entropy of the
universe increases.
This is the 2nd law of thermodynamics. 44
44 2nd Law of
Thermodynamics
A reaction is spontaneous if ∆S for the
universe is positive. ∆Suniverse > 0 for spontaneous process
If ∆Suniverse < 0, the process cannot be spontaneous;
If ∆Suniverse = 0, the system is at equilibrium. ∆Suniverse = ∆Ssystem + ∆Ssurroundings 45
45 2nd Law of
Thermodynamics
∆Suniverse = ∆Ssystem + ∆Ssurroundings
We will see that ∆Ssystem is change of
entropy created by dispersal of matter
∆Ssurround is change of entropy created by
dispersal of energy 46
46 Entropy of Surroundings
∆Suniverse = ∆Ssystem + ∆Ssurroundings
During a reaction, the change in entropy of the
system (∆Ssys ) is: ∆Ssys = ∆Srxn ∆∆SSsys
= S (products) - S (reactants)
sys = S (products) - S (reactants)
But how do we get the change of entropy of surroundings? Ssurr qΔH surr =
T surr T Ssurroundings = - = ΔHsys Hsystem
T T 47
47 Entropy of Surroundings
∆Suniverse = ∆Ssystem + ∆Ssurroundings
Ssurroundings = - Hsystem
T The change in entropy of the surroundings is
directly related to a change in the enthalpy of
the system with reversed sign.
For exothermic reactions Hsystem < 0, so Ssurroundings > 0
For endothermic reactions Hsystem > 0, so Ssurroundings < 0 48
48 2nd Law of
Thermodynamics
= ∆S
+ ∆S ∆Suniverse system surroundings Dissolving NH4NO3 in water—an entropy driven process.
Occurs because of dispersion of matter. 49
49 2nd Law of
Thermodynamics 2 H2(g) + O2(g) ---> 2 H2O(l)
∆Sosystem = -326.9 J/K qqsurroundings
H
Hsystem
surroundings
system == == TT
TT oo
surroundings
surroundings S
S Calculate ∆Hosystem = ∆Horxn = -571.7 kJ (571.7 kJ)(1000 J/kJ) == (571.7 kJ)(1000 J/kJ)
298.15 K
298.15 K oo
surroundings
surroundings S
S ∆Sosurroundings = +1917 J/K 50
50 2nd Law of
Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(l)
∆Sosystem = -326.9 J/K
∆Sosurroundings = +1917 J/K ∆Souniverse = ∆Sosystem + ∆Sosurroundings =+1590 J/K
Is the reaction spontaneous in the forward direction? • The entropy of the universe is increasing, so
the reaction is spontaneous (product-favored). 51
51 52
52 Spontaneous or Not? Remember
is proportional to –∆H˚ sys
Remember that
that ∆S˚
∆S˚surr
surris proportional to –∆H˚sys
An
> 0.
An exothermic
exothermic process
process has
has ∆S˚
∆S˚surr
surr > 0. Components of S0universe for spontaneous reactions 53
53 ∆Suniverse = ∆Ssystem + ∆Ssurroundings
exothermic
endothermic system becomes more disordered
exothermic
system becomes more disordered system becomes more ordered 54
54
Sample Problem 4
PROBLEM: Determining Reaction Spontaneity At 298K, the formation of ammonia has a negative S0sys;
N2(g) + 3H2(g) 2NH3(g) S0sys = -197 J/K Calculate S0universe, and state whether the reaction occurs
spontaneously at this temperature.
PLAN: S0universe = S0sys + S0surr
To find S0surr, first find Hsys; Hsys= Hrxn which can be calculated
using H0f values from tables in Appendix 4.
S0universe must be > 0 in order for this reaction to be spontaneous.
SOLUTION:
H0rx = [(2 mol)(H0fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0fH2)]
H0rx = -91.8 kJ
S0surr = -H0sys/T = -(-91.8x103J/298K) = 308 J/K
S0universe = S0surr + S0sys = 308 J/K + (-197 J/K) = 111 J/K S0universe > 0 so the reaction is spontaneous. 3rd Law of
Thermodynamics
Where is a reference point for entropy?
The third law of thermodynamics
says: Ssystem = 0 at 0 K A perfect crystal at a temperature of absolute
zero has no disorder (entropy = 0)
So 0 for elements in standard states 55
55 Gibbs Free Energy 56
56 ∆Suniv = ∆Ssurr + ∆Ssys
It would be more convenient
to deal with system only,
not with surroundings. Suniv = Hsys
T + Ssys Multiply through by –T :
-T∆Suniv = ∆Hsys - T∆Ssys J. Willard Gibbs
1839-1903 Gibbs defined the
free energy G for
the system as
G = H - TS Free Energy 57
57 -T∆Suniv = ∆Hsys - T∆Ssys
Since G = H – TS, then: ∆Gsys = ∆Hsys - T∆Ssys
It follows that ∆Gsystem = -T∆Suniv Change of free energy of the system
is the criterion of spontaneity
For any spontaneous
process,∆Gsystem< 0
Under standard conditions —
o
∆Gosys = ∆Hohere,
sys - T∆S sys
T is temperature in Kelvin Free Energy, G G, the change in the free energy of a
system, is a measure of the spontaneity of
the process and of the useful energy
available from it.
G0system = H0system - TS0system G < 0 for a spontaneous process G > 0 for a nonspontaneous process G = 0 for a process at equilibrium 58
58 Free Energy
Gibbs free 59
59
o ∆Go = ∆Ho -T∆S energy change = total energy change for system (∆ Ho)
- energy lost in disordering the system ( T∆So) 1) If reaction is
• exothermic (negative ∆ Ho)
(energy dispersed)
• and entropy increases (positive ∆So)
(matter dispersed)
• then ∆Go is always NEGATIVE
• reaction is spontaneous (and product-favored). Free Energy 60
60
o ∆Go = ∆Ho -T∆S 2) If reaction is • endothermic (positive ∆Ho)
• and entropy decreases (negative ∆So)
• then ∆Go must be POSITIVE • reaction is not spontaneous
favored). (reactant- 3) If signs of ∆Ho and ∆So are the same, reaction
may be spontaneous depending on temperature 61
61
o Free Energy ∆Go = ∆Ho -T∆S ∆Ho ∆So ∆Go Reaction exo(–) increase(+) – Spontaneous endo(+) decrease(-) + not spontaneous exo(–) decrease(-) ? T dependent endo(+) increase(+) ? T dependent The effect of entropy change (sign of ∆So) becomes
a dominant factor at higher temperatures. Free Energy H0 62
62
o ∆Go = ∆Ho -T∆S Description S0 G0 + + + or - Spontaneous at higher T;
nonspontaneous at lower T - - + or - Spontaneous at lower T;
nonspontaneous at higher T Calculating ∆Gorxn 63
63 Combustion of acetylene at 298 K
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate ∆Horxn = -1238 kJ
Use standard molar entropies to calculate ∆Sorxn = -97.4 J/K or -0.0974 kJ/K
∆Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is spontaneous in spite of negative
∆Sorxn.
Reaction is “enthalpy driven” Calculating ∆Gorxn NH4NO3(s) + heat ---> NH4NO3(aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven? 64
64 Calculating ∆Gorxn
NH4NO3(s) + heat ---> NH4NO3(aq)
From tables of thermodynamic data we find
∆Horxn = +25.7 kJ
∆Sorxn = +108.7 J/K or +0.1087 kJ/K
∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is spontaneous in spite of positive
∆Horxn.
Reaction is “entropy driven” 65
65 Sample Problem 5
PROBLEM: 66
66
Calculating G0 from Enthalpy and Entropy Values Potassium chlorate undergoes a disproportionation reaction when
heated: 4KClO3(s)
3KClO4(s) + KCl(s)
Use H0f and S0 values to calculate G0sys (G0rxn) at 250C for
this reaction. PLAN: Use Appendix L to find thermodynamic values (H0f and S0); place
them into the Gibbs Free Energy equation and solve. Make sure H0
and TS0 are both in the same units. SOLUTION: H0rxn = H0products - H0reactants
H0rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) (4 mol)(-397.7 kJ/mol)
H0rxn = -144 kJ Sample Problem 5 67
67
Calculating G0 from Enthalpy and Entropy Values continued
S0rxn = S0products - S0reactants
S0rxn = (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) (4 mol)(143.1 J/mol*K)
S0rxn = -36.8 J/K
G0rxn = H0rxn - T S0rxn H0rxn = -144 kJ Make sure units are consistent !!
S0rxn = -36.8 J/K = (-36.8 J/K) x (1kJ/1000J) = -0.0368 kJ/K
G0rxn = -144 kJ - (298K)(-0.0368 kJ/K) = -144 kJ + 11.0 kJ
G0rxn = -133 kJ Free Energy, G 68
68 Two methods of calculating ∆Go
a) Determine ∆Horxn and ∆Sorxn and use Gibbs equation
(∆Go = ∆Ho - T∆So)
b) Use tabulated values of free energies of formation, ∆Gfo.
oo
oo
∆∆G
= ∆G
(products) ∆G
Goorxn
= ∆G
(products) ∆G
(reactants)
ff
f f (reactants)
rxn The method b) works only at 298 K. 69
69 Free Energy of Formation
The standard free energy of formation of a compound,
(∆G˚f ) is the free energy change when forming 1 mole
of the compound from the component elements, with
all involved substances in their standard states. Note that ∆G˚f for an element = 0 70
70 oo
∆∆G
G rxn
rxn Calculating ∆Gorxn
== ∆G
∆G (products)
(products) -- ∆G
∆G (reactants)
(reactants)
oo ff oo ff Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
∆Gorxn = ∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)]
∆Gorxn = -394.4 kJ - [ 0 + 0]
Remember that free energy of formation of an
element in its standard state is 0.
∆Gorxn = -394.4 kJ
Reaction is spontaneous as expected. 71
71
Sample Problem 6 Calculating G0rxn from G0f Values PROBLEM: Use G0f values to calculate Grxn for the reaction of
disproportionation of potassium chlorate: 4KClO3(s)
3KClO4(s) + KCl(s) PLAN: Use the G summation equation. Remember to multiply
values of G0f from appendix 4 by corresponding number
of moles in the chemical equation. SOLUTION: G0rxn = G0products - G0reactants
G0f values: of KCl (s) = -408.8 kJ/mol
of KClO3 (s) = -296.3 kJ/mol
of KClO4 (s) = -303.2 kJ/mol
G0rxn = (3mol)(-303.2kJ/mol) + (1mol)(-408.8kJ/mol) (4mol)(-296.3kJ/mol)
G0rxn = -134kJ Free Energy and
Temperature 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
∆Horxn = +467.9 kJ
∆Sorxn = +560.3 J/K
∆Gorxn = +300.8 kJ at 298 K
Reaction is reactant-favored at 298 K
At what T does ∆Gorxn just change from being
(+) to being (-)?
When ∆Gorxn = 0 = ∆Horxn - T∆Sorxn
H
467.9 kJ
H
467.9 kJ = 835.1 K
rxn
rxn
T = = T = = = 835.1 K
S
0.5603 kJ/K
Srxn
0.5603 kJ/K
rxn 72
72 The effect of temperature on reaction spontaneity. 73
73 74
74
Sample Problem 7
PROBLEM: Determining the Effect of Temperature on G0 For the reaction of oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g) 2SO3(g) At 298K, G0 = -141.6kJ; H0 = -198.4kJ; and S0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 250C, and predict
how G0 will change with increasing T.
(b) Assuming H0 and S0 are constant with increasing T, is the reaction
spontaneous at 9000C?
PLAN: The sign of G0 tells us whether the reaction is spontaneous
and the signs of H0 and S0 will be indicative of the T effect.
Use the Gibbs free energy equation for part (b).
SOLUTION: (a) The reaction is spontaneous at 250C because G0 is (-).
Since H0 is (-) but S0 is also (-), G0 will become less
spontaneous as the temperature increases.
(b) G0rxn = H0rxn - T S0rxn
G0rxn ...

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