Transcript Chapter 11 Spontaneous Change and Equilibrium

Chapter 11
Spontaneous Change and
Equilibrium
• 11-1 Enthalpy and Spontaneous Change
• 11-2 Entropy
• 11-3 Absolute Entropies and Chemical
Reactions
• 11-4 The Second Law of Thermodynamics
• 11-5 The Gibbs Function
• 11-6 The Gibbs Function and Chemical
Reactions
• 11-7 The Gibbs Function and the
Equilibrium Constant
• 11-8 The Temperature Dependence of
Equilibrium Constants
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Chapter 11
Spontaneous Change and
Equilibrium
• Second Law of Thermodynamics
In a real spontaneous process the
Entropy of the universe (meaning the
system plus its surroundings) must
increase.
Suniverse > 0 (spontaneous process)
Suniverse = Ssystem + Ssurroundings > 0
• if Suniverse = 0, then everything is in
equilibrium
• The 2nd Law of Thermodynamics
profoundly affects how we look at
nature and processes
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Summarize a few Concepts
• 1st Law of Thermodynamics
– In any process, the total energy of the
universe remains unchanged: energy is
conserved
– A process and its reverse are equally
allowed
Eforward = – Ereverse
(conservation of energy)
• 2nd Law of Thermodynamics
– S, the entropy of a universe, increases
in only one of the two directions of a
reaction
– Processes that decrease ΔS are
impossible. Or improbable beyond
conception
ΔSuniv > 0
Spontaneous
ΔSuniv = 0
Equilibrium
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ΔS
univ < 0
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Non-spontaneous
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Gibbs Function
•
How are Enthalpy and
Entropy related?
G = H - T•S
•
G has several names
1. Gibbs function
2. Gibbs free energy
3. Free Enthalpy
•
For the change in the Gibbs
Energy of system, at constant
Temperature and Pressure
ΔGsys = ΔHsys - T·ΔSsys
ΔGsystem = Δ Hsys - TΔSsystem = < 0
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(spontaneous
process)
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For a change at constant temperature
and pressure
ΔGsys < 0
Spontaneous
ΔGsys = 0
Equilibrium
ΔGsys > 0
Non-spontaneous,
but the reverse is
spontaneous
From Earlier
ΔSuniv > 0
Spontaneous
ΔSuniv = 0
Equilibrium
ΔSuniv < 0
Non-spontaneous
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Typical example using Gibbs
Free energy
• Benzene, C6H6, boils at 80.1°C
and ΔHvap = 30.8 kJ
– a) Calculate ΔSvap for 1 mole of
benzene
– B) at 60°C and pressure = 1 atm
does benzene boil?
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• Benzene, C6H6, boils at 80.1°C.
ΔHvap = 30.8 kJ
– a) Calculate ΔSvap for 1 mole of benzene
Hvap
30.8 x103 J
Svap 

 87.2 JK 1
Tb
( 273.15  80.1)
– B) at 60°C and pressure = 1 atm does
benzene boil?
ΔG vap  ΔH vap - TS vap
ΔG vap  30,800J  (273K  60 C)(87.2JK1 )
ΔG vap  1749 J or  1.7kJ
ΔG vapis positive,

 benzene does not boil at 60 C and 1atm.
ΔGsys < 0 = Δ H Spontaneous
ΔG
system
sys - TΔSsystem = 0
ΔGsys = 0
Equilibrium
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process)
ΔGsys > 0 (equilibrium
Non-spontaneous
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The Gibbs Function and
Chemical Reactions
ΔG = ΔH - T·ΔS
• ΔGf° is the standard molar
Gibbs function of formation
• Because G is a State Property,
Grxno for a general reaction
aA+bB→cC+dD
ΔG  cΔG (C)  dΔG (D)
o
f
o
f
o
f
 aΔG (A)  bΔG (B)
o
f
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o
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Example 11-7
• Calculate ΔG° for the following
reaction at 298.15K. Use Appendix
D for additional information needed.
3NO(g) → N2O(g) + NO2(g)
Solution
From Appendix D
ΔGf°(N2O) = 104.18 kJ mol-1
ΔGf°(NO2) = 51.29
ΔGf° (NO) = 86.55
ΔG°= 1(104.18)+1(51.29)-3(86.55)
ΔG°= − 104.18 kJ therefore, spontaneous
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Effects of Temperature on ΔG°
For temperatures other than 298K
ΔG = ΔH - TΔS
• Typically ΔH and ΔS are almost constant
over a broad range
3NO (g) → N2O (g) + NO2 (g)
• For above reaction, as Temperature
increases ΔG becomes more positive, i.e.,
less negative.
• Next Slide
• ΔG has a strong temperature
dependence
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3NO(g) → N2O(g) + NO2(g)
G
has
a
strong
temperature
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dependence
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For temperatures other than
298K or 25C
ΔG = ΔH - T·ΔS
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For temperatures other than 298K or 25C
ΔG = ΔH - T·ΔS
B
A
D
C
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For temperatures other than 298K or 25C
ΔG = ΔH - T·ΔS
Case C
ΔH°B> 0
ΔS° < 0
A
ΔG = ΔH - T·ΔS
ΔG = (+) - T·(-) = positive
 ΔG > 0 or non-spontaneous at all Temp.
Case B
ΔH° < 0
ΔS° > 0
ΔG = ΔH - T·ΔS
D
ΔG = (-) - T·(+) = negative
C
 ΔG < 0 or spontaneous at all temp.
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For temperatures other than
298K or 25C
ΔG = ΔH - T·ΔS
The spontaneity of a reaction at different
temperatures depends on the signs of Δ H°
& ΔS°. When both signs are the same, the
temperature determines the spontaneity.
For the next two cases, it is more interesting
because there is a temperature at which
ΔG = 0
Then above or below
a temperature
the ΔG
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11
is either positive or negative
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For temperatures other than 298K or 25C
ΔG = ΔH - T·ΔS
Case D
ΔH° B< 0
A
ΔS° < 0
ΔG = ΔH - T·ΔS at a low Temp
ΔG = (-) - T·(-) = negative
 ΔG < 0 or spontaneous at low Temp.
Case A
ΔH° > 0
ΔS° > 0
ΔG = ΔH - T·ΔS
D
ΔG = (+) - T·(+) at a High Temp
C
 ΔG < 0 or spontaneous at high Temp.
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The Gibbs Function and
the Equilibrium Constant
• What about non-standard states, other
than 1 atm or a conc. [X] = 1 mol/L?
ΔG = ΔG° + RT ln Q
– Where Q is the reaction quotient

aA  bB  cC  dD
 PCc PDd 
Q   a b 
 PA PB any conditions
 PCc PDd 
K   a b 
 PA PB equilibriu m
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ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
aA+bB↔cC+dD
• If Q > K the rxn shifts towards the reactant
side
– The amount of products are too high relative to
the amounts of reactants present, and the reaction
shifts in reverse (to the left) to achieve
equilibrium
• If Q = K equilibrium
• If Q < K the rxn shifts toward the product
side
– The amounts of reactants are too high relative to
the amounts of products present, and the reaction
proceeds in the forward direction (to the right)
toward equilibrium
 PCc PDd 
Q   a b 
compare
 PA PB any conditions
 PCc PDd 
K   a b 
 PA PB equilibriu m
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ΔG = ΔG° + RT ln Q
– Where Q is the reaction quotient
aA+bB↔cC+dD
• If Q < K the rxn shifts towards the product side
• If Q = K equilibrium
• If Q > K the rxn shifts toward the reactant side
At Equilibrium conditions, ΔG = 0
- ΔG ° = RT ln K
NOTE: we can now calculate
equilibrium constants (K) for reactions
from standard ΔGf functions of
formation
c
d
[C] [D]
K
a
b
[A] [B]
o
o
o
ΔG f  cΔG f (C)  dΔG f (D)
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 aΔG (A)  bΔG (B)
o
f
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o
f
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ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
aA+bB↔cC+dD
Criteria for Spontaneity in a Chemical Reaction
Spontaneous
Processes
Equilibrium
Processes
Nonspontaneous
Processes
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGf < 0 ΔGf = 0 ΔGf > 0
Q<K
Q=K
Q>K
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Conditions
All conditions
Constant P and T
Constant P and T
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Example 11-9
• The ΔGr° for the following
reaction at 298.15K was
obtained in example 11-7.
Calculate the equilibrium
constant for this reaction at
25C.
3NO(g) ↔ N2O(g) + NO2(g)
• Strategy
Use - ΔG ° = RT ln K
Found ΔG°= - 104.18 kJ mol -1
from example 11-7
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Example 11-9
3NO(g) ↔ N2O(g) + NO2(g)
• Solution
Use
[ N 2O ]1[ NO2 ]1
18
K

1.8x10
[ NO ]3
- ΔG ° = RT ln K
Rearrange
 G 
ln K 
RT
Use ΔGr°= - 104.18 kJ mol-1
from Ex. 11-7
 (104,180J mol1 )
ln K 
 42.03
1
1
(8.3145J K mol )(298.15K)
K  antiln42.03 e
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 1.8X10
42.03
OFP
Chapter 11
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Chapter 11
Spontaneous Change and
Equilibrium
• Examples / Exercises
11-2, 11-3, 11-4, 11-5, 11-6
11-7, 11-8, 11-9, 11-10, 11-11,
11-12
• Problems WebCT
8abc
30abcd
44a
50abc
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