Chapter 19: Chemical Thermodynamics
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Transcript of Chapter 19: Chemical Thermodynamics
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
H2O (s) H2O (l)
at 25oC
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
CO2 (s) → CO2 (g)
at 25oC
‘dry ice’
at 25oC
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
rust
at 25oC
Fe (s) + O2 (g)Fe2O3 (s)
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
at 25oC 5 H2O (g) + 6 CO2 (g)C6H10O5 (s) + 6 O2 (g)
Burning Paper
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
H2O (s) → H2O (l) CO2 (s) → CO2 (g)
4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
at 25oC
…happen “on their own”although they sometimes require an initial push to get going
…are “product favored”occur in a definite direction: towards the formation of product
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
H2O (s) H2O (l)
below 0oC
The direction of a spontaneous processes may depend on temperature
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
H2O (l) → H2O (s)
CO2 (s) → CO2 (g)
2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)
at 25oC
• At a given temperature and pressure, processes arespontaneous only in one direction
non-spontaneous
spontaneous
non-spontaneous
• If a processes is spontaneous in one direction it is non-spontaneous in the other direction
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Which reactions are spontaneous?
• many spontaneous reactions are exothermic (H < 0)
… but not all!
• Some reactions are endothermic (H > 0) and still spontaneous
NH4NO3 (s) → NH4+ (aq) + NO3
- (aq)
H > 0
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Entropy can be thought of as a measure of disorder
Ludwig Boltzmann (1844-1906)
S = k log W
k = 1.38 x 10-23 J / K
W = Wahrscheinlichkeit (probability)
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
The change in entropy for any process is:
S = Sfinal - Sinitial
What is the sign of S for the following processes at 25oC ?
H2O (s) → H2O (l)
CO2 (g) → CO2 (s)
S > 0 (positive)
S < 0 (negative)
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Second Law of Thermodynamics
For any spontaneous process, the entropy of the universe increases
Souniverse = So
system + Sosurroundings > 0
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Ssolid < Sliquid < Sgas
gasliquidsolid
Of all phase states, gases have the highest entropy
In a gas, molecules are more randomly distributed
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Larger molecules/atoms generally have a larger entropy
Larger molecules have more internal motion
Ssmall < Smedium < Slarge
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Often, dissolving a solid or liquid will increase the entropy
dissolves
more disordered arrangementhigher entropy
lower entropy
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Dissolving a gas in a liquid decreases the entropy
dissolves
overall more disordered arrangement:
higher entropy
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
The entropy of a substance increases with temperature
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
} size of molecules increases
} Sgas > Sliquid
} dissolving a gas in a liquid is accompanied bya lowering of the entropy
} dissolving a liquid in another liquid is accompanied by an increase in entropy
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
What is the sign of S for the following reactions?
FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g)solid solidgas gas
1 mol 2 mol
S > 0
Ba(OH)2 (s) → BaO (s) + H2O (g)
solid solid gas
S > 0
2 SO2 (g) + O2 (g) → 2 SO3 (g)
gas gas gas
2 mol 1 mol 2 mol
S < 0
Ag+ (aq) + Cl- (aq) → AgCl (s) S < 0
in solution insoluble
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
For each of the following pairs, which substance has a highermolar entropy at 25oC ?
HCl (l) HCl (s)
C2H2 (g) C2H6 (g)
Li (s) Cs (s)
Pb2+ (aq) Pb (s)
O2 (g) O2 (aq) HCl (l) HBr (l)
CH3OH (l) CH3OH (aq)N2 (l) N2 (g)
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Sorxn = Σ n So(products) – Σ m So(reactants)
If you know the standard molar entropies of reactants andproducts, you can calculate
S for a reaction:
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
substance So (J/K-mol)
H2 130.6
C2H4 (g) 219.4
C2H6 (g) 229.5
What is So for the following reaction?
C2H4 (g) + H2 (g) → C2H6 (g)
Sorxn = [229.5] – [219.4 + 130.6] = -120.5 J/K
Do you expect S to be positive or negative?
So for elements are NOT zero
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Which reactions are spontaneous?
We need to know the magnitudes of both S and H !
Go = Gibbs free energy…
… is a measure of the amount of “useful work” a system can perform
Go = Ho - TSo
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
A reaction is spontaneous if Go is negative
J. Willard Gibbs(1839 – 1903)
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)
The reaction of sodium metal with water:
Is the reaction spontaneous?
What is the sign of Go?
What is the sign of Ho?
What is the sign of So?
Yes
Go = negative
Ho = negative (exothermic!)
So = positive
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go < 0 => reaction is spontaneous(“product favored”)
Go > 0 => reaction is non-spontaneous
Go = 0 => reaction is at equilibrium
“exergonic”
“endergonic”
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
Ho So
+ -
Go
- + -
- - sign depends on T !low T => Go is negativehigh T => Go is positve
+ + sign depends on T !low T => Go is positivehigh T => Go is negative
+
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
e. I have no idea how to even start thinking about this
endothermic H > 0
Spontaneity: G must be negative - think about the signs of H and S
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
e. I have no idea how to even start thinking about this
endothermic H > 0
number of gas molecules doubles S > 0
Spontaneity: G must be negative - think about the signs of H and S
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
Ho So
+ -
Go
- + -
- - sign depends on T !low T => Go is negativehigh T => Go is positve
+ + sign depends on T !low T => Go is positivehigh T => Go is negative
+
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
e. I have no idea how to even start thinking about this
Spontaneity: G must be negative - think about the signs of H and S
endothermic H < 0
number gas molecules doubles S > 0
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Consider the following reaction:
2 H2 (g) + O2 (g) → 2 H2O (l)
S = -326.3 J/K, H = -571.7 kJ , and G = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?
If the decomposition of water is spontaneous, thenthe formation of water is non-spontaneous:
2 H2 (g) + O2 (g) → 2 H2O (l) G > 0
i.e. is there a temperature at which G becomes > 0 (non-spontaneous)?
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Consider the following reaction:
2 H2 (g) + O2 (g) → 2 H2O (l)
S = -326.3 J/K, H = -571.7 kJ , and G = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?
G = H - TS
H < 0S < 0
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
Ho So
+ -
Go
- + -
- - sign depends on T !low T => Go is negativehigh T => Go is positve
+ + sign depends on T !low T => Go is positivehigh T => Go is negative
+
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Consider the following reaction:
2 H2 (g) + O2 (g) → 2 H2O (l)
S = -326.3 J/K, H = -571.7 kJ , and G = -475.3 kJ at 25oC. Does the the decomposition of water ever become spontaneous?
G = H - TSH < 0S < 0
The decomposition of water becomes spontaneous at high temperatures!
The formation of water becomes non-spontaneous at high temperatures
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
A diamond left behind in a burning house reacts according to
2 C (s) + O2 (g) → 2 CO (g)
What is the value of Go for the reaction at 298K ?
substance Hfo (kJ/mol) Gf
o (kJ/mol) So (J/K-mol)
O2 0 0 205.0
C (diamond, s) 1.88 2.84 2.43
C (graphite, s) 0 0 5.69
CO (g) -110.5 -137.2 197.9
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
There are two possible ways to calculate Go:
I) Go = Ho - TSo
II) Go = Σ n Gfo (products) – Σ m Gf
o (reactants)
calculate Go from Ho and So :
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
I) calculate Go from Ho and So : Go = Ho - TSo
substance Hfo (kJ/mol) Gf
o (kJ/mol) So (J/K-mol)
O2 0 0 205.0
C (diamond, s) 1.88 2.84 2.43
C (graphite, s) 0 0 5.69
CO (g) -110.5 -137.2 197.9
Ho =[2 x (-110.5] – [2 x 1.88 + 0]
2 C (s) + O2 (g) → 2 CO (g)
= -224.8 kJ
So = [2 x 197.9] – [2 x 2.43 + 205.0] = 185.9 J/K
Go = -224.8 kJ - 298K x 185.9 J/K1000J
1kJ = - 280.2 kJ
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
substance Hfo (kJ/mol) Gf
o (kJ/mol) So (J/K-mol)
O2 0 0 205.0
C (diamond, s) 1.88 2.84 2.43
C (graphite, s) 0 0 5.69
CO (g) -110.5 -137.2 197.9
2 C (s) + O2 (g) → 2 CO (g)
II) Go = Σ n Gfo (products) – Σ m Gf
o (reactants)
Go = [2 x (-137.2)] – [2 x 2.84 + 0] = -280.1 kJ