Chapter 16 1

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C H E M I S T R Y
Chapter 16
Thermodynamics: Entropy, Free Energy,
and Equilibrium
Spontaneous Processes
Spontaneous Process: A process that, once started, proceeds on its
own without a continuous external influence.
Spontaneous process

Spontaneity reaction always moves a system
toward equilibrium
◦ Both forward and reverse reaction depends on
 Temperature
 Pressure
 Composition of reaction mixture
Q < K; reaction proceeds in the forward
direction
 Q>K; reaction proceeds in the reverse direction
 Spontaneity of a reaction does not identify the
speed of reaction

Spontaneous Processes
Enthalpy, Entropy, and Spontaneous
Processes
DS = Sfinal  Sinitial
Enthalpy, Entropy, and Spontaneous
Processes
Enthalpy, Entropy, and Spontaneous
Processes
Entropy and Probability
S = k ln W
k = Boltzmann’s constant
= 1.38 x 1023 J/K
W = The number of ways that the
state can be achieved.
Entropy and Temperature
Third Law of Thermodynamics: The entropy of a perfectly ordered
crystalline substance at 0 K is zero.
Entropy and Temperature

ΔS increases
 when increasing the
average kinetic energy of
molecules
 Total energy is distributed
among the individual
molecules in a number of
ways
 Botzman- the more way
(W) that the energy can be
distributed the greater the
randomness of the state
and higher the entropy
Standard Molar Entropies and
Standard Entropies of Reaction
Standard Molar Entropy (So): The entropy of 1 mole of a pure
substance at 1 atm pressure and a specified temperature.
Standard Molar Entropies and
Standard Entropies of Reaction
DSo = So(products) - So(reactants)
aA + bB
cC + dD
DSo = [cSo(C) + dSo (D)]  [aSo (A) + bSo (B)]
Products
Reactants
Standard Molar Entropies and
Standard Entropies of Reaction
Using standard entropies, calculate the standard entropy change for the
decomposition of N2O4.
N2O4(g)
2NO2(g)
Example

Calculate the standard entropy of reaction at 25oC for
the decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
16.6 Entropy and the Second Law of
Thermodynamics
First Law of Thermodynamics: In any process, spontaneous or
nonspontaneous, the total energy of a system and its surroundings is
constant.
• Helps keeping track of energy flow between system and the
surrounding
• Does not indicate the spontaneity of the process
Second Law of Thermodynamics: In any spontaneous process, the total
entropy of a system and its surroundings always increases.
• Provide a clear cut criterion of spontaneity
• Direction of spontaneous change is always determined by the
sign of the total entropy change
Entropy and the Second Law of
Thermodynamics
DStotal = DSsystem + DSsurroundings
or
DStotal = DSsys + DSsurr
DStotal > 0
The reaction is spontaneous.
DStotal < 0
The reaction is nonspontaneous.
DStotal = 0
The reaction mixture is at equilibrium.
Entropy and the Second Law of
Thermodynamics
DSsurr a  DH
DSsurr a
1
T
DSsurr =
 DH
T
Example
Consider the oxidation of iron metal
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
Determine whether the reaction is spontaneous at 25oC

So (J/K mol)
ΔHof (kJ/mol)
Fe(s)
27.3
0
O2(g)
205.0
0
Fe2O3(s)
87.4
-824.2
Example

Consider the combustion of propane gas:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
a. Calculate the entropy change in the surrounding
associated with this reaction occurring at 25.0oC
b. Determine the sign of the entropy change for the
system
c. Determine the sign of the entropy change for the
universe. Will the reaction be spontaneous?
So (J/K mol)
ΔHof (kJ/mol)
C3H8(g)
270.2
-103.8
O2(g)
205.0
0
CO2(g)
213.6
-393.5
H2O(g)
188.7
-241.8
16.7 Free Energy
Free Energy: a thermodynamic quantity defined by G = H - TS
D G = D H - TD S
Using:
DStotal = DSsys + DSsurr
Dssurr =
DG = -TDStotal
- DH
T
DStotal = DSsys - DH
T
Constant pressure and temperature
Free Energy
Using the second law and DG = DH - TDS = -TDStotal
DG < 0 The reaction is spontaneous.
DG > 0 The reaction is nonspontaneous.
DG = 0 The reaction mixture is at equilibrium.
16.8 Standard Free-Energy Changes for
Reactions
Thermodynamic Standard State: Most stable form of a
substance at 1 atm pressure and at a specified temperature,
usually 25 °C; 1 M concentration for all substances in solution.
DG° = DH° - TDS°
The effect of H, S and T on Spontaneity
∆H
∆S
Low Temperature
High Temperature
-
+
Spontaneous (G<0)
Spontaneous (G<0)
+
-
Nonspontaneous (G > 0)
Nonspontaneous (G > 0)
-
-
Spontaneous (G< 0)
nonSpontaneous (G>0)
+
+
Nonspontaneous (G>0)
Spontaneous (G<0)
Standard Free-Energy Changes for
Reactions
Calculate the standard free-energy change at 25 oC for the Haber synthesis of
ammonia using the given values for the standard enthalpy and standard
entropy changes:
N2(g) + 3H2(g)
2NH3(g)
DHo = 92.2 kJ
DSo = 198.7 J/K
Example

What is the standard free-energy change, ΔoG, for the following
reaction at 25oC?
2KClO3(g)  2KCl(s) + 3O2(g)
ΔHo = -78.0kJ
ΔSo =493.9J/K
Is the reaction spontaneous at standard-state condition?
Does the reaction become nonspontaneous at higher temperature?
Standard Free Energies of Formation
DGo = DGof (products)  DGof (reactants)
aA + bB
cC + dD
DGo = [cDGof (C) + dDGof (D)]  [aDGof (A) + bDGof (B)]
Products
Reactants
Standard Free Energies of
Formation
Ch
aptr
16/
28
Standard Free Energies of Formation
Using table values, calculate the standard free-energy change at
25 °C for the reduction of iron(III) oxide with carbon
monoxide:
Fe2O3(s) + 3CO(g)
Gof
(kJ/mol)
2Fe(s) + 3CO2(g)
Fe2O3(s)
3CO(g)
2Fe(s)
3CO2(g)
-743.5
-137.2
0
-394.4
16.10 Free Energy Changes and the
Reaction Mixture
DG = DG° + RT ln Q
DG = Free-energy change under nonstandard conditions.
For the Haber synthesis of ammonia:
N2(g) + 3H2(g)
P
2NH3(g)
Qp =
2
NH3
3
P
N2
P
H2
Free Energy Changes and the Reaction
Mixture
Calculate DG for the formation of ethylene (C2H4) from carbon and hydrogen at 25
°C when the partial pressures are 100 atm H2 and 0.10 atm C2H4.
2C(s) + 2H2(g)
C2H4(g)
Is the reaction spontaneous in the forward or the reverse direction?
C(s)
∆ Gof (kJ/mol) 0
H2(g)
C2H4(g)
0
61.8
16.11 Free Energy and Chemical
Equilibrium
DG = DG° + RT ln Q
•
When the reaction mixture is mostly reactants:
Q << 1
RT ln Q << 0
DG < 0
The total free energy decreases as the reaction proceeds spontaneously in
the forward direction.
•
When the reaction mixture is mostly products:
Q >> 1
RT ln Q >> 0
DG > 0
The total free energy decreases as the reaction proceeds spontaneously in
the reverse direction.
Free Energy and Chemical
Equilibrium
DG = DGo + RT ln Q
At equilibrium, DG = 0 and Q = K.
DGo = RT ln K
Free Energy and Chemical Equilibrium
Calculate Kp at 25 oC for the following reaction:
CaCO3(s)
Gof (kJ/mol)
CaO(s) + CO2(g)
CaCO3(s)
CaO(s)
CO2(g)
-1128.8
-603.5
-394.4
Example

The value of ΔGof at 25oC for gaseous mercury is 31.85 kJ/mol.
What is the vapor pressure of mercury at 25oC?
Hg(l)
Hg(g)
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