Transcript Chapter 16 Thermodynamics: Entropy, Free Energy, and

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
State Function: A function or property whose value depends only on
the present state, or condition, of the system, not on the path used to
arrive at that state.
Enthalpy Change (DH): The heat change in a reaction or process at
constant pressure;
DH = DE + PDV
Entropy (S): The amount of molecular randomness in a system
Enthalpy, Entropy, and
Spontaneous Processes
Exothermic:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
DHo = 890.3 kJ
Endothermic:
H2O(s)
H2O(l)
DHfusion = +6.01 kJ
H2O(l)
H2O(g)
DHvap = +40.7 kJ
N2O4(g)
2NO2(g)
DHo = +55.3 kJ
Na+(aq) + Cl(aq)
DHo = +3.88 kJ
NaCl(s)
H2O
Enthalpy, Entropy, and
Spontaneous Processes
DS = Sfinal  Sinitial
Enthalpy, Entropy, and
Spontaneous Processes
Enthalpy, Entropy, and
Spontaneous Processes
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Chapter
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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
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter
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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)
Fe(s)
27.3
O2(g)
205.0
Fe2O3(s)
87.4
Hof (kJ/mol)
-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?
16.7 Free Energy
Free Energy: G = H - TS
D G = D H - TD S
Using:
DStotal = DSsys + DSsurr
Dssurr =
DG = DH - TDS = -TDStotal
- DH
T
DS = DSsys
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.
The effect of H, S and T on
Spontaneity
∆H
-
∆S
+
Low Temperature
Spontaneous (G<0)
High Temperature
Spontaneous (G<0)
+
-
-
-
Spontaneous (G< 0)
nonSpontaneous
(G>0)
+
+
Nonspontaneous (G>0)
Spontaneous (G<0)
Nonspontaneous (G > 0) Nonspontaneous (G
> 0)
Free Energy
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°
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
 One of the possible initial steps in the formation of
acid rain is the oxidation of the pollutant of SO2 to SO3
by the following reaction
SO2(g) + ½ O2(g)  SO3(g)
ΔHo = -98.9 kJ
ΔSo = -94.0 J/K
Calculate the ΔGo for this reaction at 25oC
Is the reaction spontaneous at standard-state
condition?
Does the reaction become spontaneous 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
Chaptr
16/30
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)
2Fe(s) + 3CO2(g)
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?
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)
CaO(s) + CO2(g)
Example
 The value of ΔGof at 25oC for gaseous mercury is 31.85
kJ/mol. What is the vapor pressure of mercury at
25oC?