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CHEMICAL THERMODYNAMICS
The first law of thermodynamics:
Energy and matter can be neither created nor
destroyed; only transformed from one form to another.
The energy and matter of the universe is constant.
The second law of thermodynamics:
In any spontaneous process there is always an increase
in the entropy of the universe.
The entropy is increasing.
The third law of thermodynamics:
The entropy of a perfect crystal at 0 K is zero.
There is no molecular motion at absolute 0 K.
STATE FUNCTIONS
A property of a system which depends only on its
present state and not on its pathway.
H = Enthalpy = heat of reaction = qp
A measure of heat (energy) flow of a system relative
to its surroundings.
H° standard enthalpy
Hf° enthalpy of formation
H° = n Hf° (products) -  m Hf° (reactants)
H = U + PV
U represents the Internal energy of the particles, both
the kinetic and potential energy. U = q + w
HEAT
energy transfer as a
result of a temperature
difference
qp
VS
WORK
energy expanded to
move an object against
a force
w = F x d
endothermic (+q)
work on a system
(+w)
exothermic (-q)
work by the system
(-w)
qc = -qh
w = -PV
SPONTANEOUS PROCESSES
A spontaneous process occurs without outside intervention.
The rate may be fast or slow.
Entropy
A measure of randomness or disorder in a system.
Entropy is a state function with units of J/K and it can be
created during a spontaneous process.
Suniv = Ssys + Ssurr
The relationship between Ssys and Ssurr
Ssys
Ssurr
Suniv
Process spontaneous?
+
+
+
Yes
No (Rx will occur in
opposite direction)
+
?
Yes, if Ssys > Ssurr
+
?
Yes, if Ssurr > Ssys
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 S0 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, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g. bond
flexibility) relate directly to entropy.
The increase in entropy from solid to liquid to gas.
The entropy change accompanying the dissolution of a salt.
pure solid
MIX
pure liquid
solution
The large decrease in entropy when a gas dissolves in a liquid.
O2 gas
O2 gas in H2O
The small increase in entropy when ethanol dissolves in water.
Ethanol
Water
Solution of
ethanol
and water
Entropy Changes in the System
S0rxn - the entropy change that occurs when all reactants
and products are in their standard states.
S0rxn = S0products - S0reactants
The change in entropy of the surroundings is directly
related to an opposite change in the heat of the system
and inversely related to the temperature at which the
heat is transferred.
Ssurroundings = -
Hsystem
T
Entropy
S = Sf - Si
S > q/T
S = H/T
For a reversible (at equilibrium) process
H - T  S < 0
For a spontaneous reaction at constant T & P
 H - T S
If the value for  H - T S is negative for a reaction
then the reaction is spontaneous in the direction of the
products.
If the value for  H - T S is positive for a reaction
then the reaction is spontaneous in the direction of the
reactants. (nonspontaneous for products)
A spontaneous endothermic chemical reaction.
water
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
H0rxn = +62.3 kJ
APPLICATION OF THE 3RD LAW OF THERMODYNAMICS
S° = standard entropy = absolute entropy
S is usually positive (+) for Substances, S can be negative
(-) for Ions because H3O+ is used as zero
Predicting the sign of S°
The sign is positive if:
1. Molecules are broken during the Rx
2. The number of moles of gas increases
3. solid  liquid liquid  gas solid  gas
an increase in order occurs
1.
2.
3.
4.
Ba(OH)2•8H2O + 2NH4NO3(s) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)
2SO(g) + O2(g)  2SO3(g)
HCl(g) + NH3(g)  NH4Cl(s)
CaCO3(s)  CaO(s) + CO2(g)
Sample Problem
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 S0rxn, and state whether the reaction occurs
spontaneously at this temperature.
PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so
S0surroundings must be > 197 J/K. To find S 0surr, first find Hsys; Hsys =
Hrxn which can be calculated using H0f values from tables. S0universe
= S0surr + S0sys.
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.
 S°=  n S°(product)- m S°(reactant)
1.
Acetone, CH3COCH3, is a volitale liquid
solvent. The standard enthalpy of formation of
the liquid at 25°C is -247.6 kJ/mol; the same
quantity for the vapor is -216.6 kJ/mol. What
is  S when 1.00 mol liquid acetone vaproizes?
2.
Calculate  S° at 25° for:
a. 2 NiS(s) + 3 O2(g)  2 SO2(g) + 2 NiO9(s)
b. Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
Formula
Nitrogen
N2(g)
NH3(g)
NO(g)
NO2(g)
HNO3(aq)
Oxygen
O2(g)
O3(g)
OH-(aq)
H2O(g)
H2O(l)
S°,
J/(mol•K)
191.5
193
210.6
239.9
146
205.0
238.8
-10.5
188.7
69.9
Formula
Sulfur
S2(g)
S(rhombic)
S(monoclinic)
SO2(g)
H2S(g)
Fluorine
F-(aq)
F2(g)
HF(g)
Chlorine
Cl-(aq)
Cl2(g)
HCl(g)
S°,
J/(mol•K)
Formula
228.1
31.9
32.6
248.1
205.6
Bromine
Br-(aq)
Br 2(l)
Iodine
I-(aq)
I2(s)
-9.6
202.7
173.7
55.1
223.0
186.8
Silver
Ag+(aq)
Ag(s)
AgF(s)
AgCl(s)
AgBr(s)
AgI(s)
S°,
J/(mol•K)
80.7
152.2
109.4
116.1
73.9
42.7
84
96.1
107.1
114
S°,
Formula
Hydrogen
H+(aq)
H2(g)
Sodium
Na+(aq)
Na(s)
NaCl(s)
NaHCO3(s)
Na2CO3(s)
Calcium
Ca2+(aq)
Ca(s)
CaO(s)
CaCO3(s)
J/(mol•K)
0
130.6
60.2
51.4
72.1
102
139
-55.2
41.6
38.2
92.9
S°,
Formula
Carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
HCO3-(aq)
CH4(g)
C2H4(g)
C2H6(g)
C6H6(l)
HCHO(g)
CH3OH(l)
CS2(g)
CS2(l)
HCN(g)
J/(mol•K)
5.7
2.4
197.5
213.7
95.0
186.1
219.2
229.5
172.8
219
127
237.8
151.0
201.7
S°,
Formula
J/(mol•K)
Carbon (continued)
HCN(l)
112.8
CCl 4(g)
309.7
CCl4(l)
214.4
CH3CHO(g)
266
C2H5OH(l)
161
Silicon
Si(s)
18.0
SiO2(s)
41.5
SiF4(g)
285
Lead
Pb(s)
64.8
PbO(s)
66.3
PbS(s)
91.3
STANDARD FREE ENERGY OF FORMATION
G°f
The free energy change that occurs when 1 mol
of substance is formed from the elements in
their standard state.
Calculate G° for:
2 CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
Formula
Gf°
kJ/mol
Nitrogen
N2(g)
NH3(g)
NO(g)
NO2(g)
HNO3(aq)
Sulfur
0
-16
86.60
51
-110.5
Oxygen
O2(g)
O3(g)
OH-(aq)
H2O(g)
H2O(l)
Formula
Gf°
kJ/mol
S2(g)
S (rhombic)
S (monoclinic)
SO2(g)
H2S(g)
Bromine
80.1
0
0.10
-300.2
-33
Fluorine
0
163
-157.3
-228.6
-237.2
F-(aq)
F2(g)
HF(g)
Br-(aq)
Br2(l)
-102.8
0
Iodine
I-(aq)
I2(s)
-51.7
0
Silver
-276.5
0
-275
Chlorine
Cl-(aq)
Cl2(g)
HCl(g)
Formula
Gf°
kJ/mol
-131.2
0
-95.3
Ag+(aq)
Ag(s)
AgF(s)
AgCl(s)
AgBr(s)
AgI(s)
77.1
0
-185
-109.7
-95.9
-66.3
Formula
Gf°
kJ/mol
Hydrogen
H+
0
H2(g)
0
Sodium
Na+(aq)
-261.9
Na(s)
0
NaCl(s)
-348.0
NaHCO3(s) -851.9
Na2CO3(s)
-1048.1
Calcium
Ca2+(aq)
-553.0
Ca(s)
0
CaO(s)
-603.5
CaCO3(s)
-1128.8
Formula
Carbon
C (graphite)
C (diamond)
CO(g)
CO2(g)
HCO3-(aq)
CH4(g)
C2H4(g)
C2H6(g)
C6H6(l)
HCHO(g)
CH3OH(l)
CS2(g)
CS2(l)
HCN(g)
Gf°
kJ/mol
0
2.9
-137.2
-394.4
-587.1
-50.8
68.4
-32.9
124.5
-110
-166.2
66.9
63.6
125
Formula
Carbon (cont.)
HCN(l)
CCl4(g)
CCl4(l)
CH3CHO(g)
C2H5OH(l)
Silicon
Si(s)
SiO2(s)
SiF4(g)
Lead
Pb(s)
PbO(s)
PbS(s)
Gf°
kJ/mol
121
-53.7
-68.6
-133.7
-174.8
0
-856.6
-1506
0
-189
-96.7
Gibbs 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 - TS0system
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process at equilibrium
G0rxn =  mG0products -  nG0reactants
INTERPRETING G° FOR SPONTANEITY
1. When G° is very small (less than -10 KJ) the
reaction is spontaneous as written. Products dominate.
G° < 0
G°(R) > G°(P)
2. When G° is very large (greater than 10 KJ) the
reaction is non spontaneous as written. Reactants
dominate.
G° > 0
G°(R) < G°(P)
3. When G° is small (+ or -) at equilibrium then both
reactants and products are present.
G° = 0
Ba(OH2)•8 H2O(g) + 2 NH4NO3(g)  2 NH3(g)+10 H2O(l) + Ba(NO3)3(aq)
GIBBS FREE ENERGY : G
G = H - TS
describes the temperature dependence of
spontaneity
Standard conditions (1 atm, if soln=1M & 25°):
G° = H° - TS°
A process ( at constant P & T) is spontaneous
in the direction in which the free energy
decreases.
1. Calculate H°, S° & G° for
2 SO2(g) + O2(g)  2 SO3(g) at 25°C & 1 atm
G AND EQUILIBRIUM
The equilibrium point occurs at the lowest free energy
available to the reaction system.
When a substance undergoes a chemical reaction, the
reaction proceeds to give the minimum free energy at
equilibrium.
G = G° + RT 1n (Q)
at equilibrium: G = 0
G° = -RT 1n (K)
G° = 0
then K = 1
G° < 0
then K > 1
G° > 0
then K < 1
Q: Corrosion of iron by oxygen is
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
calculate K for this Rx at 25°C.
Free Energy, Equilibrium and Reaction Direction
•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (G = 0)
G = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1
and ln Q = 0 so
G0 = - RT lnK
Table 2 The Relationship Between G0 and K at 250C
G0(kJ)
K
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
-1
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
Essentially no forward reaction;
reverse reaction goes to completion
Forward and reverse reactions
proceed to same extent
Forward reaction goes to
completion; essentially no reverse
reaction
REVERSE REACTION
9x10-36
FORWARD REACTION
200
Significance
1.Calculate Gº at 25ºC
Ba SO4 (s) Ba2+(aq) + SO42-(aq)
What is the value for Ksp at 25ºC?
2.Calculate K at 25ºC for
Zn(s) + 2H+(aq) Zn2+(aq) + H2 (g).
Gº & Spontaneity is dependent on Temperature
Hº
Sº
Gº
-
+
-
Spontaneous at all T
+
-
+
Non spontaneous at all T
-
-
+/-
At Low T= Spontaneous
At High T= Nonspontaneous
+
+
+/-
At low T= Nonspontaneous
At High T= Spontaneous
Q. Predict the Spontaneity for H2O(s)  H2O(l)
at (a) -10ºc , (b) 0ºc & (c) 10ºc.
1. At what temperature is the following process
spontaneous at 1 atm?
Br2 (l)  Br2 (g)
What is the normal boiling point for Br2 (l)?
2. Calculate Gº & Kp at 35ºC
N2O4 (g)  2 No2 (g)
3. Calculate Hº, Sº & Gº at 25ºc and 650ºC.
CS2 (g) + 4H2 (g) CH4 (g) + 2H2S(g)
Compare the two values and briefly discuss the
spontaneity of the Rx at both temperature.