3/30/14
First Law of Thermodynamics
First law of thermodynamics: energy cannot
Chapter 17
be created or destroyed.
The total energy of the universe cannot change
Spontaneity, Entropy, and
(though you can transfer it from one place to
another).
Free Energy
ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings
Chapter 17
Chapter 17
2
First Law of Thermodynamics
First Law of Thermodynamics
conservation of energy
ΔE is a state function: internal energy change
independent of how it was done
exothermic rxn: releases heat to the surroundings
(system
è
surroundings)
two ways energy “lost” from a system:
converted to heat, q and/or used to do work, w
Energy conservation requires that the energy change in
the system equal the heat released plus the work done.
ΔE = q + w
ΔE = ΔH + pΔV
Chapter 17
3
Chapter 17
4
1
3/30/14
Thermodynamics and Spontaneity
Comparing Potential Energy
Thermodynamics predicts whether a process will
proceed under the given conditions (spontaneous process).
Nonspontaneous processes require energy input to go.
Spontaneity is determined by comparing the chemical
potential energy of the system before the reaction with
the free energy of the system after the reaction
If the system after the rxn has less potential energy than
before the rxn, the rxn is thermodynamically favorable.
The direction of spontaneity can be determined by
comparing the potential energy of the system at the
start and the end.
spontaneity ≠ fast or slow
Chapter 17
5
Reversibility of Process
Chapter 17
6
Thermodynamics and Spontaneity
Any spontaneous process is irreversible, i.e., it will
proceed in only one direction.
A reversible process will proceed back and forth
between the two end conditions.
- equilibrium rxns (they result in no
change in free energy;)
If a process is spontaneous in one direction, it must be
nonspontaneous in the opposite direction.
Chapter 17
7
Chapter 17
8
2
3/30/14
Thermodynamics and Spontaneity
Factors Affecting Whether a Rxn is Spontaneous
diamond vs. graphite
Reactions are spontaneous in the direction of lower
chemical potential energy.
There are two factors that determine the thermodynamic
favorability: the enthalpy change and the entropy change.
The enthalpy change, ΔH, is the difference in the sum of
the internal energy and pV work energy of the reactants
compared to the products.
more stable
conversion of diamond into graphite is spontaneous
The entropy change, ΔS, is the difference in randomness
of the reactants compared to the products.
Chapter 17
Enthalpy (review)
10
Entropy
ΔH is generally measured in kJ/mol.
- stronger bonds = more stable molecules
ΔHrxn < 0: exothermic rxn; ΔHrxn > 0: endothermic rxn
ΔΗrxn = Σ (ΔH°f
Chapter 17
9
products)
− Σ (ΔH°f
reactants)
endothermic rxn: the bonds in the products are weaker
than the bonds in the reactant;
Entropy, S, is a thermodynamic function that increases as
the number of energetically equivalent ways of arranging
the components (degree of freedom) increases.
S is generally measured in J/mol.
Random systems require less energy than ordered systems.
(read section 17.1)
exothermic rxn: if the bonds in the products are stronger
than the bonds in the reactants
Chapter 17
11
Chapter 17
12
3
3/30/14
Changes in Entropy, ΔS
Changes in Entropy, ΔS
Examples:
Entropy change is favorable when the result is a more
random system (ΔS = Sfinal – Sinit. > 0 )
ΔS > 0 if:
- products are in a more random state
- going from solid to less ordered liquid to less ordered gas
- rxns that have larger # of product molecules than
reactant molecules
- increase in temperature
- solids dissociating into ions upon dissolving
Chapter 17
13
Changes in Entropy, ΔS
Chapter 17
14
Entropy Change and State Changes
When a material changes physical state, the number of
macrostates it can have changes as well.
ΔSsystem > 0 for a process in which the final condition is
more random than the initial condition
(favorable entropy)
- the more degrees of freedom the molecules have, the more
macrostates are possible;
ΔSsystem < 0 for a process in which the final condition is
more orderly than the initial condition (unfavorable
entropy)
ΔSsystem = ΔSreaction = Σ (S°prod) − Σ (S°react)
Chapter 17
15
Chapter 17
16
4
3/30/14
Problem A
Entropy Change and State Changes
- solids have fewer macrostates than liquids, which have
fewer macrostates than gases;
Predict whether ΔSsystem is + or − for
each of the following:
•  Heating air in a balloon
•  Water vapor condensing
•  Separation of oil and vinegar salad dressing
•  Dissolving sugar in tea
•  2 HgO(s) → 2 Hg(l) + O2 (g)
•  2 NH3 (g) → N2 (g) + 3 H2 (g)
Chapter 17
•  Ag+(aq) + Cl−(aq) → AgCl(s)
17
the 2nd Law of Thermodynamics
•  The total entropy change of the universe must be
18
Temperature Dependence of ΔSsurroundings
When a system process is exothermic, it adds heat to the
surroundings, increasing the entropy of the surroundings.
positive for a process to be spontaneous.
- for reversible process, ΔSuniv = 0.
-  for irreversible (spontaneous) process, ΔSuniv >0
ΔSuniverse = ΔSsystem + ΔSsurroundings
•  If the entropy of the system decreases, then the entropy
of the surroundings must increase by a larger amount.
- when ΔSsystem < 0 , ΔSsurroundings > 0 .
Chapter 17
Chapter 17
When a system process is endothermic, it takes heat from
the surroundings, decreasing the entropy of the
surroundings.
The amount the entropy of the surroundings changes
depends on its initial temperature.
19
Chapter 17
20
5
3/30/14
Problem B
Temperature dependence of ΔSsurroundings
The higher the original temperature, the less effect
addition or removal of heat has.
Chapter 17
21
Problem C The rxn below has ΔH = +66.4 kJ at 25°C.
rxn
Chapter 17
22
Problem C The rxn below has ΔH = +66.4 kJ at 25°C.
rxn
2 O2 (g) + N2 (g) → 2 NO2 (g)
2 O2 (g) + N2 (g) → 2 NO2 (g)
i. Determine ΔSsurroundings.
Chapter 17
Calculate the ΔSsurr. at 25ºC for the rxn:
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g)
ΔHrxn = − 2044 kJ.
ii. Determine the sign of ΔSsystem.
23
Chapter 17
24
6
3/30/14
Problem C The rxn below has ΔHrxn = + 66.4 kJ at 25°C.
iii. Determine whether the process is
spontaneous
2 O2 (g) + N2 (g) → 2 NO2 (g)
Gibbs Free Energy and Spontaneity
ΔSuniv = ΔSsys + ΔSsurr
-Τ ΔSuniv = ΔGsys
ΔGsys = ΔHsys− TΔSsys
ΔSuniv = ΔSsys
ΔHsys
T
-Τ ΔSuniv = -T ΔSsys -
-T ΔHsys
T
-Τ ΔSuniv = -T ΔSsys + ΔHsys
-Τ ΔSuniv = ΔHsys - Τ ΔSsys
Chapter 17
Chapter 17
25
Gibbs Free Energy and Spontaneity
Gibbs Free Energy, ΔG
The Gibbs free energy is the maximum amount of work
energy that can be released to the surroundings by a
system.
26
-Τ ΔSuniv = ΔGsys
ΔSuniv is + when spontaneous, so ΔG is −
And therefore, when ΔG < 0 process is spontaneous
- valid at a constant temperature and pressure
- often called the chemical potential energy because it is
analogous to the storing of energy in a mechanical
system
ΔGsys = ΔHsys− T ΔSsys
- since ΔSuniv determines whether a process is
spontaneous, ΔG also determines spontaneity;
Possible ΔG values:
ΔG < 0:
ΔH < 0 and ΔS > 0
ΔGsys = ΔHsys− T ΔSsys
rxn exothermic & more random
ΔH < 0 and large and ΔS < 0 but small
ΔH > 0 but small and ΔS > 0 and large
-Τ ΔSuniv = ΔGsys
or high temperature
Chapter 17
27
Chapter 17
28
7
3/30/14
Gibbs Free Energy, ΔG
Free Energy Change and Spontaneity
ΔGsys = ΔHsys− T ΔSsys
Possible ΔG values:
ΔG > 0 :
ΔH > 0 and ΔS < 0
never spontaneous
at any temperature
ΔG = 0 : the reaction is at equilibrium
Chapter 17
Problem D
The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has
ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C.
i. Calculate ΔG for the reaction.
Chapter 17
Chapter 17
29
31
Problem D
30
continued
ii. Determine if the reaction is spontaneous.
Chapter 17
32
8
3/30/14
Problem E
The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has
ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C.
i. Calculate ΔG for the reaction.
Chapter 17
Problem F
continued
ii. Determine if the reaction is spontaneous.
33
The rxn CCl4 (g) → C (s, graphite) + 2 Cl2 (g) has
ΔH = + 95.7 kJ and ΔS = + 142.2 J/K at 25 °C.
Calculate the minimum temperature for it to
be spontaneous.
Chapter 17
Problem E
35
Problem F
Chapter 17
34
Chapter 17
36
continued
9
3/30/14
Problem G
The rxn Al(s) + Fe2O3 (s) → Fe(s) + Al2O3 (s) has
ΔH = − 847.6 kJ and ΔS = − 41.3 J/K at 25 °C.
Calculate the maximum temperature for it to
be spontaneous.
Chapter 17
Problem G
Problem G
continued
Chapter 17
37
continued
38
The Third Law of Thermodynamics
Absolute Entropy
Read pages 656 - 659
Chapter 17
39
Chapter 17
40
10
3/30/14
Problem H Calculate ΔS° for the rxn
4 NH3 (g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g)
Changes in Entropy
ΔSºreaction = ( ∑ np Sºproducts) − ( ∑ nr Sºreactants)
the standard entropy change is the difference in absolute
entropy between the reactants and products under
standard conditions;
Chapter 17
Problem H
Substance S°, J/mol⋅K
NH3 (g)
192.8
O2 (g)
NO(g)
H2O(g)
41
continued
Problem H
Chapter 17
205.2
210.8
188.8
43
Chapter 17
42
Chapter 17
44
continued
11
3/30/14
Problem I
Problem I
Calculate ΔS° for the rxn
2 H2 (g) + O2 (g) → 2 H2O(g)
continued
Substance S°, J/mol⋅K
H2 (g)
O2 (g)
130.7
205.2
H2O(g)
188.8
Chapter 17
Chapter 17
45
Problem J
Calculating ΔGo
1. ΔGorxn = Σ nΔGof (prod.) − Σ mΔGof (react.)
at temperatures other than 25 °C:
assuming the change in
ΔHoreaction
ΔSoreaction
Substance
at 25 °C
CH4 (g)
O2 (g)
CO2 (g)
H2O(g)
O3 (g)
and
negligible
  nother way:
A
 
2. ΔG°rxn = ΔH°rxn − T ΔS°rxn
46
Calculate ΔG° at 25 °C for the reaction
CH4 (g) + 8 O2 (g) → CO2 (g) + 2 H2O(g) + 4 O3 (g)
ΔG°f,
kJ/mol
−50.5
0.0
−394.4
−228.6
163.2
or,
3. ΔG°total = ΔG°rxn 1 + ΔG°rxn 2 + ...
Chapter 17
47
Chapter 17
48
12
3/30/14
Problem J
continued
Problem J
continued
Is the rxn spontaneous?
Chapter 17
Problem K
49
Determine the free energy change ΔGo in the
following rxn at 298 K:
2 H2O(g) + O2 (g) → 2 H2O2 (g)
Chapter 17
51
Problem K
Chapter 17
50
Chapter 17
52
continued
13
3/30/14
Problem K
continued
Problem K
Chapter 17
53
ΔGo Relationships
-  if a rxn can be expressed as a series of rxns, the sum of
the ΔG values of the individual rxn is the ΔG of the total
rxn (ΔG is a state function);
continued
Chapter 17
54
Problem L Determine ΔGo in the following rxn at 298 K:
2 H2O(g) + O2 (g) → 2 H2O2 (g)
substance
ΔG° (kJ/mol)
H2 (g) + O2 (g) → H2O2 (g)
−105.6
2 H2 (g) + O2 (g) → 2 H2O(g)
-  if a rxn is reversed, the sign of its ΔG value reverses;
−457.2
-  if the amount of materials is multiplied by a factor, the
value of the ΔG is multiplied by the same factor;
(the value of ΔG is an extensive property)
Chapter 17
55
Chapter 17
56
14
3/30/14
Problem L
continued
Problem M
Find ΔGºrxn for the following rxn using the
given equations: 3 C(s) + 4 H2 (g) → C3H8 (g)
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g) ΔGº = −2074 kJ
Chapter 17
Problem M
C(s) + O2 (g) → CO2 (g)
ΔGº = −394.4 kJ
2 H2 (g) + O2 (g) → 2 H2O(g)
ΔGº = −457.1 kJ
57
continued
Problem M
Chapter 17
59
Chapter 17
58
Chapter 17
60
continued
15
3/30/14
Problem M
continued
Free Energy and Reversible Reactions
•  The change in free energy is the theoretical
limit as to the amount of work that can be
done.
•  If the reaction achieves its theoretical limit, it
is a reversible reaction.
Chapter 17
61
Chapter 17
62
Real Reactions
Free Energy under Nonstandard Conditions
•  In a real reaction, some of the free energy is “lost” as
ΔG = ΔG° only when the reactants and products are in
their standard states;
heat (if not most).
•  Therefore, real reactions are irreversible.
- their normal state at that temperature
- partial pressure of gas = 1 atm
- concentration = 1 M
Under nonstandard conditions, ΔG = ΔG° + RT lnQ
( Q is the rxn quotient )
At equilibrium, ΔG = 0 , therefore
Chapter 17
63
Chapter 17
ΔG° = −RTlnK
64
16
3/30/14
Problem N
Free Energy under Nonstandard Conditions
Calculate ΔG at 298 K for the rxn under the
given conditions:
2 NO(g) + O2 (g) → 2 NO2 (g) ΔGº = −71.2 kJ
substance p (atm)
NO(g)
0.100
O2 (g)
0.100
NO2 (g)
Chapter 17
Problem N
65
continued
Problem N
Chapter 17
2.00
67
Chapter 17
66
Chapter 17
68
continued
17
3/30/14
Problem P
Calculate ΔGrxn for the given rxn at 700 K
under the given conditions:
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ
Problem P
continued
substance p (atm)
N2 (g)
33
H2 (g)
99
NH3 (g)
2.0
Problem P
Chapter 17
69
Chapter 17
70
Chapter 17
71
Chapter 17
72
continued
18
3/30/14
Problem R
substance ΔG°f (kJ/mol)
N2O4 (g)
+99.8
NO2 (g)
+51.3
continued
ΔGo from Appendix IIB
Chapter 17
Problem R
Problem R
Calculate K at 298 K for the reaction
N2O4 (g)
2 NO2 (g)
Chapter 17
73
continued
Problem S
Chapter 17
75
74
Estimate the equilibrium constant for the
given rxn at 700 K:
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔGº = + 46.4 kJ
Chapter 17
76
19
3/30/14
Problem S
continued
End of Chapter 17
Spontaneity, Entropy, and
Free Energy
Chapter 17
77
Chapter 17
20