Energetics

1
Energetics
6.1 What is Energetics? 6.2 Enthalpy Changes
Related to Breaking and Forming of
Bonds 6.3 Standard Enthalpy Changes 6.4 Experiment
al Determination of Enthalpy Changes by
Calorimetry 6.5 Hesss Law 6.6 Calculations
involving Standard Enthalpy Changes of Reactions
2
What is energetics?
6.1 What is energetics? (SB p.136)
Energetics is the study of energy changes
associated with chemical reactions.
Thermochemistry is the study of heat changes
associated with chemical reactions.
3
Internal Energy (U)
U kinetic energy potential energy
4
? T (K)
translational rotational vibrational
Energy
Relative position among particles
Bond breaking ? P.E. ? Bond forming ? P.E. ?
5
Bond breaking -
Bond forming -
Ionization -
Na(g) ? Na(g) e? P.E. ?
6
Q.1
4H(g) 2O(g)
7
Internal energy and enthalpy
6.1 What is energetics? (SB p.137)
H U PV
8
Internal energy and enthalpy
6.1 What is energetics? (SB p.137)
e.g. Mg(s) 2HCl(aq) ? MgCl2(aq) H2(g)
qv ?U -473 kJ mol?1
qp ?H -470 kJ mol?1
Mg
Mg
9
qv qp 3 qp w qp - P?V
P?V ? (Nm?2)(m3) Nm
Force ? displacement
10
Internal energy and enthalpy
6.1 What is energetics? (SB p.138)
?H ?U ?PV ?U P?V at constant P
qp qv P?V
Heat change at fixed P
Heat change at fixed V
Work done
11
Internal energy and enthalpy
6.1 What is energetics? (SB p.138)
qp qv P?V
On expansion, P?V gt 0 Work done by the system
against the surroundings System gives out less
energy to the surroundings qp is less negative
than qv (less exothermic)
12
Internal energy and enthalpy
6.1 What is energetics? (SB p.138)
qp qv P?V
On contraction, P?V lt 0 Work done by the
surroundings against the system System gives out
more energy to the surroundings qp is more
negative than qv (more exothermic)
13

• ?H is more easily measured than ?H as
• most reactions happen in open vessels.
• i.e. at constant pressure.

• The absolute values of H and H cannot
• be measured.

14
Exothermic and endothermic reactions
6.1 What is energetics? (SB p.138)
An exothermic reaction is a reaction that
releases heat energy to the surroundings. (?H
-ve)
15
Exothermic and endothermic reactions
6.1 What is energetics? (SB p.139)
An endothermic reaction is a reaction that
absorbs heat energy from the surroundings. (?H
ve)
16
Law of conservation of energy
6.1 What is energetics? (SB p.136)
The law of conservation of energy states that
energy can neither be created nor destroyed, but
can be exchanged between a system and its
surroundings
17
Exothermic - P.E. of the system ? K.E. of the
surroundings
Endothermic - K.E. of the surroundings ? P.E.
of the system
18
Enthalpy changes related to breaking and forming
of bonds
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.140)
CH4 2O2? CO2 2H2O
19
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.140)
In an exothermic reaction, E absorbed to break
bonds lt E released as bonds are formed.
20
Enthalpy changes related to breaking and forming
of bonds
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.140)
N2(g) 2O2(g) ? 2NO2(g)
21
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.140)
In an endothermic reaction, E absorbed to break
bonds gt E released as bonds are formed.
22
For non-gaseous reactions, P?V ? 0 ?H ?U P?V
? ?U For gaseous reactions, ?H ?U P?V ?U
(?n)RT (PV nRT)
23
Q.2
CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
Given R 8.314 J K?1 mol?1, T 298 K
?U ?H (?n)RT
?885 kJ mol?1
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C(g) 4H(g) 4O(g)
25
?H / kJ mol?1
At 298K
CH4(g) 2O2(g) ? CO2(g) 2H2O(l) ?890
At 373K
CH4(g) 2O2(g) ? CO2(g) 2H2O(g) ?802
2H2O(l) ? 2H2O(g) 88 kJ
26
C(g) 4H(g) 4O(g)
298 K
H1
H2
27
C(g) 4H(g) 4O(g)
Enthalpy
373 K
H1
H2
Assume constant ?H
Reaction coordinate
28
C(g) 4H(g) 4O(g)
373 K
In fact, ?H depends on T
29
C(g) 4H(g) 4O(g)
373 K
30
Standard Enthalpy Changes
31
Standard enthalpy changes
6.3 Standard enthalpy changes (SB p.141)
32
Standard enthalpy changes
6.3 Standard enthalpy changes (SB p.141)
As enthalpy changes depend on temperature and
pressure, it is necessary to define standard
conditions
1. elements or compounds in their normal
physical states2. a pressure of 1 atm (101325
Nm-2) and3. a temperature of 25oC (298 K)
33
Standard enthalpy change of reaction
The enthalpy change when the molar quantities of
reactants as stated in the equation react under
standard conditions.
per mole of O2
34
Standard enthalpy change of reaction
2H2(g) O2(g) ? 2H2O(l)
?
?H ?572 kJ mol?1
per mole of O2
per mole of H2 or H2O
35
Standard enthalpy change of formation
The enthalpy change when one mole of the
substance is formed from its elements under
standard conditions.
Q.3
O2(g) ? O2(g)
36
C(graphite) ? C(diamond)
37
Q.4
(i) C(graphite) O2(g) ? CO2(g)
(ii) C(graphite) 2H2(g) ? CH4(g)
(v) 2C(graphite) 2H2(g) O2(g) ? CH3COOH(l)
38
Q.5
qv ?U ?140.3 kJ per g of H2
?n 0 0.496 0.248 -0.744 mol
39
Q.5
?H ?U ?nRT
-142.1 kJ
Heat released for the formation of 0.496 mol of
water
Molar ?HfH2O
40
Standard enthalpy change of combustion
The enthalpy change when one mole of the
substance undergoes complete combustion under
standard conditions.
C2H5OH(l) 3O2(g) ? 2CO2(g) 3H2O(l)
41
6.3 Standard enthalpy changes (SB p.147)
?Hc (kJ mol-1)
Substance
ø
-395.4 -393.5
C (diamond) C (graphite)
42
Q.6
Incomplete combustion
(b) 2H2(g) O2(g) ? 2H2O(l)
43
Q.6
(c) C(graphite) O2(g) ? CO2(g)
(d) CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
44
Standard enthalpy changes of neutralization
6.3 Standard enthalpy changes (SB p.142)
45
Standard enthalpy changes of neutralization
6.3 Standard enthalpy changes (SB p.142)
Enthalpy level diagram for the neutralization of
a strong acid and a strong alkali
46
6.3 Standard enthalpy changes (SB p.142)
H(aq) OH-(aq) Cl?(aq) ? H2O(l) Cl?(aq)
?H2 -57.3
NH3(aq) HCl(aq) ? NH4Cl (aq)
47
6.3 Standard enthalpy changes (SB p.142)
H(aq) OH-(aq) Na(aq) ? H2O(l) Na(aq)
?H2 -57.3
HF(aq) NaOH(aq) ? NaF(aq) H2O(l)
48
Standard enthalpy change of solution
6.3 Standard enthalpy changes (SB p.142)
49
Standard enthalpy change of solution
6.3 Standard enthalpy changes (SB p.142)
50
Standard enthalpy change of solution
6.3 Standard enthalpy changes (SB p.143)
51
Standard enthalpy change of solution
6.3 Standard enthalpy changes (SB p.143)
52
Standard enthalpy change of solution
6.3 Standard enthalpy changes (SB p.143)
53
Experimental Determination of Enthalpy Changes by
Calorimetry
54
Experimental determination of enthalpy changes by
calorimetry
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.148)
Calorimeter is any set-up used for the
determination of ?H. By temperature measurement.
?H qp (m1c1 m2c2)?T
55
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.148)
?H qp (m1c1 m2c2)?T where m1 is the mass
of the reaction mixture, m2 is the mass of the
calorimeter, c1 is the specific heat capacity of
the reaction mixture, c2 is the specific heat
capacity of the calorimeter, ?T is the
temperature change of the reaction mixture.
56
Determination of enthalpy change of neutralization
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.149)
57
If the reaction is fast enough, T1 ? T2
?H ? (m1c1 m2c2)(T2 T0)
?H (m1c1 m2c2)(T1 T0)
T0
58
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.150)
Determination of enthalpy change of combustion
The Philip Harris calorimeter used for
determining the enthalpy change of combustion of
a liquid fuel
59
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.151)
Determination of enthalpy change of combustion
A simple apparatus used to determine the enthalpy
change of combustion of ethanol
60
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.151)
Heat evolved (m1c1 m2c2) ?T Where m1 is the
mass of water in the calorimeter, m2 is the mass
of the calorimeter, c1 is the specific heat
capacity of the water, c2 is the specific heat
capacity of the calorimeter, ?T is the
temperature change of the reaction
61
Q.7(Example)
heat given out
62
Hesss Law
63
Hesss Law
6.5 Hesss law (SB p.153)
Hesss law of constant heat summation states that
the total enthalpy change accompanying a chemical
reaction is independent of the route by which
the chemical reaction takes place and depends
only on the difference between the total enthalpy
of the reactants and that of the products.
64
Hesss Law
6.5 Hesss law (SB p.153)
?H1 HB HA
?H2 ?H3
?H4 ?H5
65
Importance of Hesss law
6.5 Hesss law (SB p.155)

• The enthalpy change of some chemical reactions
  cannot be determined directly because

• the reactions cannot be performed/controlled in
  the laboratory
• the reaction rates are too slow
• the reactions may involve the formation of side
  products

But the enthalpy change of such reactions can be
determined indirectly by applying Hesss Law.
66
Enthalpy change of formation of CO(g)
6.5 Hesss law (SB p.153)
due to further oxidation of CO to CO2 The
reaction cannot be controlled.
67
Enthalpy change of formation of CO(g)
6.5 Hesss law (SB p.153)
-393 - (-283 )
-110 kJ mol-1
68
Enthalpy cycle (Born-Haber cycle)
6.5 Hesss law (SB p.155)

• Relate the various equations involved in a
  reaction

69
Steps for drawing Born-Haber cycle
6.5 Hesss law (SB p.153)
1. Give the equation for the change being
considered.
70
Steps for drawing Born-Haber cycle
6.5 Hesss law (SB p.153)
2. Complete the cycle by giving the equations for
the combustion reactions of reactants and
products.
71
Steps for drawing Born-Haber cycle
6.5 Hesss law (SB p.153)
2. Complete the cycle by giving the equations for
the combustion reactions of reactants and
products.
72
Calculation of standard enthalpy change of
formation from standard enthalpy changes of
combustion
6B
73
Q.8
4O2(g)
4(-393) 5(-286) (?2877) kJ mol?1 ?125 kJ
mol?1
74
Q.8 Method B By addition and/or subtraction of
equations with known
(1) C(graphite) O2(g) ? CO2(g) ?393
Overall reaction 4?(1) 5?(2) (3)
?125 kJ mol?1
75
Enthalpy level diagram
6.5 Hesss law (SB p.154)

• Relate substances together in terms of enthalpy
  changes of reactions

76
Steps for drawing enthalpy level diagram
1. Draw the enthalpy level of elements.
77
Steps for drawing enthalpy level diagram
2. Enthalpies of elements are arbitrarily taken
as zero.
C(graphite) O2(g)
78
Steps for drawing enthalpy level diagram
3. Higher enthalpy levels are drawn above that of
elements
C(graphite) O2(g)
79
Steps for drawing enthalpy level diagram
4. Lower enthalpy levels are drawn below that of
elements
C(graphite) O2(g)
Route 1
80
Steps for drawing enthalpy level diagram
4. Lower enthalpy levels are drawn below that of
elements
C(graphite) O2(g)
?
Route 1
Route 2
81
6.5 Hesss law (SB p.158)
Check Point 6-5
82
6.5 Hesss law (SB p.158)
2C(graphite) H2(g) ?? C2H2(g)
2O2(g)
0.5O2(g)
2.5O2(g)
H2O(l)
By Hesss law,
2(?393.5) (?285.8) (?1299) kJ mol?1
226.2 kJ mol?1
83
(iii). Draw an enthalpy level diagram for the
reaction using the enthalpy changes in (ii)
Route 1
Route 2
84
Calculations involving Standard Enthalpy Changes
of Reactions
85
Calculation of standard enthalpy change of
reaction from standard enthalpy changes of
formation
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.159)
86
Q.9
0.5H2(g) 0.5Cl2(g)
By Hesss law,
?314 (?46) (?92) kJ mol?1 ?176 kJ mol?1
87
Q.9
0.5H2(g) 0.5Cl2(g)
By Hesss law,
88
Q.10
?
?H
4CH3NHNH2(l) 5N2O4(l) 4CO2(g)
9N2(g) 12H2O(l)
By Hesss law,
4(?393) 12(?286) - 4(53) 5(?20) kJ
?5116 kJ
Highly exothermic/ignites spontaneously Used as
rockel fuel in Apollo 11
89
Q.11
(1) C3H6(g) 4.5O2(g) ? 3CO2(g) 3H2O(l)
?2058
(2) H2(g) 0.5O2(g) ? H2O(l) ?286
(3) C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(l)
?2220
Overall reaction (1) (2) (3)
(?2058) (?286) (?2220) kJ mol?1 ?124 kJ
mol?1
90
Q.11
?
?H
C3H6(g) H2(g) C3H8(g)
By Hesss law,
(?2058) (?286) (?2220) kJ mol?1 ?124 kJ
mol?1
91
Q.11
?
?H
C3H6(g) H2(g) C3H8(g)
H2O(l)
92
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93
?
H2(g) O2(g) ? H2O2(l)
?Hf H2O2(l) ?188 kJ mol?1
94
H2O2 is energetically stable w.r.t. H2 and O2
H2O2 is energetically unstable w.r.t. H2O and
0.5O2
H2(g) O2(g)
H2O2(l)
H2O(l) 0.5O2(g)
95
H2 and O2 are energetically unstable w.r.t. H2O2
and H2O
H2(g) O2(g)
H2O2(l)
H2O(l) 0.5O2(g)
96
Energetically unstable
C(diamond) O2(g)
?
C(graphite) O2(g)
CO2(g)
97
Kinetically stable
The rate of conversion is extremely low
C(diamond) O2(g)
?
C(graphite) O2(g)
CO2(g)
98
C(diamond) O2(g)
C(graphite) O2(g)
?Hcgraphite ?393 kJ
CO2(g)
99
Rate of reaction depends on the ease of bond
breaking in reactants (e.g. diamond) The minimum
energy required for a reaction to start is known
as the activation energy, Ea
C(diamond) O2(g)
?
?H ?2 kJ
C(graphite) O2(g)
?Hcgraphite ?393 kJ
CO2(g)
100
The extremely low rate is due to high Ea
C(diamond) O2(g)
?
C(graphite) O2(g)
CO2(g)
101
Ea is always gt 0 as bond breaking is endothermic
C(diamond) O2(g)
?
C(graphite) O2(g)
CO2(g)
102
Ea tells how fast a reaction can proceed.
Rate of reaction / kinetics / kinetic stability
Equilibrium / energetics / energetic stability
103
Ea ? kinetic stability (rate of reaction) Higher
Ea ? higher kinetic stability of reactants
w.r.t. products
? lower rate to give products
Lower Ea ? lower kinetic stability of reactants
w.r.t. products
? higher rate to give products
104
Diamond ? energetically unstable w.r.t.
graphite ? kinetically stable w.r.t.
graphite Graphite ? stable w.r.t. diamond
energetically and kinetically
105
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106
6.7 Entropy change (SB p.164)
Spontaneity of Changes
107

• Spontaneity The state of being spontaneous
• Spontaneous
• self-generated
• natural
• happening without external influence

internal
108
6.7 Entropy change (SB p.164)

• A process is said to be spontaneous
• If no external forces are required to keep the
  process going,
• although external forces may be required to
  get the process started (Ea).
• The process may be a physical change or a
  chemical change

109
6.7 Entropy change (SB p.164)

• Spontaneous physical change
• E.g. condensation of steam at 25C
• Spontaneous chemcial change
• E.g. burning of wood once the fire has started

Exothermic ? spontaneous ? Endothermic ? not
spontaneous ?
Q.12
110
Spontaneous ? (Yes / No)
Exothermic ? (Yes / No)
Change
Yes
Burning of CO at 25C
Yes
Condensation of steam at 25C
No
Melting of ice at 25C
No
Dissolution of NH4Cl in water at 25C
111
6.7 Entropy change (SB p.164)

• Exothermicity is NOT the only reason for the
  spontaneity of a process
• Some spontaneous changes are endothermic
• E.g. Melting of ice
• Dissolution of NH4Cl in water

112
6.7 Entropy change (SB p.164)
113
6.7 Entropy change (SB p.165)
Entropy

• Entropy is a measure of the randomness or the
  degree of disorder (freedom) of a system

114
6.7 Entropy change (SB p.166)
Entropy change (?S)

• Entropy change means the change in the degree of
  disorder of a system
• ?S Sfinal - Sinitial

115
6.7 Entropy change (SB p.166)
Positive entropy change (?S gt 0)

• Increase in entropy
• Sfinal gt Sinitial
• Example
• Ice (low entropy) ?? Water (high entropy)
• ?S Swater Sice ve

116
6.7 Entropy change (SB p.166)
Negative entropy change (?S lt 0)

• Decrease in entropy
• Sinitial gt Sfianl
• Example
• Water (high entropy) ?? Ice (low entropy)
• ?S Sice Swater -ve

Q.13
117
Changes ?S
CO(g) O2(g) ? CO2(g)
H2O(g) ? H2O(l)
H2O(s) ? H2O(l)
NH4Cl(s) ? NH4Cl(aq)
118
Changes ?S
2CO(g) O2(g) ? 2CO2(g) -ve
H2O(g) ? H2O(l) -ve
H2O(s) ? H2O(l) ve
NH4Cl(s) ? NH4Cl(aq) ve
119
Changes ?S
C(s) O2(g) ? CO2(g)
SO2(g) O2(g) ? SO3(g)
Diffusion of a drop of ink in water
diamond ? graphite
120
Changes ?S
C(s) O2(g) ? CO2(g) ve
2SO2(g) O2(g) ? 2SO3(g) -ve
Diffusion of a drop of ink in water ve
diamond ? graphite ve
121
Consider an isolated system which has no exchange
of energy and matter with its surroundings
122
?S S2 S1 gt 0
Which one is spontaneous ?
?S S1 S2 lt 0
123
Spontaneous, ?S S2 S1 gt 0
Not spontaneous, ?S S1 S2 lt 0
124
A molecular statistical interpretation
Free adiabatic expansion of an ideal gas into a
vacuum.
125
A molecular statistical interpretation
Free adiabatic expansion of an ideal gas into a
vacuum.
Slit is open
126
A spontaneous process taking place in an isolated
system is always associated with an increase in
entropy (I.e. ?S gt 0)
6B
In closed system, The spontaneity of a process
depends on both ?H and ?S. The driving force of a
process is a balance of ?H and ?S.
127
6.7 Entropy change (SB p.166)
Ice (less entropy) ?? Water (more entropy) ?H is
ve ? not favourable ?S is ve ?
favourable Considering both ?S ?H , the process
is spontaneous
Q.14
128
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C
H2O(g) ? H2O(l) at 110C
H2O(s) ? H2O(l) at 25C
H2O(s) ? H2O(l) at -10C
129
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
130
Favourable
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
131
Not favourable
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
132
Spontaneity depends on temperature
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
133
Favourable
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
134
Not favourable
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
135
Spontaneity depends on temperature
Changes ?H ?S Spontaneous (Yes / No)
H2O(g) ? H2O(l) at 25C -ve -ve Yes
H2O(g) ? H2O(l) at 110C -ve -ve No
H2O(s) ? H2O(l) at 25C ve ve Yes
H2O(s) ? H2O(l) at -10C ve ve No
136
Spontaneity of a process depends on ?H, ?S T ?G
?H T?S G is the (Gibbs) free energy
137
?G ?H T?S Spontaneity depends on ?G Free
means the energy free for work
138
A spontaneous process is always associated with a
decrease in the free energy of the system. ?G lt 0
? spontaneous process ?G gt 0 ? not spontaneous
process
Q.15
139
?G ?H T?S
?H ?S T ?G Results
ve ve high
ve ve low
-ve -ve high
-ve -ve low
140
?G ?H T?S
Results
?G
T
?S
?H
Spontaneous
-ve
high
ve
ve
Not spontaneous
ve
low
ve
ve
-ve
-ve
-ve
-ve
141
?G ?H T?S
?H ?S T ?G Results
-ve ve high
-ve ve low
ve -ve high
ve -ve low
142
?G ?H T?S
Results
?G
T
?S
?H
Spontaneous
-ve
high
ve
-ve
low
ve
-ve
Not spontaneous
ve
high
-ve
ve
low
-ve
ve
143
?G ?H T?S
Q.16
144
?H lt 0 ?S gt 0
?G ?H T?S lt 0
The process is spontaneous, although activation
energy is required to start the conversion.
145
Spontaneous ?S S2 S1 gt 0
The entropy of a system can be considered as a
measure of the availability of the system to do
work. Before expansion, the system is available
to do work. After expansion, the system is not
available to do work.
146
Spontaneous ?S S2 S1 gt 0
The lower the entropy of a system(before
expansion), the more available is the system to
do work. Thus, entropy is considered as a measure
of the useless energy of a system.
147
?G ?H T?S G H TS H G TS
148
If the universe is an isolated system H is a
constant and ?H is always zero
H G TS
149
cosmic background radiation 4K
Useful energy
Useless energy
H G TS
?H ?G ?TS 0
?G T?S 0
?S always gt 0,
S always ?,
useless energy always ?
?G always lt 0,
G always ?,
useful energy always ?
150
In an isolated system, entropy will only increase
with time, it will not decrease with time.
The second law of thermodynamics
151
If the universe is an isolated system,
?Suniverse ?Ssystem ?Ssurroundings gt 0 the
total entropy (randomness) of the universe will
tend to increase to a maximum the total free
energy of the universe will tend to decrease to a
minimum.
152
As time increases, the universe will always
become more disordered. Entropy is considered as
a measure of time. Entropy can be used to
distinguish between future and past.
153
Time can only proceed in one direction that
results in an increase in the total entropy of
the universe. This is known as the arrow of
time.
154
The history of the universe
T ? 1.4?1032 K
H G TS
T ? 0 K
155
h Plancks constant G gravitational
constant c speed of light in vacuum k
Boltzmann constant
Plancks units
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
156
Absolute hot beyond which all physical laws break
down
Plancks units
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
157
Plancks units
Physical significance not yet known
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
158
Plancks units
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
159

• 7.3?10-37 m

Plancks units
?22 times of Mr Chios wavelength
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
160
It is the time required for light to travel, in a
vacuum, a distance of 1 Planck length.
Plancks units
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
161
10-15 s ? femtosecond(??) 10-18 s ?
attosecond(???)
Plancks units
Time taken for light to travel the length of 3 H
atoms
1.416785(71) 1032 K
Plancks temperature
Plancks length
Plancks time
162
Q.17(a)
The drop in temperature of the system is
accompanied by the rise in temperature of its
surroundings. ?Ssystem lt 0 ?Ssurroundings gt
0 ?Suniverse ?Ssystem ?Ssurroundings gt 0
163
Q.17(b)
The drop in entropy of the system is at the cost
of the rise in entropy of its surroundings.
?Ssystem lt 0 ?Ssurroundings gt 0 ?Suniverse
?Ssystem ?Ssurroundings gt 0
164
6.8 Free energy change (SB p.170)
165
The END
166
6.1 What is energetics? (SB p.140)
Back
Check Point 6-1
State whether the following processes are
exothermic or endothermic. (a) Melting of
ice. (b) Dissolution of table salt. (c)
Condensation of steam.
Answer
167
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.141)
Check Point 6-2

• State the difference between exothermic and
  endothermic reactions with respect to
• (i) the sign of ?H
• (ii) the heat change with the surroundings
• (iii) the total enthalpy of reactants and
  products.

Answer

• (i) Exothermic reactions ?H -ve endothermic
  reactions ?H ve
• (ii) Heat is given out to the surroundings in
  exothermic reactions whereas heat is taken in
  from the surroundings in endothermic reactions.
• (iii) In exothermic reactions, the total
  enthalpy of products is less than that of the
  reactants. In endothermic reactions, the total
  enthalpy is greater than that of the reactants.

168
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.141)
Check Point 6-2

• Draw an enthalpy level diagram for a reaction
  which is
• (i) endothermic, having a large activation
  energy.
• (ii) exothermic, having a small activation
  energy.

Answer
169
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.141)
Check Point 6-2
170
6.2 Enthalpy changes related to breaking and
forming of bonds (SB p.141)
Check Point 6-2
Back
171
6.3 Standard enthalpy changes (SB p.147)
Check Point 6-3

1. Why must the condition burnt completely in
   oxygen be emphasized in the definition of
   standard enthalpy change of combustion?

Answer

1. If the substance is not completely burnt in
   excess oxygen, other products such as C(s) and
   CO(g) may be formed. The enthalpy change of
   combustion measured will not be accurate.

172
6.3 Standard enthalpy changes (SB p.147)
Check Point 6-3

• (b) The enthalpy change of the following reaction
  under standard conditions is 566.0 kJ.
• 2CO(g) O2(g) ?? 2CO2(g)
• What is the standard enthalpy change of
  combustion of carbon monoxide?

Answer
173
6.3 Standard enthalpy changes (SB p.147)
Check Point 6-3
(c) ½ Enthalpy change of combustion of nitrogen
or enthalpy change of formation of nitrogen
dioxide.
Answer
Back
174
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.149)
Example 6-4A
Determine the enthalpy change of neutralization
of 25 cm3 of 1.25 M hydrochloric acid and 25 cm3
of 1.25 M sodium hydroxide solution using the
following data Mass of calorimeter 100
g Initial temperature of acid 15.5 oC (288.5
K) Initial temperature of alkali 15.5 oC
(288.5 K) Final temperature of the reaction
mixture 21.6 oC (294.6 K) The specific heat
capacities of water and calorimeter are 4200 J
kg-1 K-1 and 800 J kg-1 K-1 respectively.
Answer
175
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.149)
Example 6-4A
Assume that the density of the reaction mixture
is the same as that of water, i.e. 1 g cm-3. Mass
of the reaction mixture (25 25) cm3 ? 1 g
cm-3 50 g 0.05 kg Heat given out (m1c1
m2c2) ?T (0.05 kg ?
4200 J kg-1 K-1 0.1 kg ? 800 J kg-1 K-1) ?
(294.6 288.5) K
1769 J H(aq) OH-(aq) ?? H2O(l) Number of
moles of HCl 1.25 mol dm-3 ? 25 ? 10-3 dm3
0.03125 mol Number of moles of NaOH 1.25 mol
dm-3 ? 25 ? 10-3 dm3 0.03125 mol Number of
moles of H2O formed 0.03125 mol
176
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.149)
Example 6-4A
Back
177
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.151)
Example 6-4B
Determine the enthalpy change of combustion of
ethanol using the following data Mass of spirit
lamp before experiment 45.24 g Mass of spirit
lamp after experiment 44.46 g Mass of water in
copper calorimeter 50 g Mass of copper
calorimeter without water 380 g Initial
temperature of water 18.5 oC (291.5 K) Final
temperature of water 39.4 oC (312.4 K) The
specific heat capacities of water and copper
calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1
K-1 respectively.
Answer
178
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.151)
Example 6-4B
179
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.151)
Example 6-4B
Back
180
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.152)
Example 6-4C
0.02 mol of anhydrous ammonium chloride was added
to 45 g of water in a polystyrene cup to
determine the enthalpy change of solution of
anhydrous ammonium chloride. It is found that
there was a temperature drop from 24.5 oC to
23.0 oC in the solution. Given that the specific
heat capacity of water is 4200 J kg-1 K-1
and NH4Cl(s) aq ?? NH4Cl(aq) Calculate the
enthalpy change of solution of anhydrous ammonium
chloride. (Neglect the specific heat capacity of
the polystyrene cup.)
Answer
181
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.152)
Example 6-4C
Back
182
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4
(a) A student tried to determine the enthalpy
change of neutralization by putting 25.0 cm3 of
1.0 M HNO3 in a polystyrene cup and adding 25.0
cm3 of 1.0 M NH3 into it. The temperature rise
recorded was 3.11 oC. Given that the mass of the
polystyrene cup is 250 g, the specific heat
capacities of water and the polystyrene cup are
4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively.
Determine the enthalpy change of neutralization
of nitric acid and aqueous ammonia. (Density of
water 1 g cm-3)
Answer
183
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4

• Heat evolved m1c1?T m2c2 ?T
• 0.050 kg ? 4200 J
  kg-1 K-1 ? 3.11 K 0.25 kg ?
  800 J kg-1 K-1 ? 3.11 K
• (653.1 622) J
• 1275.1 J
• No. of moles of HNO3 used 1.0 M ? 25 ? 10-3
  dm3
• 
  0.025 mol

184
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4
Back

• No. of moles of NH3 used 1.0 M ? 25 ? 10-3 dm3
• 
  0.025 mol
• No. of moles of H2O formed 0.025 mol

51.004 kJ mol-1 The enthalpy change of
neutralization of nitric acid and aqueous ammonia
is 51.004 kJ mol-1.
185
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4

• When 0.05 mol of silver nitrate was added to 50 g
  of water in a polystyrene cup, a temperature drop
  of 5.2 oC was recorded. Assuming that there was
  no heat absorption by the polystyrene cup,
  calculate the enthalpy change of solution of
  silver nitrate.
• (Specific heat capacity of water 4200 J kg-1
  K-1)

Answer
186
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4
Back
187
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4
(c) A student used a calorimeter as shown in
Fig. 6-15 to determine the enthalpy change of
combustion of methanol. In the experiment, 1.60 g
of methanol was used and 50 g of water was
heated up, raising the temperature by 33.2 oC.
Given that the specific heat capacities of water
and copper calorimeter are 4200 J kg-1 K-1 and
2100 J kg-1 K-1 respectively and the mass of the
calorimeter is 400 g, calculate the enthalpy
change of combustion of methanol.
Answer
188
6.4 Experimental determination of enthalpy
changes by calorimetry (SB p.153)
Check Point 6-4

• Heat evolved m1c1?T m2c2 ?T
• 50 g ? 4.18 J g-1
  K-1 ? 33.2 K 400g ? 2.10 J g-1 K-1 ?
  33.2 K
• (6939 27888) J
• 34827 J
• No. of moles of methanol used
• 
  0.05 mol
• Heat evolved per mole of methanol used
• 
  696.5 kJ mol-1
• The enthalpy change of combustion of methanol is
  696.5 kJ mol-1.

189
6.5 Hesss law (SB p.158)
Check Point 6-5
Answer
190
6.5 Hesss law (SB p.158)
Check Point 6-5
191
6.5 Hesss law (SB p.158)
Check Point 6-5
Answer
192
6.5 Hesss law (SB p.158)
Check Point 6-5
193
6.5 Hesss law (SB p.158)
Check Point 6-5
Answer
194
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.159)
Example 6-6A
Answer
195
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.159)
Example 6-6A
Back
196
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.160)
Example 6-6B
Answer
197
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.160)
Example 6-6B
Back
198
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.160)
Example 6-6C
Answer
199
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.160)
Example 6-6C
Back
200
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.161)
Example 6-6D
Answer
201
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.161)
Example 6-6D
Back
202
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.162)
Example 6-6E
Answer
203
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.162)
Example 6-6E
204
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.163)
Back
Example 6-6E
205
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.163)
Example 6-6F
Answer
206
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.163)
Example 6-6F
207
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.163)
Back
Example 6-6F
208
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.164)
Check Point 6-6
Answer
209
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.164)
Check Point 6-6
210
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.164)
Check Point 6-6
Answer
211
6.6 Calculations involving standard enthalpy
changes of reactions (SB p.164)
Back
Check Point 6-6
212
6.7 Entropy change (SB p.167)
Back
Check Point 6-7

• Predict whether the following changes or
  reactions involve an increase or a decrease in
  entropy.
• Dissolving salt in water to form salt solution
• Condensation of steam on a cold mirror
• Complete combustion of carbon
• Complete combustion of carbon monoxide
• Oxidation of sulphur dioxide to sulphur
  trioxide

Answer
213
6.8 Free energy change (SB p.170)
Back
Let's Think 1
In the process of changing of ice to water, at
what temperature do you think ?G equals 0?
Answer
?G equals 0 means that neither the forward nor
the reverse process is spontaneous. The system is
therefore in equilibrium. Melting point of ice is
0 oC (273 K) at which the process of changing ice
to water and the process of water turning to ice
are at equilibrium. At 0 oC, ?G of the processes
equals 0.
214
6.8 Free energy change (SB p.170)
Check Point 6-8

• At what temperatures is the following process
  spontaneous at 1 atmosphere?
• Water ?? Steam
• What are the two driving forces that determine
  the spontaneity of a process?

Answer
215
6.8 Free energy change (SB p.170)
Check Point 6-8
Back

• State whether each of the following cases is
  spontaneous at all temperatures, not spontaneous
  at any temperature, spontaneous at high
  temperatures or spontaneous at low temperatures.
• (i) positive ?S and positive ?H
• (ii) positive ?S and negative ?H
• (iii) negative ?S and positive ?H
• (iv) negative ?S and negative ?H

1. Spontaneous at high temperatures
2. Spontaneous at all temperatures
3. Not spontaneous at any temperature
4. Spontaneous at low temperatures

Answer
216
Enthalpy change of formation of CaCO3(s)
6.5 Hesss law (SB p.153)
217
Enthalpy change of hydration of MgSO4(s)
6.5 Hesss law (SB p.153)
aq
ø