Download Energetics

6
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
Experimental Determination of Enthalpy
Changes by Calorimetry
6.5
Hess’s Law
6.6
Calculations involving Standard Enthalpy
Changes of Reactions
6.1 What is energetics? (SB p.136)
What is energetics?
Energetics is the study of energy changes
associated with chemical reactions.
Thermochemistry is the study of heat changes
associated with chemical reactions.
2
Internal Energy (U)
U = kinetic energy + potential energy
3
Kinetic Energy  T (K)
translational
rotational
vibrational
heat
Energy
Potential Energy
Relative position among particles
Bond breaking  P.E. 
Bond forming  P.E. 
4
Bond breaking : -
H – H(g)  H(g)
+
H(g)
P.E. 
Bond forming : -
H(g)
+
H(g) 
H – H(g)
Ionization : Na(g)  Na+(g) + e
5
P.E. 
P.E. 
Q.1
4H(g) + 2O(g)
Internal
energy
U1
bond breaking
bond
forming
2H2(g) + O2(g)
U = U2 – U1 = -(y-x) kJ
U2
2H2O(l)
Reaction coordinate
6
6.1 What is energetics? (SB p.137)
Internal energy and enthalpy
H = U + PV
enthalpy
7
Internal
energy
6.1 What is energetics? (SB p.137)
Internal energy and enthalpy
e.g.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
qv = U = -473 kJ mol1
Mg
8
qp = H = -470 kJ mol1
Mg
qv = U = -473 kJ mol1
qv
= qp – 3
= qp – w
= qp - PV
qp = H = -470 kJ mol1
PV  (Nm2)(m3)
= Nm
Force  displacement
Work done against the surroundings
9
6.1 What is energetics? (SB p.138)
Internal energy and enthalpy
H = U + PV
= U + PV at constant P
qp
=
qv
Heat
Heat
change at change at
fixed P
fixed V
10
+
PV
Work
done
6.1 What is energetics? (SB p.138)
Internal energy and enthalpy
qp
=
qv
+
PV
On expansion, PV > 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)
11
6.1 What is energetics? (SB p.138)
Internal energy and enthalpy
qp
=
qv
+
PV
On contraction, PV < 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)
12
• 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.
13
6.1 What is energetics? (SB p.138)
Exothermic and endothermic reactions
An exothermic reaction is a reaction that
releases heat energy to the surroundings.
(H = -ve)
14
6.1 What is energetics? (SB p.139)
Check Point 6-1
Exothermic and endothermic reactions
An endothermic reaction is a reaction that
absorbs heat energy from the surroundings.
(H = +ve)
15
6.1 What is energetics? (SB p.136)
Law of conservation of energy
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
16
Exothermic : P.E. of the system  K.E. of the surroundings
Endothermic : K.E. of the surroundings  P.E. of the system
17
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
Enthalpy changes related to breaking and
forming of bonds
CH4 + 2O2 CO2 + 2H2O
18
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
In an exothermic reaction,
E absorbed to break bonds < E released as bonds are formed.
19
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
Enthalpy changes related to breaking and
forming of bonds
N2(g) + 2O2(g)  2NO2(g)
20
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
In an endothermic reaction,
E absorbed to break bonds > E released as bonds are formed.
Check Point 6-2
21
For non-gaseous reactions,
PV  0
H = U + PV  U
For gaseous reactions,
H = U + PV = U + (n)RT (PV = nRT)
22
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
 8.314 
 890  (1  3)
298 
 1000 
= 885 kJ mol1
23
C(g) + 4H(g) + 4O(g)
Enthalpy
H1
bond breaking
bond
forming
CH4(g) + 2O2(g)
H = 890 kJ mol1
H2
CO2(g) + 2H2O(l)
Reaction coordinate
24
At 298K
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
At 373K
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(l)
373K
25

2H2O(g)
373K
+88 kJ
H / kJ mol1
890
802
C(g) + 4H(g) + 4O(g)
Enthalpy
H1
CH4(g) + 2O2(g)
298 K
H = 890 kJ mol1
H2
CO2(g) + 2H2O(l)
Reaction coordinate
26
C(g) + 4H(g) + 4O(g)
Enthalpy
H 1’
CH4(g) + 2O2(g)
373 K
H = 890 kJ mol1
H2’
CO2(g) + 2H2O(l)
Assume constant H
Reaction coordinate
27
C(g) + 4H(g) + 4O(g)
Enthalpy
CH4(g) + 2O2(g)
373 K
H = 890 kJ mol1
CO2(g) + 2H2O(l)
In fact, H depends on T
Reaction coordinate
28
C(g) + 4H(g) + 4O(g)
Enthalpy
CH4(g) + 2O2(g)
373 K
H = 802 kJ mol1
CO2(g) + 2H2O(g)
CO2(g) + 2H2O(l)
H = +88 kJ mol1
Reaction coordinate
29
6.3
Standard Enthalpy
Changes
30
6.3 Standard enthalpy changes (SB p.141)
Standard enthalpy changes
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ mol-1 at 373 K
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -890 kJ mol-1 at 298 K
31
6.3 Standard enthalpy changes (SB p.141)
Standard enthalpy changes
As enthalpy changes depend on temperature
and pressure, it is necessary to define
standard conditions:
1.
2.
3.
elements or compounds in their normal
physical states;
a pressure of 1 atm (101325 Nm-2); and
a temperature of 25oC (298 K)
Enthalpy change under standard conditions
denoted by symbol: H
32
Standard enthalpy change of reaction
The enthalpy change when the molar quantities of
reactants as stated in the equation react under
standard conditions.
2H2(g) + O2(g)  2H2O(l)
H = 572 kJ mol1
33
per mole of O2
Standard enthalpy change of reaction
2H2(g) + O2(g)  2H2O(l)
H = 572 kJ mol1
per mole of O2
1
H2(g) + O2(g)  H2O(l)
2
H = 286 kJ mol1
per mole of H2 or H2O
4H2(g) + 2O2(g)  4H2O(l)
H = 1144 kJ
H depends on the equation
34
Standard enthalpy change of formation
Hf
The enthalpy change when one mole of the
substance is formed from its elements under
standard conditions.
1
H2(g) + O2(g)  H2O(l)
2
35
Hf [H2O] = 286 kJ mol1
Q.3
Hf [element] = 0 kJ mol1
O2(g)  O2(g)
Hf [O2] = 0 kJ mol1
C(graphite)  C(diamond)
Most stable
allotrope
36
Hf [diamond] = +1.9 kJ mol1
Q.4
(i)
C(graphite) + O2(g)  CO2(g)
(ii)
C(graphite) + 2H2(g)  CH4(g)
(iii)
1
Mg(s) + 2 O2(g)  MgO(s)
1
3
Na(s) + 2 H2(g) + C(graphite) + O2(g)  NaHCO3(s)
2
(iv)
(v)
37
2C(graphite) + 2H2(g) + O2(g)  CH3COOH(l)
Q.5
1
H2(g) + O2(g)  H2O(l)
2
qv = U = 140.3 kJ per g of H2
no.of moles of H2 (g) used 
1.00
 0.496 mol
2  1.008
0.496
no.of moles of O2 (g) used 
mol  0.248 mol
2
n = 0 – 0.496 – 0.248 = -0.744 mol
38
Q.5
H = U + nRT
 8.31 kJ mol -1 K -1 
(298 K)
 140.3 kJ  (-0.744 mol) 
1000


= -142.1 kJ
Heat released for the formation of 0.496 mol of water
 142.1 kJ
-1

-286.5
kJ
mol
Molar Hf[H2O] =
0.496 mol
39
Standard enthalpy change of combustion
Hc
The enthalpy change when one mole of the
substance undergoes complete combustion
under standard conditions.
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
Hc [C2H5OH(l)] = -1368 kJ mol1
40
6.3 Standard enthalpy changes (SB p.147)
Substance
Hc (kJ mol-1)
C (diamond)
C (graphite)
-395.4
-393.5
Enthalpy
C(diamond) + O2(g)
1.9
395.4
C(graphite) + O2(g)
393.5
CO2(g)
Reaction coordinate
41
Hf [diamond]
= +1.9 kJ mol1
Q.6
(a)
1
C(graphite) + O2(g)  CO(g)
2
Incomplete
combustion
H = Hf [CO(g)]  Hc [graphite]
(b)
2H2(g) + O2(g)  2H2O(l)
H = 2  Hc [H2(g)]
= 2  Hf [H2O(l)]
42
Q.6
(c)
C(graphite) + O2(g)  CO2(g)
H = Hc [graphite] = Hf [CO2(g)]
(d)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = Hc [CH4(g)]
 Hf [CO2(g)]
 2  Hf [H2O(l)]
Not formed from
elements
Check Point 6-3
43
6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy changes of
neutralization
Standard enthalpy change of neutralization
(Hneut) is the enthalpy change when one mole of
water is formed from the neutralization of an
acid by an alkali under standard conditions.
e.g. H+(aq) + OH-(aq)  H2O(l)
Hneut = -57.3 kJ mol-1
44
6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy changes of
neutralization
Enthalpy level diagram for the neutralization of a
strong acid and a strong alkali
45
6.3 Standard enthalpy changes (SB p.142)
Acid
HCl
HCl
HCl
HF
NH3(aq) + H2O(l)
Alkali
NaOH
KOH
NH3
NaOH
Hneu
-57.1
-57.2
-52.2
-68.6
NH4+(aq) + OH(aq)
H1 > 0
H+(aq) + OH-(aq) + Cl(aq)  H2O(l) + Cl(aq) H2 = -57.3
NH3(aq) + HCl(aq)  NH4Cl (aq)
Hneu = H1 + H2 = 52.2 kJ mol1
46
6.3 Standard enthalpy changes (SB p.142)
Acid
HCl
HCl
HCl
HF
HF(aq)
Hneu
-57.1
-57.2
-52.2
-68.6
Alkali
NaOH
KOH
NH3
NaOH
H+(aq) + F(aq)
H1 < 0
H+(aq) + OH-(aq) + Na+(aq)  H2O(l) + Na+(aq) H2 = -57.3
HF(aq) + NaOH(aq)  NaF(aq) + H2O(l)
Hneu = H1 + H2 = 68.6 kJ mol1
47
6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy change of solution
Standard enthalpy change of solution (Hsoln) is
the enthalpy change when one mole of a solute
is dissolved in a specified number of moles of
solvent (e.g. water) under standard conditions.
NaCl(s) + 10H2O(l)  Na+(aq) + Cl-(aq)
Hsoln[NaCl(s)]= +2.008 kJ mol-1
NaCl(aq)
48
dilution
NaCl(aq) H > 0
6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy change of solution
Standard enthalpy change of solution (Hsoln) is
the enthalpy change when one mole of a solute
is dissolved to form an infinitely dilute solution
under standard conditions.
concentration  0
NaCl(s) + water  Na+(aq) + Cl-(aq)
Hsoln= +4.98 kJ mol-1
49
6.3 Standard enthalpy changes (SB p.143)
Standard enthalpy change of solution
e.g. NaCl(s) + water  Na+(aq) + Cl-(aq)
Hsoln= +4.98 kJ mol-1
+ 4.98 kJ mol1
Enthalpy level diagram
for the dissolution of
NaCl
50
6.3 Standard enthalpy changes (SB p.143)
Standard enthalpy change of solution
e.g. LiCl(s) + water  Li+(aq) + Cl-(aq)
Hsoln= -37.2 kJ mol-1
Enthalpy level diagram
for the dissolution of
LiCl in water
51
6.3 Standard enthalpy changes (SB p.143)
Standard enthalpy change of solution
52
Salt
Hsoln(kJ mol-1)
NH3
NaOH
HCl
H2SO4
LiCl
NaCl
NaNO3
NH4Cl
35.5
43.1
73.0
74.0
37.2
+4.98
+21.0
+22.6
6.4
Experimental
Determination of
Enthalpy Changes
by Calorimetry
53
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)
Experimental determination of enthalpy
changes by calorimetry
Calorimeter is any set-up used for the
determination of H.
By temperature measurement.
H = qp = (m1c1 + m2c2)T
54
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.
55
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Determination of enthalpy change of
neutralization
56
If the reaction is fast enough, T1  T2
T1
T2
H  (m1c1 + m2c2)(T2 – T0)
H = (m1c1 + m2c2)(T1 – T0)
T0
t1
57
t2
Check Point 6-4(a)
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
58
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
59
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Heat evolved = (m1c1 + m2c2) ΔT
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,
ΔT is the temperature change of the reaction
60
Q.7(Example)
qp  (800 g)(4.18 J g-1K1 )(13 K)  43.5 kJ
1.5 g
no. of moles of ethanol 
 0.326 mol
46.0 g
ΔHc [ethanol] 
- 43.5 kJ
 - 1330 kJ mol -1
0.0326 mol
heat given out
Check Point 6-4(c)
61
6.5
Hess’s Law
62
6.5 Hess’s law (SB p.153)
Hess’s Law
Hess’s 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.
63
6.5 Hess’s law (SB p.153)
Hess’s Law
H2
A(HA)
Route 2
C
Route 1
H1
H4
H3
B(HB)
H5
D
Route 3
H1 = HB – HA = H2 + H3 = H4 + H5
64
6.5 Hess’s law (SB p.155)
Importance of Hess’s law
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 Hess’s Law.
65
6.5 Hess’s law (SB p.153)
Enthalpy change of formation of CO(g)
C(graphite) + ½O2(g)
Hf [CO(g)]
CO(g)
Hf [CO(g)] cannot be determined directly
due to further oxidation of CO to CO2
The reaction cannot be controlled.
66
6.5 Hess’s law (SB p.153)
Enthalpy change of formation of CO(g)
Hc [graphite] = -393 kJ mol-1
Hc [CO(g)] = -283.0 kJ mol-1
C(graphite) + ½O2(g)
+ ½O2(g)
Hf [CO(g)]
+ ½O2(g)
H1
CO2(g)
H2
Hf [CO(g)] + H2 = H1
Hf [CO(g)] = H1 - H2
= -393 - (-283 )
= -110 kJ mol-1
67
CO(g)
6.5 Hess’s law (SB p.155)
Enthalpy cycle (Born-Haber cycle)
• Relate the various equations involved in a
reaction
C(graphite) + ½O2(g)
+ ½O2(g)
68
Hf [CO(g)]
CO(g)
+ ½O2(g)
H1
CO2(g)
H2
6.5 Hess’s law (SB p.153)
Steps for drawing Born-Haber cycle
1. Give the equation for the change being considered.
C(graphite) + ½O2(g)
69
Hf [CO(g)]
CO(g)
6.5 Hess’s law (SB p.153)
Steps for drawing Born-Haber cycle
2. Complete the cycle by giving the equations for the
combustion reactions of reactants and products.
C(graphite) + ½O2(g)
+ ½O2(g)
H1
Hf [CO(g)]
CO(g)
+ ½O2(g)
CO2(g)
H2
Hf [CO(g)]= Hc[graphite] - Hc[CO(g)]
70
6.5 Hess’s law (SB p.153)
Steps for drawing Born-Haber cycle
2. Complete the cycle by giving the equations for the
combustion reactions of reactants and products.
C(graphite) + ½O2(g)
+ ½O2(g)
H1
Hf [CO(g)]
CO(g)
+ ½O2(g)
CO2(g)
H2
Hf [CO(g)]= Hc[reactant] - Hc[product]
71
Calculation of standard enthalpy change of
formation from standard enthalpy changes
of combustion
6B
Hf =  Hc [reactants] -  Hc [product]
72
Q.8
4C(graphite) + 5H2(g)
Hf [C4H10(g)]
+ 4O2(g) 5Hc [H2(g)]
+ 2.5 O2(g)
4Hc [graphite]
C4H10(g)
Hc [C4H10(g)]
+ 6.5 O2(g)
4CO2(g) + 5H2O(l)
Hf [C4H10(g)]
= 4Hc[C(graphite)] + 5Hc[H2(g)] - Hc[C4H10(g)]
= [4(-393)+ 5(-286) – (2877)] kJ mol1
= 125 kJ mol1
73
Q.8
Method B :
By addition and/or subtraction of equations with known Hc
Hc /kJ mol1
(1) C(graphite) + O2(g)  CO2(g)
1
(2) H2(g) + O2(g)  H2O(l)
2
(3) C4H10(g) + 6
393
286
1
O2(g)  4CO2(g) + 5H2O(l) 2877
2
Overall reaction : 4(1) + 5(2) – (3)
Hf [C4H10(g)]
4C(graphite) + 5H2(g)
C4H10(g)
Hf [C4H10(g)] = [4(-393) + 5(-286)  (2877)] kJ mol1
= 125 kJ mol1
74
6.5 Hess’s law (SB p.154)
Enthalpy level diagram
• Relate substances together in terms of
enthalpy changes of reactions
Enthalpy level
diagram for the
oxidation of
C(graphite) to CO2(g)
75
Steps for drawing enthalpy level diagram
Enthalpy / kJ mol1
1. Draw the enthalpy level of elements.
76
C(graphite) + O2(g)
Steps for drawing enthalpy level diagram
Enthalpy / kJ mol1
2. Enthalpies of elements are arbitrarily taken as zero.
77
C(graphite) + O2(g)
Steps for drawing enthalpy level diagram
Enthalpy / kJ mol1
3. Higher enthalpy levels are drawn above that of elements
78
C(graphite) + O2(g)
Steps for drawing enthalpy level diagram
Enthalpy / kJ mol1
4. Lower enthalpy levels are drawn below that of elements
C(graphite) + O2(g)
Hc[graphite]
= 393 kJ
CO2(g)
Route 1
79
Steps for drawing enthalpy level diagram
Enthalpy / kJ mol1
4. Lower enthalpy levels are drawn below that of elements
C(graphite) + O2(g)
Hf[CO(g)] = 110 kJ
Hc[graphite]
= 393 kJ
1
O (g)
2 2
Hc[CO(g)] = 283 kJ
CO2(g)
Route 1
80
CO(g) +
Route 2
6.5 Hess’s law (SB p.158)
(c) The formation of ethyne (C2H2(g) can be represented by
the following equation:
2C(graphite) + H2(g)  C2H2(g)
(i) Draw an enthalpy cycle relating the above
equation to carbon dioxide and water.
(ii) Calculate the standard enthalpy change of formation
of ethyne.
(Given: Hc [C(graphite)] = -393.5 kJ mol-1;
Hc [H2(g)] = -285.8 kJ mol-1;
Hc [C2H2(g)] = -1299 kJ mol-1)
81
6.5 Hess’s law (SB p.158)
Hf
2C(graphite) + H2(g)  C2H2(g)
2Hc[graphite]
+ 2O2(g)
Hc[H2(g)]
+ 0.5O2(g)
Hc[C2H2(g)]
+ 2.5O2(g)
2CO2(g) + H2O(l)
By Hess’s law,
Hf
+ Hc[C2H2(g)] = 2Hc[graphite] + Hc[H2(g)]
Hf = 2Hc[graphite] + Hc[H2(g)] - Hc[C2H2(g)]
= [2(393.5) + (285.8) –(1299)] kJ mol1
= +226.2 kJ mol1
82
Enthalpy / kJ mol1
(iii).
Draw an enthalpy level diagram for the reaction using
the enthalpy changes in (ii)
C2H2(g) + 2.5O2(g)
Hf[C2H2(g)]
2C(graphite) + H2(g) + 2.5O2(g)
2Hc[graphite]
Hc[C2H2(g)]
2CO2(g) + H2(g) + 0.5O2(g)
Hc[H2(g)]
2CO2(g) + H2O(l)
Route 1
83
Route 2
6.6
Calculations involving
Standard Enthalpy
Changes of Reactions
84
6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
Calculation of standard enthalpy change of
reaction from standard enthalpy changes
of formation
H from Hf
85
Q.9
NH3(g) + HCl(g)
H
Hf[HCl(g)]
Hf[NH3(g)]
NH4Cl(s)
Hf[NH4Cl(s)]
0.5N2(g) + 1.5H2(g) + 0.5H2(g) + 0.5Cl2(g)
By Hess’s law,
Hf[NH3(g)] + Hf[HCl(g)] + H
= Hf[NH4Cl(s)]
H = Hf[NH4Cl(s)] - Hf[NH3(g)] - Hf[HCl(g)]
= [314 –(46) –(92)] kJ mol1 = 176 kJ mol1
86
Q.9
NH3(g) + HCl(g)
H
Hf[HCl(g)]
Hf[NH3(g)]
NH4Cl(s)
Hf[NH4Cl(s)]
0.5N2(g) + 1.5H2(g) + 0.5H2(g) + 0.5Cl2(g)
By Hess’s law,
Hf[NH3(g)] + Hf[HCl(g)] + H
= Hf[NH4Cl(s)]
H = Hf[NH4Cl(s)] - Hf[NH3(g)] - Hf[HCl(g)]
Hreaction =  Hf [products] -  Hf [reactants]
87
Q.10
4CH3NHNH2(l) + 5N2O4(l)
H
5Hf[N2O4(l)]
4Hf[CH3NHNH2(l)]
4CO2(g) + 9N2(g) + 12H2O(l)
4Hf[CO2(g)]
12Hf[H2O(l)]
4C(graphite) + 12H2(g) + 4N2(g) + 5N2(g) + 10O2(g)
By Hess’s law,
H = [4(393) + 12(286) - 4(+53) – 5(20)] kJ
= 5116 kJ
Highly exothermic/ignites spontaneously
Used as rockel fuel in Apollo 11
88
Q.11
C3H6(g) + H2(g)
H
C3H8(g)
H /kJ mol1
(1)
C3H6(g) + 4.5O2(g)  3CO2(g) + 3H2O(l)
(2)
H2(g) + 0.5O2(g)  H2O(l)
(3)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Overall reaction : (1) + (2) – (3)
H = [(2058) + (286) – (2220)] kJ mol1
= 124 kJ mol1
89
2058
286
2220
Q.11
C3H6(g) + H2(g)
H
Hf[H2O(l)]
Hc[C3H6(g)]
C3H8(g)
Hc[C3H8(g)]
3H2O(l) + 3CO2 (g) + H2O(l)
By Hess’s law,
H + (2220) = (2058) + (286)
H = [(2058) + (286) – (2220)] kJ mol1
= 124 kJ mol1
90
Q.11
C3H6(g) + H2(g)
H
Hc[H2(g)]
Hc[C3H6(g)]
C3H8(g)
Hc[C3H8(g)]
3H2O(l) + 3CO2 (g) + H2O(l)
Hf =  Hc [reactants] -  Hc [product]
Hrx =  Hc [reactants] -  Hc [products]
91
Relative stability of compounds and Hf
Hf indicates the energetic stability of the compound
with respect to its elements
Elements  Compound
Hf
Hf
< 0  energetically more stable than its elements
Hf
> 0  energetically less stable than its elements
92
H2(g) + O2(g)  H2O2(l)
Hf [H2O2(l)] = 188 kJ mol1
Enthalpy / kJ mol1
H2O2(l)  H2O(l) + 0.5O2(g)
93
Hrx = 98 kJ mol1
H2(g) + O2(g)
Hf[H2O2(l)] = 188 kJ
H2O2(l)
Hrx= 98 kJ
H2O(l) + 0.5O2(g)
H2O2 is energetically stable w.r.t. H2 and O2
Enthalpy / kJ mol1
H2O2 is energetically unstable w.r.t. H2O and 0.5O2
94
H2(g) + O2(g)
Hf[H2O2(l)] = 188 kJ
H2O2(l)
Hrx= 98 kJ
H2O(l) + 0.5O2(g)
Enthalpy / kJ mol1
H2 and O2 are energetically unstable w.r.t. H2O2 and H2O
95
H2(g) + O2(g)
Hf[H2O2(l)] = 188 kJ
Hc[H2(g) ]
= 286 kJ
H2O2(l)
Hrx= 98 kJ
H2O(l) + 0.5O2(g)
C(diamond)  C(graphite)
H = 2 kJ mol1
Enthalpy / kJ mol1
Energetically
unstable
96
C(diamond) + O2(g)
H = 2 kJ
C(graphite) + O2(g)
Hc[diamond]
= 395 kJ
CO2(g)
Hc[graphite]
= 393 kJ
C(diamond)  C(graphite)
H = 2 kJ mol1
Enthalpy / kJ mol1
Kinetically
The rate of conversion is extremely low
stable
97
C(diamond) + O2(g)
H = 2 kJ
C(graphite) + O2(g)
Hc[diamond]
= 395 kJ
CO2(g)
Hc[graphite]
= 393 kJ
C(g) + O(g) + O(g)
Enthalpy / kJ mol1
bond breaking
98
bond forming
C(diamond) + O2(g)
C(graphite) + O2(g)
Hc[diamond]
= 395 kJ
CO2(g)
Hc[graphite]
= 393 kJ
Rate of reaction depends on the ease of bond breaking in
reactants (e.g. diamond)
Enthalpy / kJ mol1
The minimum energy required for a reaction to start is
known as the activation energy, Ea
99
C(diamond) + O2(g)
H = 2 kJ
C(graphite) + O2(g)
Hc[diamond]
= 395 kJ
CO2(g)
Hc[graphite]
= 393 kJ
C(diamond)  C(graphite)
H = 2 kJ mol1
Enthalpy / kJ mol1
The extremely low rate is due to high Ea
100
C(diamond) + O2(g)
H = 2 kJ
C(graphite) + O2(g)
Hc[diamond]
= 395 kJ
CO2(g)
Hc[graphite]
= 393 kJ
C(diamond)  C(graphite)
H = 2 kJ mol1
Enthalpy / kJ mol1
Ea is always > 0 as bond breaking is endothermic
101
C(diamond) + O2(g)
H = 2 kJ
C(graphite) + O2(g)
Hc[diamond]
= 395 kJ
CO2(g)
Hc[graphite]
= 393 kJ
Ea tells how fast a reaction can proceed.
Rate of reaction / kinetics / kinetic stability
Hf tells how far a reaction can proceed
Equilibrium / energetics / energetic stability
102
Hf
 Energetic stability
Hf
< 0  higher energetic stability w.r.t. its elements
Hf
> 0  lower energetic stability w.r.t. its elements
Ea  kinetic stability (rate of reaction)
Higher Ea
Lower Ea
103

higher kinetic stability of reactants
w.r.t. products

lower rate to give products

lower kinetic stability of reactants
w.r.t. products

higher rate to give products
diamond
graphite
Diamond  energetically unstable w.r.t. graphite
 kinetically stable w.r.t. graphite
Graphite  stable w.r.t. diamond
energetically and kinetically
104
diamond
graphite
diamond
105
extremely slow
extremely slow
graphite
6.7 Entropy change (SB p.164)
6.7
Spontaneity of Changes
106
Spontaneity : The state of being spontaneous
Spontaneous :
- self-generated
- natural
- happening without external influence
external
external
boundary
internal
external
external
107
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
108
6.7 Entropy change (SB p.164)
• Spontaneous physical change:
E.g. condensation of steam at 25°C
• Spontaneous chemcial change:
E.g. burning of wood once the fire has started
Exothermic  spontaneous ?
Endothermic  not spontaneous ?
Q.12
109
110
Change
Exothermic ?
(Yes / No)
Spontaneous ?
(Yes / No)
Burning of CO at
25°C
Yes
Yes
Condensation of
steam at 25°C
Yes
Yes
Melting of ice at
25°C
No
Yes
Dissolution of
NH4Cl in water
at 25°C
No
Yes
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
111
6.7 Entropy change (SB p.164)
Melting of ice
Dissolution of ammonium
nitrate in water
112
6.7 Entropy change (SB p.165)
Entropy
• Entropy is a measure of the randomness or the
degree of disorder (freedom) of a system
Solid
Liquid
Entropy increases
113
Gas
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
114
6.7 Entropy change (SB p.166)
Positive entropy change (S > 0)
• Increase in entropy
• Sfinal > Sinitial
• Example:
Ice (low entropy)  Water (high entropy)
S = Swater – Sice = +ve
115
6.7 Entropy change (SB p.166)
Negative entropy change (S < 0)
• Decrease in entropy
• Sinitial > Sfianl
• Example:
Water (high entropy)  Ice (low entropy)
S = Sice – Swater = -ve
Q.13
116
Changes
CO(g) + O2(g)  CO2(g)
H2O(g)  H2O(l)
H2O(s)  H2O(l)
NH4Cl(s)  NH4Cl(aq)
117
S
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
Changes
C(s) + O2(g)  CO2(g)
SO2(g) + O2(g)  SO3(g)
Diffusion of a drop of ink in
water
diamond  graphite
119
S
Changes
2 C(s) + O2(g)  2 CO2(g)
120
S
+ve
2SO2(g) + O2(g)  2SO3(g)
-ve
Diffusion of a drop of ink in
water
+ve
diamond  graphite
+ve
Consider an isolated system which has
no exchange of energy and matter with
its surroundings
121
S1
S2
S = S2 – S1 > 0
Which one is spontaneous ?
S2
S1
S = S1 – S2 < 0
122
S1
S2
Spontaneous, S = S2 – S1 > 0
S2
S1
Not spontaneous, S = S1 – S2 < 0
123
A molecular statistical interpretation
Free adiabatic expansion of an ideal gas into a vacuum.
vacuum
Slit is open
Probability
 1  1  1  1  1
      
 2  2  2  2  16
124
A molecular statistical interpretation
Free adiabatic expansion of an ideal gas into a vacuum.
1 mole
vacuum
Slit is open
Probability
1
 
 2
125
61023

1
1.81023
10
A spontaneous process taking place in an
isolated system is always associated with
an increase in entropy (I.e. S > 0)
In closed system,
6B
The spontaneity of a process depends on
both H and S.
The driving force of a process is a balance
of H and S.
126
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
127
Changes
H2O(g)  H2O(l) at 25°C
H2O(g)  H2O(l) at 110°C
H2O(s)  H2O(l) at 25°C
H2O(s)  H2O(l) at -10°C
128
H
S
Spontaneous
(Yes / No)
129
Changes
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve
-ve
Yes
H2O(g)  H2O(l) at 110°C
-ve
-ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Favourable
Changes
130
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve -ve
Yes
H2O(g)  H2O(l) at 110°C
-ve -ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Not favourable
Changes
131
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve -ve
Yes
H2O(g)  H2O(l) at 110°C
-ve -ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Spontaneity depends on temperature
Changes
132
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve -ve
Yes
H2O(g)  H2O(l) at 110°C
-ve -ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Favourable
Changes
133
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve -ve
Yes
H2O(g)  H2O(l) at 110°C
-ve -ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Not favourable
Changes
134
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve -ve
Yes
H2O(g)  H2O(l) at 110°C
-ve -ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Spontaneity depends on temperature
Changes
135
H
S
Spontaneous
(Yes / No)
H2O(g)  H2O(l) at 25°C
-ve -ve
Yes
H2O(g)  H2O(l) at 110°C
-ve -ve
No
H2O(s)  H2O(l) at 25°C
+ve +ve
Yes
H2O(s)  H2O(l) at -10°C
+ve +ve
No
Spontaneity of a process depends on
H, S & T
G = H –TS
G is the (Gibbs’) free energy
J. Willard Gibbs (1839 – 1903)
136
G = H –TS
Spontaneity depends on G
‘Free’ means the energy free for work
137
A spontaneous process is always associated
with a decrease in the free energy of the
system.
G < 0  spontaneous process
G > 0  not spontaneous process
Q.15
138
G = H –TS
139
H
S
T
+ve
+ve
high
+ve
+ve
low
-ve
-ve
high
-ve
-ve
low
G
Results
G = H –TS
140
H
S
T
G
Results
+ve
+ve
high
-ve
Spontaneous
+ve
+ve
low
+ve
Not spontaneous
-ve
-ve
high
+ve
Not spontaneus
-ve
-ve
low
-ve
Spontaneous
G = H –TS
141
H
S
T
-ve
+ve
high
-ve
+ve
low
+ve
-ve
high
+ve
-ve
low
G
Results
G = H –TS
H
S
T
-ve
+ve
high
-ve
+ve
low
+ve
-ve
high
+ve
142
-ve
low
G
Results
-ve
Spontaneous
+ve
Not spontaneous
G = H –TS
Q.16
143
Diamond
Graphite
H < 0
S > 0
G = H –TS < 0
The process is spontaneous, although
activation energy is required to start
the conversion.
144
S1
S2
Spontaneous S = S2 – S1 > 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.
145
S1
S2
Spontaneous S = S2 – S1 > 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.
146
G = H –TS
G = H – TS
H = G + TS
Total energy
Useless energy
Useful energy
147
If the universe is an isolated system
H is a constant and H is always zero
H = G + TS
Total energy
Useless energy
Useful energy
148
cosmic background radiation = 4K
Useful Useless
energy energy
H = G + TS
H = G + TS = 0
= G + TS = 0
S always > 0, S always , useless energy always 
G always < 0, G always , useful energy always 
149
In an isolated system, entropy will
only increase with time, it will not
decrease with time.
The second law of thermodynamics
150
If the universe is an isolated system,
Suniverse = Ssystem + Ssurroundings > 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.
151
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.
152
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.
153
The history of the universe
minimum entropy,
maximum free energy
(singularity)
H = G +TS
T  1.41032 K
expanding
maximum entropy,
minimum free energy
154
Big bang
Big chill
T0K
Planck’s units
Planck’s
temperature
Planck’s length
Planck’s time
155
h = Planck’s constant
G = gravitational constant
c = speed of light in vacuum
k = Boltzmann constant
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s units
Planck’s
temperature
Planck’s length
Planck’s time
156
Absolute hot beyond
which all physical laws
break down
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s units
Planck’s
temperature
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s length
hG
lP 
3
2c
1.616252(81) × 10−35 m
Planck’s time
157
Physical significance
not yet known
Planck’s units
Planck’s
temperature
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s length
hG
lP 
3
2c
1.616252(81) × 10−35 m
Planck’s time
158
1
20 the diameter of proton
10
Planck’s units
Planck’s
temperature
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s length
hG
lP 
3
2c
1.616252(81) × 10−35 m
Planck’s time
159
 = 7.310-37 m
22 times of Mr Chio’s
wavelength
Planck’s units
Planck’s
temperature
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s length
hG
lP 
2c 3
1.616252(81) × 10−35 m
hG
tp 
2c 5
5.39124(27) × 10−44 s
Planck’s time
160
It is the time required for
light to travel, in a vacuum,
a distance of 1 Planck length.
Planck’s units
10-15 s  femtosecond(飛秒)
10-18 s  attosecond(阿托秒)
Time taken for light to travel the length of 3 H atoms
Planck’s
temperature
hc 5
TP 
2
2Gk
1.416785(71) × 1032 K
Planck’s length
hG
lP 
2c 3
1.616252(81) × 10−35 m
hG
tp 
2c 5
5.39124(27) × 10−44 s
Planck’s time
161
Q.17(a)
The drop in temperature of the system is
accompanied by the rise in temperature of
its surroundings.
Ssystem < 0
Ssurroundings > 0
Suniverse = Ssystem + Ssurroundings > 0
162
Q.17(b)
The drop in entropy of the system is at the
cost of the rise in entropy of its
surroundings.
Ssystem < 0
Ssurroundings > 0
Suniverse = Ssystem + Ssurroundings > 0
163
6.8 Free energy change (SB p.170)
Check Point 6-8
164
The END
165
6.1 What is energetics? (SB p.140)
Back
State whether the following processes are exothermic or
endothermic.
(a) Melting of ice.
(a) Endothermic
(b) Dissolution of table salt.
(c) Condensation of steam.
166
(b) Endothermic
(c) Exothermic
Answer
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
(a) State the difference between exothermic and endothermic
reactions with respect to
(i) the sign of H;
(ii) the heat change with the surroundings;
Answer
(iii) the total enthalpy of reactants and products.
(a) (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.
167
(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.
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
(b) 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
168
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
(b) (i)
169
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
Back
(ii)
170
6.3 Standard enthalpy changes (SB p.147)
(a) Why must the condition “burnt completely in oxygen” be
emphasized in the definition of standard enthalpy change
of combustion?
(a) 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.
171
Answer
6.3 Standard enthalpy changes (SB p.147)
(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
(b) Standard enthalpy change of combustion of CO
1
=
 (-566.0) kJ
2
= -283.0 kJ
172
6.3 Standard enthalpy changes (SB p.147)
(c) What terms may be given for the enthalpy change of the
following reaction?
1
N2(g) + O2(g)  NO2(g)
2
(c) ½ Enthalpy change of combustion of nitrogen or
enthalpy change of formation of nitrogen dioxide.
Back
173
Answer
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
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
174
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
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
175
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Heat given out per mole of H2O formed
1769 J
=
0.03125 mol
= 56608 J mol-1
The enthalpy change of neutralization is –56.6 kJ mol-1.
Back
176
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
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.
177
Answer
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Heat evolved by the combustion of ethanol
= Heat absorbed by the copper calorimeter
= (m1c1 + m2c2) T
= (0.05 kg  4200 J kg-1 K-1 + 0.38 kg  2100 J kg-1 K-1) 
(312.4 – 291.5)K
= 21067 J
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
Mass of ethanol burnt = (45.24 – 44.46) g = 0.78 g
0.78 g
Number of moles of ethanol burnt =
= 0.017 mol
1
46.0 g mol
178
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Heat given out per mole of ethanol
21067 J
=
0.017 mol
= 1239235 J mol-1
Back
= 1239 kJ mol-1
The enthalpy change of combustion of ethanol is –1239 kJ mol-1.
There was heat loss by the system to the surroundings, and incomplete
combustion of ethanol might occur. Also, the experiment was not carried out
under standard conditions. Therefore, the experimentally determined value
(-1239 kJ mol-1) is less than the theoretical value of the standard enthalpy
change of combustion of ethanol (-1371 kJ mol-1).
179
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)
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
180
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)
Heat absorbed = m1c1T ( c2  0)
= 0.045 kg  4200 J kg-1 K-1  (297.5 – 296) K
= 283.5 J (0.284 kJ)
Heat absorbed per mole of ammonium chloride =
0.284 kJ
0.02 mol
= 14.2 kJ mol-1
The enthalpy change of solution of anhydrous ammonium chloride is
+14.2 kJ mol-1.
Back
181
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(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
182
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(a) 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
183
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
Back
(a) 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
Heat evolved per mole of H2O formed
1275.1 J
=
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.
184
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(b) 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
185
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
Back
(b) Energy absorbed = mcT
= 0.05 kg  4200 J kg-1 K-1  5.2 K
= 1092 J
No. of moles of AgNO3 used = 0.05 mol
1092 J
0.05 mol
= 21.84 kJ mol-1
Energy absorbed per mole of AgNO3 used =
The enthalpy change of solution of silver nitrate is +21.84 kJ mol-1.
186
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(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
187
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(c) 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
1.60 g
No. of moles of methanol used =
32.0 g mol -1
= 0.05 mol
34827 J
Heat evolved per mole of methanol used =
0.05 mol
= 696.5 kJ mol-1
The enthalpy change of combustion of methanol is –696.5 kJ mol-1.
188
6.5 Hess’s law (SB p.158)
(a) Given the following thermochemical equation:
2H2(g) + O2(g)  2H2O(l)
(i) Is the reaction endothermic or exothermic?
(ii) What is the enthalpy change for the following
reactions?
(1) 2H2O(l)  2H2(g) + O2(g)
1
(2) H2(g) + O2(g)  H2O(l)
2
(iii) If the enthalpy change for the reaction H2O(l) 
H2O(g) is +41.1 kJ mol-1, calculate the H for 2H2(g) +
O2(g)  2H2O(g).
Answer
189
6.5 Hess’s law (SB p.158)
(a) (i) Exothermic
(ii) (1) +571.6 kJ mol-1
(2) –285.8 kJ mol-1
(iii)
H = [-571.6 + 2  (+41.1)] kJ mol-1 = -489.4 kJ mol-1
190
6.5 Hess’s law (SB p.158)
(b) Given the following information about the enthalpy
change of combustion of allotropes of carbon:
Hc [C(graphite)] = -393.5 kJ mol-1
Hc [C(diamond)] = -395.4 kJ mol-1
(i) Which allotrope of carbon is more stable?
(ii) What is the enthalpy change for the following process?
C(graphite)  C(diamond)
191
Answer
6.5 Hess’s law (SB p.158)
(b) (i) Graphite
(ii)
H = [-393.5 – (-395.4)] kJ mol-1 = +1.9 kJ mol-1
192
6.5 Hess’s law (SB p.158)
(c) The formation of ethyne (C2H2(g) can be represented by
the following equation:
2C(graphite) + H2(g)  C2H2(g)
(i) Draw an enthalpy level diagram relating the above
equation to carbon dioxide and water.
(ii) Calculate the standard enthalpy change of formation
of ethyne.
(Given: Hc [C(graphite)] = -393.5 kJ mol-1;
Hc [H2(g)] = -285.8 kJ mol-1;
Hc [C2H2(g)] = -1299 kJ mol-1)
193
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
Given the following information, find the standard enthalpy
change of the reaction:
C2H4(g) + H2(g)  C2H6(g)
Hf [C2H4(g)] = +52.3 kJ mol-1
Hf [C2H6(g)] = -84.6 kJ mol-1
194
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
Back
Note:
H1 = [Hf (reactants)] = Hf [C2H4(g)] + Hf [H2(g)]
H2 = [Hf (products)] = Hf [C2H6(g)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= Hf [C2H6(g)] – (Hf [C2H4(g)] + Hf [H2(g)])
= [-84.6 – (+52.3 + 0)] kJ mol-1 =-136.9 kJ mol-1
The
195
standard enthalpy change of the reaction is –136.9 kJ mol-1.
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Given the following information, find the standard enthalpy
change of the reaction:
6PbO(s) + O2(g)  2Pb3O4(s)
Hf [PbO(g)] = -220.0 kJ mol-1
Hf [Pb3O4(g)] = -737.5 kJ mol-1
196
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Back
Note:
H1 = [Hf (reactants)] = 6  Hf [PbO(s)] + Hf [O2(g)]
H2 = [Hf (products)] = 2  Hf [Pb3O4(s)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= 2  Hf [Pb3O4(s)] – (6  Hf [PbO(s)] + Hf [O2(g)])
= [2  (-737.5) – 6  (-222.0) – 0] kJ mol-1 =-155.0 kJ mol-1
The
197
standard enthalpy change of the reaction is –155.0 kJ mol-1.
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Given the following information, find the standard enthalpy
change of the reaction:
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Hf [Fe2O3(s)] = -822.0 kJ mol-1
Hf [CO(g)] = -110.5 kJ mol-1
Hf [CO2(g)] = -393.5 kJ mol-1
198
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Back
H1 = [Hf (reactants)] = Hf [Fe2O3(s)] + 2  Hf [CO(g)]
H2 = [Hf (products)] = 2  Hf [Fe(s)] + 3  Hf [CO2(g)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= 2  Hf [Fe(s)] + 3  Hf [CO2(g)] - Hf [Fe2O3(s)] - 3 
Hf [CO(g)]
= [2  (0) + 3  (-393.5) –(-822.0) – 3  (-110.5)] kJ mol-1
=-27.0 kJ mol-1
199
The standard enthalpy change of the reaction is –27.0 kJ mol-1.
Note:
6.6 Calculations involving standard enthalpy changes of reactions (SB p.161)
Given the following information, find the standard enthalpy
change of the reaction:
4CH3 · NH · NH2(l) + 5N2O4(l)
 4CO2(g) + 12H2O(l) + 9N2(g)
Hf [CH3 · NH · NH2(l)] = +53 kJ mol-1
Hf [N2O4(l)] = -20 kJ mol-1
Hf [CO2(g)] = -393.5 kJ mol-1
Hf [H2O(l)] = -285.8 kJ mol-1
200
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.161)
Back
Note:H1 = [Hf (reactants)] = 4  Hf [CH3·NH ·NH2(l)] + 5  Hf [N2O4(l)]
H2 = [Hf (products)] = 4  Hf [CO2(g)] + 12  Hf [H2O(l)] + 9  Hf [N2(g)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= (4  Hf [CO2(g)] + 12  Hf [H2O(l)] + 9  Hf [N2(g)] – (3 
Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]
= [4  (-393.5) + 12  (-285.8) + 9  (0) – 4  (+53) – 5  (-20)] kJ mol-1
201
=- 5115.6 kJ mol-1
The standard enthalpy change of the reaction is –5115.6 kJ mol-1.
6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)
Given the following information, find the standard enthalpy
change of formation of methane gas.
C(graphite) + O2(g)  CO2(g)
Hc [C(graphite)] = -393.5 kJ mol-1
1
H2(g) + O2(g)  H2O(l)
Hc [H2(g)] = -285.8 kJ mol-1
2
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Hc [CH4(g)] = -20 kJ mol-1
Answer
202
6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)
Direct measurement of ΔHf [CH4(g)] is impossible because carbon(graphite)
and hydrogen do not combine directly, and methane does not decompose
directly to form carbon(graphite) and hydrogen. Since methane contain
carbon and hydrogen only, they can be related to carbon dioxide and water
by the combustion of methane and its constituent elements as shown in the
diagram below.
203
6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
Back
Note: H1 = Hc [C(graphite)]
H2 = 2  Hc [H2(g)]
H3 = Hc [CH4(g)]
Applying Hess’s law,
Hf [CH4(g)] + H3 = H1 + H2
Hf [CH4(g)] = H1 + H2 - H3
= Hc [C(graphite)] + 2  Hc [H2(g)] - Hc [CH4(g)]
= [-393.5 + 2  (-285.8) –(-890.4)] kJ mol-1
= -74.7 kJ mol-1
The standard enthalpy change of formation of methane gas is –74.7 kJ mol-1.
204
6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
Given the following information, find the standard enthalpy
change of formation of methanol.
C(graphite) + O2(g)  CO2(g)
Hc [C(graphite)] = -393.5 kJ mol-1
1
H2(g) + O2(g)  H2O(l)
Hc [H2(g)] = -285.8 kJ mol-1
2
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
Hc [C2H5OH(l)] = -1371 kJ mol-1
Answer
205
6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
206
6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
Back
Note: H1 = 2  Hc [C(graphite)]
H2 = 3  Hc [H2(g)]
H3 = Hc [C2H5OH(l)]
Applying Hess’s law,
Hf [C2H5OH(l)] + H3 = H1 + H2
Hf [C2H5OH(l)] = H1 + H2 - H3
= 2  Hc [C(graphite)] + 3  Hc [H2(g)] - Hc [C2H5OH(l)]
= [2  (-393.5) + 3  (-285.8) –(-1371)] kJ mol-1
= -273.4 kJ mol-1
The standard enthalpy change of formation of ethanol is –273.4 kJ mol-1.
207
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
(a) Find the standard enthalpy change of formation of
butane gas (C4H10(g)).
Given: Hc [C(graphite)] = -393.5 kJ mol-1
Hc [H2(g)] = -285.8 kJ mol-1
Hc [C4H10(g)] = -2877 kJ mol-1
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Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
Hf [C4H10(g)]
= Hc [C(graphite)]  4 + Hc [H2(g)]  5 - Hc [C4H10(g)]
= [(-393.5)  4 + (-285.8)  5 – (-2877)] kJ mol-1
=209-126 kJ mol-1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
(b) Find the standard enthalpy change of the reaction:
Br2(l) + C2H4(g)  C2H4Br2(l)
Given: Hf [C2H4(g)] = +52.3 kJ mol-1
Hf [C2H4Br2(l)] = -80.7 kJ mol-1
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Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
Back
H
= [Hf (products)] - [Hf (reactants)]
= [-80.7 – (+52.3) – 0)] kJ mol-1
= -133 kJ mol-1
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6.7 Entropy change (SB p.167)
Back
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
(a) Increase
(b) Decrease
(c) Increase
• Complete combustion of carbon monoxide
(d) Decrease
• Oxidation of sulphur dioxide to sulphur trioxide
(e) Decrease
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Answer
6.8 Free energy change (SB p.170)
Back
In the process of changing of ice to water, at what
temperature do you think G equals 0?
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.
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Answer
6.8 Free energy change (SB p.170)
(a) At what temperatures is the following process
spontaneous at 1 atmosphere?
Water  Steam
(a) 100 oC
(b) What are the two driving forces that determine the
spontaneity of a process? (b) Enthalpy and entropy
Answer
214
6.8 Free energy change (SB p.170)
Back
(c) 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
(i)
Spontaneous at high temperatures
(ii) Spontaneous at all temperatures
(iii) Not spontaneous at any temperature
(iv) Spontaneous at low temperatures
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Answer
6.5 Hess’s law (SB p.153)
Enthalpy change of formation of CaCO3(s)
Ca(s) + C(graphite) + 3 O2
2
Hf [CaCO3(s)]
CaCO3(s)
H2
H1
CaO(s) + CO2(g)
Hf [CaCO3(s)] = H1 + H2
= -1028.5 kJ mol-1 + (-178.0) kJ mol-1
= -1206.5 kJ mol-1
216
6.5 Hess’s law (SB p.153)
Enthalpy change of hydration of MgSO4(s)
MgSO4(s) + 7H2O(l)
ΔH1
ΔH
aq
MgSO4·7H2O(s)
aq
ΔH2
Mg2+(aq) + SO42-(aq) + 7H2O(l)
ΔH = enthalpy change of hydration of MgSO4(s)
ΔH1 = molar enthalpy change of solution of anhydrous
magnesium sulphate(VI)
ΔH2 = molar enthalpy change of solution of magnesium
sulphate(VI)-7-water
ΔH = ΔH1 - ΔH2
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