C9 THERMOCHEMISTRY

Answer

c = q ΔT
mx

= 204.75 J
15 g x (60°C - 25°C)

= 0.39 Jg-1°C-1

51

Heat Change, q

• The equation for calculating the heat
change, q is given by :

q = mc∆T or q = C∆T

q = heat released by substance
m = mass of substance
c = specific heat capacity
C = heat capacity
∆T = temperature change

52

 Heat is a form of energy which flows from a
hot body to a colder body when the two are
placed in thermal contact.

 The heat released is equal the heat absorbed.
qreleased + qabsorbed = 0
- qreleased = qabsorbed

53

Exercise 2

Calculate the amount of heat needed to
increase the temperature of 250 g of water
from 20oC to 56oC.
(specific heat capacity of water = 4.18 Jg-1°C-1)

Aqns=wemrcΔT 54
= 250 g x 4.18 Jg-1°C-1 x (56-20) °C

= 37620 J
= 37.62 kJ

Simple calorimeter:
(Constant-pressure Calorimeter)

 The outer Styrofoam cup insulate
the reaction mixture from the
surroundings (it is assumed that
no heat is lost to the
surroundings)

 It is suitable for measuring
enthalpy change for reactions
that take place in aqueous
solution.

 Heat release by the reaction(qrxn)
is absorbed by solution(qs)
and the calorimeter(qcal). 55

qsystem = qsolution + qcalorimeter + qreaction = 0
qrxn = - (qs + qcal)

= - (mscs ∆T + Ccal ∆T )

ms = mass of solution
cs = specific heat capacity of solution
Ccal = heat capacity of calorimeter
∆T = temperature change

56

KEEP IN MIND

 For aqueous solutions, the density and
specific heat capacity are usually assumed to
be the same as for water
(1.00 g mL-1 and 4.184 Jg-1°C-1 )

57

Exercise 3

Two different solutions were reacted in a
constant-pressure calorimeter that has a heat
capacity of 335 J°C-1. The initial temperatures
of the both solutions are the same, 24.6 °C and
the final temperature of the mixed solution is
27.0 °C. Calculate the heat released by the
reaction if the total volume of the mixed
solution is 200 mL.

Assume that the densities and specific heats of
the solutions are the same as for water (1.00 g
mL-1 and 4.184 J g-1°C-1 respectively)

58

Answer

Given:

Ccal = 335 J°C-1 ΔT= 27°C – 24.6°C = 2.4 °C

V of solution = 200 ml ρ of solution = 1 g/ml

mass of solution = 200 g c of solution = 4.18 Jg-1°C-1

qreaction = -(qs + qcal)
= - [(mscs ∆T) + (Ccal ∆T )]
= - [(200 g x 4.18 Jg-1 C-1 x 2.4 °C) +

(335 J°C-1 x 2.4 °C )]

= - 2.812 x 103 J

Heat released by the reaction is 2.812 x 103 J

59

Exercise 4

100 mL of 0.5 M HCl is mixed with 100 mL of
0.5 M NaOH in a coffee-cup calorimeter, both
at initial temperature of 22.50oC. The final
temperature of the reaction mixture is 25.90oC.
Calculate the enthalpy of neutralization for the
reaction.

(answer : -56.9 kJ mol-1)

60

Answer

Given:

ΔT= 25.90°C – 22.50°C = 3.4 °C

V of solution = 200 ml mass of solution = 200 g

qreaction = -(qs + qcal)
= - (mscs ∆T)
= - [(200 g x 4.18 Jg-1 C-1 x 3.4 °C)

= - 2.8424 103 J

= - 2.8424 kJ

61

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

100 ml 100 ml
0.5 M 0.5 M
0.05 mol 0.05 mol

Mole of water produced = 0.05 mol
0.05 mol H2O ≡ -2.8424 kJ
1 mol H2O ≡ -56.85 kJ

ΔHn = qrxn
Mole of H2O

= - 2.8424 kJ
0.05 mol

= - 56.85 kJ/mol

62

Bomb Calorimeter
(Constant-Volume Calorimetry)

• A bomb calorimeter is made
up of a strong steel vessel
filled with pure oxygen at a
pressure of about 30 atm and
immersed in a known mass of
water.
• A known mass of the
substance is then introduced
into the steel vessel and
ignited electrically.
• A bomb calorimeter is an
instrument used to measure
the heat of a combustion
reaction.

63

qsystem = qwater + qbomb + qreaction = 0
qrxn = - (qw + qbomb)

= - (mwcw ∆T + Cbomb∆T )

Reaction occurs at constant volume,
so no heat enters or leaves.

64

Exercise 5

1.435 g of naphthalene,C10H8 was burned in a
constant-volume bomb calorimeter. The
temperature of the water increases from
20.17oC to 25.84oC. If the mass of water
surrounding the calorimeter is 2000 g and the
heat capacity of the bomb calorimeter is 1.80
kJoC-1, calculate the heat of combustion of
naphthalene.

(answer : - 5.14 x 103 kJ mol-1)

65

Answer

Given:

ΔT= 25.84°C – 20.17°C = 5.67 °C

Cbomb = 1.8 kJ°C-1 mass of water = 2000 g

qreaction = -(qwater + qbomb)
= - [(mwcs∆T) + (Cbomb ∆T)]
= - [(2000 g x 4.18 Jg-1 C-1 x 5.67°C)

+ (1.8 x 103 J C-1 x 5.67 °C)]

= - 5.761 104 J

= - 57.61 kJ 66

C10H8(s) + 12O2(g) → 10CO2(g) + 4H2O(l)

Mole of C10H8 = 1.435 g
128 g/mol

= 1.121 x 10-2 mol

0.05 mol C10H8 ≡ -57.61 kJ
1 mol C10H8 ≡ -5.14 x 103 kJ
qrxn
ΔHc = Mole of C10H8

= - 57.61 kJ
1.121 x 10-2 mol

= - 5.14 x 103 kJ/mol 67

CHAPTER 9
THERMOCHEMISTRY

9.3 : HESS’S LAW

68

LEARNING OUTCOME

a) State Hess’s law

b) Apply Hess’s law to calculate enthalpy
changes using the algebraic method and
the energy cycle method.

c) Illustrate the dissolution process of ionic
solids

69

Hess’s Law

 Hess’s Law states that when reactants are
converted to products, the change in
enthalpy is the same whether the
reaction takes place in one step or in the
series of steps.

AC

B

From Hess’s Law:

70

Application of Hess’s Law

1. To calculate the enthalpy of reaction.
2. To construct Born-Haber cycle.
• The enthalpy of reaction can be calculated by:

i. Algebraic method
ii. Energy cycle method

71

Algebraic Method

Given: ∆Ho = -187.9 kJ
∆Ho = -285.8 kJ
H2(g) + O2(g)  H2O2(l)
H2(g) + ½O2(g)  H2O(l)

Calculate ∆Ho for the decomposition of

hydrogen peroxide:

2H2O2(g)  2H2O(l) + O2(g)

72

[H2O2(l) H2(g) + O2(g) ∆Ho = +187.9 kJ] x 2
[H2(g) + ½O2(g) H2O(l) ∆Ho = -285.8 kJ] x 2

2H2O2(l)  2H2(g) + 2O2(g) ∆Ho = +375.8 kJ
2H2(g) + O2(g)  2H2O(l) ∆Ho = - 571.6 kJ
2H2O2(l)  2H2O(l) + O2(g) ∆Ho = - 195.8 kJ

73

Energy Cycle Method

Calculate the standard enthalpy of formation
of ethane C2H6(g). Given the following data:

Hoc C(graphite)  393kJmol-1
Hoc H2(g)  - 286kJmol-1
ΔHoc C2H6(g)  -1560kJmol-1

74

Target equation :
2C(graphite) + 3H2(g)  C2H6(g) ∆Hf = ?

C(graphite) + O2(g)  CO2(g) ∆Ho = -393 kJ
H2(g) + ½O2(g)  H2O(g) ∆Ho = -286 kJ

C2H6(g) + 7/2 O2(g)  2CO2(g) + 3H2O(g) ∆Ho = -1560 kJ

75

2C(graphite) + 3H2(g) ∆Hf = ? C2H6(g)

2O2(g) 3/2 O2(g)

∆H1 = 2(-393) ∆H2 = 3(-286) 7/2 O2(g)

2CO2(g) + 3H2O(g) ∆H3 = -1560

∆Hf = ∆H1 + ∆H2 + ∆H3 76
∆Hf = [2(-393) + 3(-286) + (+1560)]

= - 84 kJ mol-1

Exercise 1

Calculate the enthalpy for the following
reaction:
2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s)
Given:

Hof [Al2O3(s)]  -1601kJmol-1
Hof [Fe2O3(s)]  - 821kJmol-1

[answer : -780 kJ ]

77

Answer

2Al(s) + 3/2O2(g)  Al2O3(s) ∆Ho = -1601 kJ
2Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆Ho = -821 kJ

Fe2O3(s)  2Fe(s) + 3/2O2(g) ∆Ho = +821 kJ
2Al(s) + 3/2O2(g)  Al2O3(s) ∆Ho = -1601 kJ

2 Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) ∆Ho = - 780 kJ

78

Exercise 2

Calculate the enthalpy of formation of
benzene, C6H6 using energy cycle method.
Given:

Hof [CO2(g)]  - 393.3kJmol-1
Hof [H2O(l)]  - 285.5kJmol-1
Hoc [C6H6(l)]  - 3265.3kJmol-1

(answer : + 49.0 kJ mol-1 )

79

Answer

Target equation :

6C(graphite) + 3H2(g) C6H6(g) ∆Hf = ?

C(graphite) + O2(g)  CO2(g) ∆Ho = -393.3 kJ

H2(g) + ½O2(g)  H2O(g) ∆Ho = -285.5 kJ

C6H6(g) + 15 O2(g)  6CO2(g) + 3H2O(g) ∆Ho = -3265.3 kJmol-1
2

80

6C(graphite) + 3H2(g) ∆Hf = ? C6H6(g)

6O2(g) 3 O2(g)
2

∆H1 = 6(-393.3) ∆H2 = 3(-285.5) 15 O2(g)
2

6CO2(g) + 3H2O(g) ∆H3 = -3265.3

∆Hf = ∆H1 + ∆H2 + ∆H3
∆Hf = [6(-393.3) + 3(-285.5) + (+3265.3)]

= + 49.0 kJ mol-1

81

Dissolution of Ionic Solids

 Ionic crystals are made up of positive and
negative ions attracted to each other by
electrostatic forces.

 When an ionic compound dissolves in water,
there is an enthalpy change.
MX(s)  Mp+(aq) + Xq-(aq) ΔHsolution

82

 The overall process of solution can be

divided into two stages:
a) The breaking down of the crystal lattice

into isolated gaseous ions.(lattice
dissociation energy)

b) The hydration of separated gaseous ions
by water molecules.

 The enthalpy of solution of an ionic crystal in
water is the sum of the lattice energy and
the total enthalpy of hydration of the ions.

ΔHsoln = ΔHlattice + ΔHhyd 83

Example

Na+ and Cl- ion in
the gaseous state

Heat of Solution

Na+ and Cl- ion in Hydrated Na+ and Cl- ion
the solid state
84

Exercise 1

The lattice energy of NaCl is –776 kJ mol–1
and the enthalpy change when one mole of
solid NaCl dissolved in water is +4 kJ mol–1.
If the enthalpy change of hydration of Na+
ion is –390 kJmol–1, calculate the enthalpy
of hydration of Cl- ion in the solution process
using the energy cycle method.

85

Answer

Na+(g) + Cl-(g)  NaCl(s) ∆H = -776 kJ
NaCl(s)  Na+(g) + Cl-(g) ∆H = +776 kJ
NaCl(s)  Na+(aq) + Cl-(aq) ∆H = +4 kJ

Na+(g)  Na+(aq) ∆H = -390 kJ

Cl-(g)  Cl-(aq) ∆H = ? kJ

NaCl(s) ∆Hsoln= +4 Na+(aq) + Cl-(aq)

∆HLE = +776 ∆HHyd = -390 ∆HHyd = ?
Na+(g) + Cl-(g)
86

ΔHsoln = ΔHlattice + ΔHhyd(Na+ + Cl-)

+4 = +776 + (-390) + ΔHhyd of Cl-
ΔHhyd of Cl- = -382 kJ mol-1

87









CHAPTER 9
THERMOCHEMISTRY

9.4 : Born-Haber Cycle

92

LEARNING OUTCOME

(a) Define lattice energy and electron affinity.
(b) Explain the following effects to the magnitude of

lattice energy.
i. ionic charges
ii. ionic radii

(c) Construct Born-Haber cycle for simple ionic solids
using energy cycle diagram and energy level
diagram.

(d) Calculate enthalpy changes using Born-Haber
cycle.

93

Lattice Energy, Hlattice

 Can be defined as:
i) Lattice dissociation energy
ii) Lattice formation energy

 Lattice dissociation energy is the heat
absorbed to completely separate one
mole of a solid ionic compound into
gaseous ions

Example:
NaCl(s) → Na+(g) + Cl-(g) ΔHlattice = +771 kJ

(lattice dissociation)

94

•Lattice formation energy is the heat released
when one mole of ionic crystal is formed
from its gaseous ions.
Example:
Na+(g) + Cl-(g) → NaCl(s) ΔHlattice = -771 kJ

(lattice formation)

95

The magnitude of the lattice energy depends
on:

(a) The charges on the ions
•The magnitude of lattice energy is
proportional to the ionic charges, i.e the
force of attraction between ions increases
as the charge on the ion increases.
• Highly negative lattice energy indicate a
strong ionic bond, e.g LE NaCl (-787 kJ)
>-ve NaBr (-747 kJ)
• Electrostatic force between Na+ and Cl-
are stronger than that of Na+ and Br-

96

(b) Ionic radii
• The magnitude of lattice energy
increases as the ionic radii decrease.
• Small ions are close together and have
smaller distance of separation thus form
stronger attractive force between the
ions.
• Therefore lattice energy becomes more
negative and the ionic compound
becomes more stable.

97

 The ionic lattice becomes more stable
as the magnitude of lattice energy increases.

 It is because the electrostatic attraction
between the positive and negative ions
becomes stronger, so more energy is needed
to break apart the ions.

98

Example

Given:

Compound Lattice energy (kJ/mol)
MgO 3890
Na2O 2570

ΔHolattice (MgO) > ΔHolattice (Na2O)
• because Mg2+ is smaller in size and has

higher charge than Na+ .

• Attractive forces between Mg2+ and O2- are
stronger than between Na+ and O2-

99

Consider the following data:

Ionic Lattice Energy Cation / Anion / Size
Compounds (kJ mol-1)
826 Increasing size of
LiCl 771 cation
NaCl 701
KCl 771 Li+ < Na+ < K+
NaCl 733
NaBr 684 Increasing size of
NaI anion

Cl- < Br- < I-

100