Energy = force x distance (Joules)
In chemical reactions, we need energy usually in the
form of heat.
Energy is absorbed to break the bonds of the
reactants and energy is given out when new bonds
are formed in the products.
Exothermic reactions
Endothermic reactions
Heat is the energy transferred between
objects that are at different temperatures.
 The amount of heat transferred depends on
the amount of the substance.
◦ Energy is measured in units called joules
(J).

Temperature is a measure of “hotness” of a
substance and represent the average kinetic
energy of the particles in a substance.
 It does not depend on the amount of the
substance.

Do both beakers contain the same amount of heat?



All chemical reactions are accompanied by some
form of energy change
Exothermic
Energy is given out
Endothermic
Energy is absorbed

Activity : observing exothermic and endothermic reactions


Enthalpy (H) is the heat content that is stored in a
chemical system.
We measure the change in enthalpy ∆H i.e. the amount
of heat released or absorbed when a chemical reaction
occurs at constant pressure, measured in
kilojoules per mole (kJmol-1).
∆H = H(products) – H(reactants)

For exothermic reactions, the reactants have more
energy than the products, and the enthalpy change,
∆H = H(products) - H(reactants)
Enthalpy

∆H is negative since H(products) < H(reactants)

There is an enthalpy decrease and heat is released to the
surroundings


Self-heating cans
◦ CaO (s) + H₂O (l)  Ca(OH)₂ (aq)
Combustion reactions
◦ CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l)


neutralization (acid + base)
◦ NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l)
Respiration
◦ C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)

For endothermic reactions, the reactants have less
energy than the products, and the enthalpy change,
∆H = H(products) - H(reactants)
Enthalpy

∆H is positive since H(products) < H(reactants)

There is an enthalpy increase and heat is absorbed from
the surroundings



Self-cooling beer can
◦ H ₂O (l)  H₂O (g)
Thermal decomposition
 CaCO₃ (s)  CaO (s) + CO ₂ (g)
Photosynthesis
 6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)


Amount of heat required to raise the temperature of
a unit mass of a substance by 1 degree or 1 kelvin.
Uint : Jg-1 0C-1
The specific heat capacity of alminium is 0.90 Jg-1 0C-1 .
If 0.90J of energy is put into 1g of aluminium, the temperature
will be raised by 10C.
Calculating heat absorbed and released
q = c × m × ΔT
q = heat absorbed or released
c = specific heat capacity of substance
m = mass of substance in grams
ΔT = change in temperature in Celsius
Heat given off by a process is measured
through the temperture change in another
substance (usually water).
 Due to the law of conservation of energy, any
energy given off in a process must be
absorbed by something else, we assume that
the energy given out will be absorbed by the
water and cause a temperature change.
 calculate the heat through the equation Q =
mcΔT

How much heat is required to increase the
temperature of 20 grams of nickel (specific heat
capacity 440Jkg-1 0C-1) from 500C to 700C?
The standard enthalpy change of combustion for a
substance is the heat released when 1 mole of a pure
substance is completely burnt in excess oxygen under
standard conditions.
 Example,
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔHƟc=-698 kJmol-1



The heat given out is used to heat another
substance,e.g. water with a known specific heat
capacity.
The experiment set-up can be used to determine the
enthalpy change when 1 mole of a liquid is burnt.
Example : refer to page 185



Loss of heat to the surroundings (exothermic
reaction); absorption of heat from the surroundings
(endothermic reaction). This can be reduced by
insulating the calorimeter.
Using incorrect specific heat capacity in the calculation
of heat change. If copper can is used, the s.h.c. of
copper must be accounted for.
Others include – e.g incomplete combustion. Some of
the ethanol could be used to produce CO & soot &
water (less heat is given out)
Use bomb calorimeter – heavily insulated & substance is ignited
electronically with good supply of oxygen

If 1g of methanol is burned to heat 100g of water,
raising its temperature by 42K, calculate the
enthalpy change when 1 mole of methanol is burnt.
Note: Specific heat capacity of water is 4.18 Jg-1 0C-1
Practice questions page 187 #1-4
Enthalpy change of neutralisation (ΔHn)
The standard enthalpy change of neutraisation is the enthalpy change that
takes place when 1 mole of H+ is completely neutralised by an alkali under
standard conditions.
Example,
NaOH(g) + HCl(g)  NaCl(g) + H2O(l)
ΔHƟ=-57 kJmol-1



The enthalpy change of neutralisation of a strong acid and a strong alkali is
almost the same as they undergo complete ionisationof ions in water.
Reaction between strong acid and strong base involves
H+(aq) + OH-(aq)  H2O(l)
ΔHƟ=-57 kJmol-1
For sulfuric acid, the enthalpy of neutralisation equation is
½ H2SO4(aq) + KOH(aq)  ½K2SO4(aq) + H2O(l)
ΔHƟ=-57 kJmol-1
Example : refer to page 188
Enthalpy change of neutralisation (ΔHn)


The standard enthalpy change of neutraisation is the enthalpy
change that takes place when 1 mole of H+ is completely
neutralised by an alkali under standard conditions.
Example,
NaOH(g) + HCl(g)  NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1
The enthalpy change of neutralisation of a strong acid and a
strong alkali is almost the same as they undergo complete
ionisationof ions in water.
Enthalpy change of solution (ΔHsol)
-
The enthalpy change when 1 mol of solute is dissolved in excess solvent to
form a solution of ‘infnite dilution’ under standard conditions.
NH4 NO3(s) in excess water  NH4 + (aq)+ NO 3 -(aq)
Example : refer to page 188
For neutralisation between a weak acid, a weak
base or both, the enthalpy of neutraisation
will be smaller than -57 kJmol-1 (less exothermic)
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)

ΔHƟ=-55.2 kJmol-1
Some of the energy released is used to ionise
the acid.

200.0cm3 of 0.150 M HCl is mixed with 100.0cm3 of 0.350 M
NaOH. The temperature rose by 1.360C. If both solutions
were originally at the same temp, calculate the enthalpy
change of neutralisation.
Assume that the density of the solution is 1 gcm-3 and the
specific heat capacity is 4.18J Jg-1 0C-1.
-56.8kJmol-1
The experimental change of neutralisation is
-56.8 kJmol-1
The accepted literature value is -57.2 kJmol-1
(1) Heat loss to the environment.
(2) Assumptions that
(a) the denisty of NaOH and HCl solutions are the same as
water.
(b) the specific heat capacity of the mixture are the same as that
of water
When 3 g of sodium carbonate are added to 50 cm3 of 1.0 M HCl,
the temperature rises from 22.0 °C to 28.5°C. Calculate the
enthalpy change for the reaction. Assume that the density of the
solution is 1 gcm-3 and the specific heat capacity is
4.18J Jg-1 0C-1.
Example : refer to page 189 dissolving
ammonium chloride
The experimental change of solution is
+13.8 kJmol-1
The accepted literature value is 15.2 kJmol-1
(1) Absorption of heat from the environment.
(2) Assumptions that the specific heat capacity of the solution is
the same as that of water
(3) The mass of ammonium chloride is not taken into
consideration when working out the heat energy released.
Example
100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a
styrofoam cup. 1.30 g of powdered zinc is added and a single replacement
reaction occurs. The temperature of the solution over time is shown in the
graph below. Determine the enthalpy value for this reaction.
First step
Make sure you understand the
graph.
Extrapolate to determine the
change in temperature.
The extrapolation is necessary to compensate for heat loss while the reaction
is occurring. Why would powdered zinc be used?
100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a
styrofoam cup. 1.30 g of powdered zinc is added and a single replacement
reaction occurs. The temperature of the solution over time is shown in the
graph below. Determine the enthalpy value for this reaction.
Determine the limiting reactant
Calculate Q
Calculate the enthalpy for the reaction.
The following measurements are taken:



Mass of cold water (g)
Temperature rise of the water (0C)
The loss of mass of the fuel (g)
We know that it takes 4.18J of energy to raise the
temperature of 1g of water by 10C. This is called the
specific heat capacity of water, c, and has a value of
4.18Jg-1K-1.
Hence, energy transferred can be calculated using: Energy
transfer = mcΔT (joules)
 If one mole of the fuel has a mass of M grams, then:
 Enthalpy transfer = m x 4.18 x T x M/y
 where y is mass loss of fuel.
Given that:
Vol of water = 100 cm3
Temp rise = 34.50C
Mass of methanol burned = 0.75g
Specific heat capacity of water = 4.18 Jg-10C-1
Calculate the molar enthalpy change of the combustion of
methanol.
What is the big assumption made with this type
of experiment?
States that
 If a reaction consists of a number of steps, the
overall enthalpy change is equal to the sum of
enthalpy of individual steps.
 the overall enthalpy change in a reaction is
constant, not dependent on the pathway take.

measured under standard conditions: pressure of 1
atmosphere (1.013 x 105 Pa), temperature of 250C
(298K) and concentration of 1 moldm-1.
e.g.
N2(g) + 3H2(g)  2NH3(g)
ΔHƟ = -92 kJmol-1
The enthalpy change of reaction is -92 kJmol-1
92 kJ of heat energy are given out when 1 mol of nitrogen reacs
with 3 mols of hydrogen to form 2 mols of ammonia.
Calculate the enthalpy change for the formation of
sodium chloride solution from solid sodium hydroxide.
Indirect path
+ H2O(l)
NaOH(s)
NaOH(aq)
ΔH2
ΔH1
Direct path
+ HCl(aq)
+ HCl(aq)
NaCl(s) + H2O(l)
1. Indirect path: NaOH(s) + (aq)  NaOH(aq)
ΔHƟ1=-43kJmol-1
2.
NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) ΔHƟ1=-57kJmol-1
3.
NaOH(s) + HCl (aq)  NaCl(aq) + H2O(l).
Calculate the enthalpy change for the combustion of
carbon monoxide to form carbon dioxide.
C(s) + O2(g)  CO2(g)
ΔHƟ =-394 kJmol-1
2C(s) + O2(g)  2CO(g)
ΔHƟ = -222kJmol-1
2CO(s) + O2(g)  2CO2(g)
2CO(g) + O2(g)
ΔHƟ
= -222kJmol-1
ΔHƟ
2C(s)+O2(g)+O2(g)
2CO2(g)
ΔHƟ
= -394kJmol-1
ΔHƟ = -(-222)+2(-394) = -566kJmol-1
Example : refer to page 196 evaporation
of water & 197 formation of ethanol
from ethene
Calculate the enthalpy change for the thermal
decomposition of calcium carbonate.
CaCO3(s)  CaO(s) + CO2(g)
CaCO3(s) +2HCl(aq)  CaCl2(aq) + H2O(l)
CaO(s) +2HCl(aq)  CaCl2(aq) + H2O(l)
CaCO3(s)
-17 kJmol-1
ΔH
Direct path
ΔHƟ1=-17 kJmol-1
ΔHƟ1=-195kJmol-1
CaO(s) +CO2(g)
+ 2HCl(aq)
-195kJmol-1
CaCl2(aq) + H2O(l) +CO2(g)
Indirect path
Calculate the enthalpy of hydration of anhydrous
copper(II)sulfate change.
CuSO4(s) +5H2O(l)  CuSO4.5H2O (s)
CuSO4(s) +5H2O(l)
ΔH
Direct pathway
ΔH1
CuSO4.5H2O (s)
ΔH2
Cu2+(aq) + SO42- (aq)
Indirect pathway
From the following data at 250C and 1 atmosphere
pressure:
Eqn 1: 2CO2(g) 2CO(g) + O2(g) ΔHƟ=566 kJmol-1
Eqn 2: 3CO(g) + O3(g) 3CO2(g) ΔHƟ=-992 kJmol-1
Calculate the enthalpy change calculated for the
conversion of oxygen to 1 mole of ozone,i.e. for the
reaction 3 O2(g) O3 (g)
2
Calculate the enthalpy change for the conversion of
graphite to diamond under standard thermodynamic
conditions.
C (s,graphite) + O2(g) CO2 (g)
ΔHƟ=-393 kJmol-1
C (s, diamond) + O2(g) CO2(g) ΔHƟ=-395 kJmol-1
Practice questions page 199 #7-9



Enthalpy changes can also be calculated
directly from bond enthalpies.
The bond enthalpy is the amount of energy
required to break one mole of a specified
covalent bond in the gaseous state.
For diatomic molecule the bond enthalpy is
defined as the enthalpy change for the
process X-Y(g)
X(g) + Y(g) [gaseous state]
Bond enthalpy can only be calculated for
substances in the gaseous state.
atomisation
Br2(l)  2Br(g)
ΔHƟ= 224 kJmol-1

Br2(l)
2 x ΔH Ɵat
ΔH Ɵvap
2Br(g)
Br-Br
bond enthalpy
enthalpy change
of vaporisation
Br2(g)
Energy must be supplied to break the van der Waals’ forces between the Bromine
molecules and to break the Br-Br bonds. Endothermic process

Ave bond enthalpies are enthalpies calculated
from a range of compounds,eg C-H bond
enthalpy is based on the ave bond energies in
CH4 , alkanes and other hydrocarbons.
Bond
Ave bond enthalpy,
ΔHƟ (Kjmol-1)
Bond length
(nm)
H-H
436
0.07
C-C
348
0.15
C-H
412
0.11
O-H
463
0.10
N-H
388
0.10
N-N
163
0.15
C=C
612
0.13
O=O
496
0.12
C ΞC
837
0.12
NΞN
944
Refer to page 201
When a hydrocarbon e.g. methane (CH4) burns,
CH4 + O2  CO2 + H2O
What happens?
Enthalpy Level (KJ)
Bond Breaking
O
ENERGY
H
+
C
H
H
O
O
O
O
H
H
O
H
O
ENERGY
H
Bond Forming
H
4 C-H
O
C
2 O=O
H
O
O
C
O
H
4 H-O
CH4 + 2O2  CO2 + 2H2O
Progress of Reaction
O
H
H
2 C=O
CH4 + 2O2  CO2 + 2H2O
H
+
C
H
H
H
O
O
O
O
H
O
O
C
H
O
H
H
Why is this an exothermic reaction (produces heat)?
O
Break
Form

CH4 + 2O2  CO2 + 2H2O
H
H
C
H
O O
+
H O O
O H
H
H
O C O
O
H
Energy absorbed when bonds are broken
= (4 x C-H + 2 x O=O)
= 4 x 412 + 2 x 496
= 2640 kJ/mol
Energy given out when bonds are formed
= ( 2 x C=O + 4 x H-O)
= 2 x 803 + 4 x 464
= 3338 kJ/mol
Bond
Ave Bond
Enthalpy (kJ/mol)
C-H
412
H-O
463
O=O
496
C=O
743
Energy absorbed when bonds are broken (a)
= 2640 kJ/mol
Energy released when bonds are formed (b)
= 3338 kJ/mol
Enthalpy change, ΔH
= ∑(bonds broken) - ∑(bonds made)
= a + (-b)
= 2640 – 3338
= -698 kJ/mol
Why is this an exothermic reaction (produces heat)?
What can be said about the hydrogenation
reaction of ethene?
H
H
C=C
(g) + H-H (g) 
H
H
H H
H-C-C-H (g)
H H
What can be said about the combustion of
hydrazine in oxygen?
H
H
N-N
H
(g) + O=O (g) 
H
NΞN (g) + 2 O
H
(g)
H
Example
Calculate the mean Cl-F bond enthalpy given that
Cl2(g) + 3F2(g)  2ClF3(g) ΔHƟ= -164 kJmol-1
Bond enthalpy for Cl-Cl = 242 kJmol-1 and F-F = 158 kJmol-1
Standard enthalpy change of atomisation (ΔH Ɵat ) is the
enthalpy change when 1 mole of gaseous atoms is
formed from the element under standard conditions.
Example
C(s)  C(g)
ΔH Ɵat = 715 kJmol-1
Calculate the enthalpy change for the process
3 C(s) + 4H2(g)  C3H8(g) ΔHƟ= -164 kJmol-1
Bond enthalpy for C-H = 412 kJmol-1 , H-H = 436 kJmol-1 and C-C = 348 kJmol-1
Practice questions page 206 #10,12,13
The combustion of both C and CO to form CO2 can
be measured easily but the combustion of C to CO
cannot. This can be represented by the energy
cycle.
C(s)+ ½O2(g)
-393kJmol-1
½O2(g)
CO2(g)
ΔHx = -393 – (-283)
= - 110 kJmol-1
CO(g)
ΔHx
½O2(g)
-283kJmol-1
Calculate the standard enthalpy change when one
mole of methane is formed from its elements in
their standard states. The standard enthalpies of
combustion of carbon, hydrogen and methane are
-393, -286 and -890 kJmol-1 respectively.