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Industrial Applications of
Chemical reactions
Unit 26
Enthalpy Changes and Hess’s Law
Principles of energetics
• Law of conservation of energy
• Energy cannot be created or destroyed but only
changed from one form into another.
• Chemical changes
• Chemical reactions are accompanied by energy
changes which are usually in the form of heat. Where
heat is released to the surroundings is called
EXOTHERMIC and when heat is taken in from the
surroundings is called ENDOTHERMIC
Enthalpy change
• It is impossible to measure the total heat
energy of a system but we can measure
changes in the heat energy when a reaction
occurs.
• This is called the enthalpy change. It can be
positive or negative.
Enthalpy Change
• Usually there is energy change involved in a chemical
reaction somehow. In chemistry this change in heat
energy is represented as enthalpy change using the
following symbol:
• So that this figure is the same everywhere and fair, it is
taken at standard conditions which are 298K (250C),100
kPa pressure (1atmosphere) and 1 moldm-3.
Standard conditions
Accurate comparisons of enthalpy changes
should be measured under standard
conditions and these are set as being 25oC
(298K) and one standard atmosphere pressure
with any solutions involved at a concentration
of 1.00moldm-3 and with all other reacting
species in their standard states.
Two examples:
For the first rection, the enthalpy change is negative this is
because the reaction is exothermic meaning it gives out
heat. The reason it is negative is energy is leaving the
reaction.
The second reaction has a positive enthalpy change, this
is because it is endothermic and heat is entering the
system.
Enthalpy of Reaction, Combustion,
and Formation
• ΔHr is standard enthalpy of reaction this is the energy change of a
reaction in the quantities that are expressed. It uses the symbol r.
• ΔHc is the standard enthalpy of combustion is the enthalpy
change when 1 mole of substance is completely burned in excess
oxygen under standard conditions, all reactants in their standard
states. The enthalpies of combustion can be determined using a
calorimeter which is explained below; it uses the symbol c.
• ΔHf
is the standard enthalpy of formation is the enthalpy change
when 1 mole of compound is produced from its elements in
standard conditions, all products and reactants in the standard
state. The standard state is important, it means the way that the
element is, at standard conditions (see above). So you would have
H2(g) NOT H on its own. Therefore you sometimes need to do
fractions of elements, as with the standard enthalpy of formation of
ethanol:
• C (s) + 2H2 (g) + ½O2 (g)
CH3OH (l)
Using Calorimetry to measure enthalpy change
•
This is a method we can use to find out what the enthalpy changes of reactions
are. Typically you do this using some sort of alcohol heating up some water. To do
it correctly, you would have it in a closed system so no heat escapes, but this is not
usually possible in a school or college lab. Once the water has been heated up you
must use the following equation:
specific heat capacity is the energy required to raise the temperature of one
gram of substance by 1 kelvin (or celcius depending on what you're using). It is
measured in Jg-1K-1 and water has a specific heat capacity of 4.18 Jg-1K-1.
Problem:
• 1g of ethanol is burned to heat 100g of water,
raising its temperature by 42K. Calculate the
standard enthalpy of combustion.
• First we calculate the energy transferred:
Q = m c ∆T
Q=
100g x 4.18 Jg-1K-1 x 42K
Q = 17556 J or 17.6 kJ
To find the kJ per mole divide the energy
released by the number of moles of ethanol:
Mr ethanol (C2H5OH) = (12*2) + (1*6) + (16*1)
= 46
Number of moles = mass/Mr = 1/46 = 0.02174
So ∆Hѳc = -17.6/0.02174 = -809.6 kJmol-1
Note: the figure obtained experimentally is lower than the textbook value as
energy is lost from the system during the experiment.
Standard enthalpy change of neutralisation, ∆Hѳneut
• This is the enthalpy change that takes place when
1mol of hydrogen ions (H+(aq)) is neutralised under
standard conditions:
H+ (aq) + OH- (aq)
H2O (l) ∆Hѳneut = -57.5 kJmol-1
• The method of determining ∆Hѳneut involves mixing
equal volumes of 1 mol dm-3 solutions of a strong
acid(e.g. hydrochloric acid) and a strong base (e.g.
sodium hydroxide) and measuring the maximum
temperature rise produced.
• The heat produced (Q) can then be
determined using the formula:
Q (J) = total mass of solution(g) x 4.2 (JK-1g-1) x ∆T (oC)
The value in kJ mol-1 can then be calculated.
e.g. If 25cm3 of 1M HCl is reacted with 25cm3 1M
NaOH and a temperature rise of 6.7oC is obtained:
Q = 50 x 4.2 x 6.7 = 1407 J = 1.4kJ
No. moles HCl in 25cm3 of 1M solution = 25/1000 = 1/40 moles. So 1.4k J is
produced by producing 1/40 moles of water. To work out how much
energy would be evolved if we produced 1 mole of water we simply
multiply our answer by 40:
∆Hѳneut = 1.4 x 40 = -56kJmol-1
Problem
• In an experiment 50.0 cm3 of 1.00 moldm-3
HCl was exactly neutralised by 2.00 moldm-3
KOH. The initial temperature of both solutions
was 15.3oC and the final temperature was
24.3oC. Calculate the molar enthalpy change
of neutralisation of HCl.
Answer
KOH + HCl
KCl + H2O
ΔH =
mCΔT
ΔH = 75.0 x
4.18 x 9.0
= 2882J = 2.882kJ
The temperature rise shows the reaction must
be exothermic hence ΔH = -2.882kJ
n =
M x
1000
V
=
50.0 x 1.00
= 0.005mol
1000
Hence molar enthalpy change for HCl =
-2.822 x 1
0.005
= -56.4kJmol-1
Problems
1. Iron metal will react with copper sulphate solution in
an exothermic reaction

Fe(s) + CuSO4(aq)
FeSO4(aq) + Cu(s)
When 0.5g of powdered iron is added to 100cm3 of
0.200 mol dm-3 copper sulphate solution the
temperature rises by 3.3oC
(a) calculate the enthalpy change of reaction
(b) how many moles of iron were used
(c) what is the molar enthalpy change for the
reaction
ANSWER
(a) ΔH =
mCΔT
= 100 x 4.18 x 3.3
= - 1400J = - 1.4kJ
(b)
n = 0.50
55.85
= 0.009mol
(c)
ΔH = 1.4
0.009
= -156 kJ mol-1 for iron
Problem
• 2. When 1.00g of methanol was completely burned under a
container of water the temperature of the water changed
from 21.0oC to 45.5oC. The mass of water in the container was
220g.
• (a) how much energy was absorbed by the water?
• (b) how much energy is given out by the methanol?
• (c) is the combustion of methanol endothermic?
• (d) how many moles of the methanol were used?
• (e) calculate the molar enthalpy change for the reaction
•
when methanol is completely burned.
ANSWER
(a)
(b)
ΔH = 220 x 4.18 x 24.5
=
+22.5kJ
> 22.5kJ as some heat will be lost to the surrounding air and container
(c)
Exothermic
(d)
1.00
32.0
= 0.0312 mol
(e) At least 22.5
0.0312
= -721 kJ
Standard Enthalpy of Formation, ∆Hѳf
• This is the enthalpy change that occurs when 1 mol
of a compound is formed from its elements in their
standard states and under standard conditions.
• E.g.
• C (graphite) + 2H2(g)
CH4 (g) ∆Hѳf = -75kJmol-1
• It is not possible to measure the standard enthalpy
change of formation directly. So they have to be
calculated indirectly from enthalpy changes of
combustion using enthalpy cycles.
Note: The standard enthalpy of formation of an
element in its standard state is zero.
Use of standard enthalpy of formation
• The standard molar enthalpy of reaction ΔHθr
can be calculated if the standard molar
enthalpy of formation ΔHθf of the reactants
and products are known.
• i.e. ΔHθr = ΔHθf(products)- ΔHθf(reactants)
Hess’s Law
This law states that the overall enthalpy change
that accompanies a chemical reaction is
independent of the route by which the
reaction takes place, provided the initial and
final states are the same. .
• In practical terms this means that you can do
other reactions to find the enthalpy change
for the main one.
Hess’s Law (continued)
For any given overall chemical reaction, the
overall energy change is the same whether
the reaction takes place in a single step or via
a series of steps.
A ΔH1 D
ΔH
ΔH
ΔH1= ΔH2 + ΔH3 + ΔH4
B
C
2
4
ΔH3
Hess’ Law Triangle
• Hess’ law is useful for working out reactions
which are either hard or impossible to do in
the laboratory.
• Suppose we wanted to measure ΔHθf for
carbon monoxide, CO. The reaction:• C(graphite) + ½O2(g)
CO(g)
• is very hard to control. However the following
combustions are known:-
C(graphite)
CO(g)
+
+
O2(g)
½ O2(g)
CO2(g)
ΔHc = -394kJmol-1
CO2(g)
ΔHr = -283kJmol-1
All the information is then placed into the Hess’ triangle as shown below:-
C(graphite) + ½ O2(g)
O2(g)
ΔH1
CO(g)
½O2(g)
ΔH2
ΔH3
CO2(g)
Hence ΔH1 = ΔHof = ΔH3 - ΔH2 = -394 - (-283) = 111kJmol-1
PROBLEM
Calculate the standard molar enthalpy of formation for butane:4C(graphite) +
5H2(g)
C4H10(g)
given the following enthalpy changes of combustion below:6½O2(g)
C(graphite) +
O2(g)
CO2(g)
ΔHc = -394kJmol-1
H2(g)
½O2(g)
H2O(l)
ΔHc = -286kJmol-1
+
4CO2(g) +
5H2O(g)
ΔHc = -2877kJmol-1
C4H10(g) +
ANSWER
Setting up the triangle, remembering to take into account the amounts
present in each of the equations:-
4C(graphite) +
4 x ΔH3
5H2(g)
ΔH1
ΔH2
5 x ΔH4
4CO2(g)
+
C4H10(g)
5H2O(l)
From Hess’ Law:ΔH1 = ΔHof = ( 4 x ΔH3 ) + ( 5 x ΔH4 ) - ΔH2
= ( 4 x -394 ) + ( 5 x -286 ) - ( -2877 )
= -1576 -1430 + 2877 = 129kJmol-1
Stability of metal oxides
• The standard molar enthapies of formation of the
metal oxides show how stable they are and hence
the ease of extraction of their corresponding metals.
The more exothermic (negative) is the value then the
more stable the oxide and hence the more difficult it
is to extract the metal.
• Ag2O (-31kJmol-1) Fe2O3 (-824kJmol-1)
• Cr2O3 ( -1128kJmol-1) Al2O3 (-1669kJmol-1)
Extraction of the metal
• Ag2O is broken up on heating
• Fe2O3 needs to be heated with carbon or carbon
monoxide
• Cr2O3 needs to be heated with another metal
• Al2O3 can only be extracted by electrolysis
Use of enthalpy cycles
These can also be used to find different enthalpy changes, not just those of
formation.
The Enthalpy Cycle below provides an alternative route for the
combustion of 1 mole of methanol. The methanol is first converted to the
elements from which it was formed (the reverse of the enthalpy of
formation), and these are then burned to form the products (enthalpy of
combustion).
Example problem
• Calculate the standard enthalpy change for the
reaction:
• C2H4 (g) + H2 (g)
C2H6 (g)
Given the following standard enthalpy changes of
combustion are as follows:
∆Hѳc (C2H4(g) = -1411 kJmol-1
∆Hѳc (H2(g)) = -286 kJmol-1
∆Hѳc (C2H6(g)) = -1560 kJmol-1
You are given the equation so write it down with
∆H above the arrow:
∆H
C2H4 (g) + H2 (g)
-1411
-286
C2H6 (g)
-1560
2CO2 (g) + 2H2O (l) + H2O (l)
Find two routes around the equation without going against the flow of any
arrows.
Using Hess’s law:
∆H + –1560 = -1411 + -286
∆H = -1411 -286 +1560 = -137kJmol-1
Bond Enthalpy
• To break a bond needs an input of energy – an
endothermic change.
• When a bond is made energy is evolved – an
exothermic change.
The amount of energy needed to break a bond is
called the bond energy or bond enthalpy.
For a simple diatomic molecule (containing only
two atoms) the energy needed to break one
mole of the bond is called the bond
dissociation energy.
• For a compound with more than two atoms
e.g. CH4 an average value is taken – this is
called the mean bond enthalpy. These are
often used in calculations.
Average bond enthalpies
Knowing the average molar bond enthalpies and
that bond breaking is endothermic (positive)
and bond forming is exothermic (negative) we
can calculate the overall enthalpy change for a
reaction.
e.g.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
By drawing out the full graphical formula we can see which bonds need to be
broken and which bonds need to be formed.
H
H
C
H
+
2O
O
O
C
O + 2H
O
H
H
Bonds broken
Bonds formed
4xC
H = (4 x +413) = +1652
2xC
2xO
O = (2 x + 498) = + 996
4 x O H = (4 x 464) = -1856
Total = + 2548kJ
Hence:-
O = (2 x -805) = -1610
Total = -3466kJ
ΔHr = +2548 -3466 = -918kJmol-1
PROBLEM
Calculate the molar enthalpy of combustion of ethanol:C2H5OH(g) +
3O2(g)
Bonds broken
2CO2(g)
+
3H2O(g)
Bonds formed
1x C
C = +347
4x C
O = -3320
5xC
H = +2065
6x O
H = -2784
1 x C O = +358
Total
= -6004kJmol-1
1 x O H = +464
3 x O O = +1494
Total = +4728kJmol-1
Hence ΔHc = +4728 -6004 =
-1276kJmol-1