University of QLD Foundation Year
THERMOCHEMISTRY
11.1
INTRODUCTION
In order to understand thermochemistry correctly, you have to first have a good idea
of what energy is. Energy is defined as the ability to do work.
There are two main types of energy:I.
Kinetic (movement) energy:This includes(amongst others):
 Thermal - sub-microscopic particles in motion
 Mechanical - macroscopic particles in motion
 Electrical - energy that is found in the movement of an electric current.
II. Potential (stored) energy:
 Gravitational
- energy that is stored in a gravitational field
 Chemical - energy that is bound up in the bonds of a chemical substance.
 Nuclear
- the stored energy of the atomic nucleus
Thermochemistry is the study of the thermal energy changes that occur during
chemical reactions and physical changes of state.
The Law of the Conservation of Energy states that in any chemical or physical
process, energy is neither created nor destroyed. This obviously means that the
amount of energy put into a system must equal the amount of energy coming out of
that system. Energy however, can be changed from one form to another.
11.2
UNITS
Energy is measured in the SI unit of Joules (J) named after the English physicist
James Prescott Joule (1818-1889) and is defined as the amount of energy needed to
raise the temperature of 1g of pure water by 0.239C. Another way of measuring
energy is to measure the amount of energy it takes to raise the temperature of that
body by 1C.
The unit of measurement here is the calorie, where one calorie is the quantity of heat
that raises the temperature of lg of pure water by 1C.
Food nutritionists also use the Calorie as a measure of energy, but here the term
Calorie refers to the amount of energy the food contains. One dietary Calorie is
actually = 1 000 calories (or one kilocalorie).
eg: 10g of sugar has 41 Calories, means that 10g of sugar releases 41 kilocalories of
heat when it is burnt up completely.
The relationship between these 3 units above is shown in the flow diagram
below:
1 Calorie = 1 000cal = 1 kcal = 4186J
or
1J = 0.239cal
11.3
HEAT CAPACITY
Two other measurements in thermochemistry must be identified here:
a)
Heat capacity. The heat capacity of an object is the amount of heat it takes to
change that object's temperature by 1C. Thus, for example, a cup of water
has a much greater heat capacity than a drop of water. Also, it is obvious, that
the greater the mass of the object, the greater is its heat capacity. The heat
capacity of an object also depends on its chemical composition. For this
reason, different substances of the same mass may have different heat
capacities.
b)
Specific heat capacity/specific heat. The quantity of heat (q) needed to raise
the temperature of 1g of that substance by 1C is known as the specific heat of
that substance. It is measured in J / gC.
e.g.: Metals have low specific heat capacity values, like iron that only
requires one calorie of heat to raise the temperature of 1g of iron by 9C.
Heat affects the temperature of objects with a high specific heat capacity much
less than the temperature of those with a low specific heat capacity. e.g.:
water has a high specific heat capacity and it requires a lot of heating before a
volume of water can have its temperature raised by 1C per 1g!
To calculate the specific heat of a substance, use the following formula
specific heat (C) =
q
m  T
=
heat (J)
mass (g) x change in
temperature (C)
For example:
The temperature of a piece of copper with a mass of 95.4g changes from 25.0C to
48.0C when the metal absorbs 849J of heat. What is the specific heat of copper?
25.0C is the initial temperature and 48.0C is the final temperature.
Thus:
C
=
q
m  T
=
849J
95.4g x Tf - Ti
=
849J
95.4g x 23.0C
=
0.387 J/gC
11.4
THERMOCHEMICAL EQUATIONS
The term heat content or enthalpy is used to indicate the stored energy in a chemical
system at a given temperature and pressure. When the heat content of the products is
less than the heat content of the reactants, the reaction is said to be exothermic and
heat energy is given off. ie: a process that loses heat to the surroundings is called an
exothermic process. When the heat content of the reactants is less than that of the
products, the reaction if said to be endothermic. i.e.: A process that absorbs heat
from the surroundings is called an endothermic process.
(A system is the specific part of the universe on which you focus your attention. The
surroundings include everything outside the system).
Exothermic and endothermic reactions can be shown with the following equation:H
=
H(products) - H(reactants)
H means the heat of reaction.
If the H is negative, then there was heat given off and the reaction was exothermic.
If the H is positive, then there was heat absorbed from the surroundings and the
reaction was endothermic.
Products
Reactants
Enthalpy
 H = -ve
Products
Reaction Path
Enthalpy
 H = +ve
Reactants
Reaction Path
Endothermic
Exothermic
The physical
state of the reactants and productsReaction
in a thermochemical reaction must be
Reaction
stated. To see why, compare the following two equations for the decomposition of
water:H2(g) + ½O2(g)
H = +285.8kJ
H2(g) + ½O2(g)
H = +241.8kJ
difference = 44.0kJ
In one case the reactant, water is a gas. In the other case the reactant is a liquid. The
vaporisation of the liquid water to water vapour at 25C requires an extra 44.0kJ of
heat.
H2O(l)
H2O(g)


When most substances are burnt, an amount of energy is given off. In these cases the
heat of reaction (or the differences in the amount of energy of the reactant and
product) is very negative and the reaction is obviously an exothermic reaction. We
call this type of heat of reaction, the heat of combustion, which is the heat of reaction
for the complete burning of one mole of a substance.
Example:
When one mole of methane is burnt, the heat of reaction is -890.4kJ. But because
this reaction is an exothermic reaction and 1 mol of the substance was completely
burned, we refer to this heat of reaction as the heat of combustion. This is the reaction
that occurs in many homes in Australia that use gas for cooking and heating.
11.5
HEAT AND CHANGES OF STATE
All solids absorb heat in melting to liquids. The heat absorbed by one mole of a
substance in melting from a solid to a liquid at a constant temperature is called the
molar heat of fusion (Hfus.).
The heat lost when one mole of a liquid changes to a solid at a constant temperature
is the molar heat of solidification (Hsolid. )
Because energy is conserved in all physical and chemical changes, the quantity of
heat absorbed by a melting solid is exactly the same as the quantity of heat lost when
the liquid solidifies, or Hfus = Hsolid
When liquids absorb heat at their boiling points, they become vapours. Vaporisation
of a liquid, through boiling or evaporation cools the environment around the liquid as
heat flows from the surroundings to the liquid. The amount of heat absorbed by one
mole of a liquid that is undergoing evaporation is called the molar heat of
vaporisation (Hvap) and conversely the condensation of 1 mole of vapour releases
heat as the molar heat of condensation (Hcond).
Vapou
r
 Hcond = -ve
 Hvap = +ve
Enthal
py
Liquid
 Hsolid = -ve
 Hfus = +ve
Solid
Refer to the text for the Standard Heats of Physical Change of some substances.
Heat changes can occur when a substance is dissolved in a solvent. The heat change
caused by dissolution of one mole of substance is the molar heat of solution, 
Hsoln.
Examples:
1.
How many grams of ices at 0C could be melted by the addition of 2.25kJ
of
heat?
Refer to the table of molar heats and identify that the molar heat of fusion of
water is 6.01kJ/mol.
Thus: 1 mol of water requires 6.01kJ of heat to melt
x mol of water requires 2.25kJ heat to melt
x
=
2.25kJ x 1 mol
6.01kJ
x
=
0.37 mol
1 mol water = 18.0g
0.37 mol
= xg
x
2.
= 6.74g
How much heat is absorbed when 24.8g of liquid water at 100C is
converted
to steam at 100C?
For water, Hvap = -Hcond = 40.7kJ/mol
1 mol water
x mol
x
x
Thus:
3.
= 18.0g
= 24.8g
= 24.8g x 1 mol
18.0g
= 1.38 mol
1 mol water requires 40.7kJ/mol
1.38mol
requires x kJ/mol
x
=
40.7kJ/mol x 1.38mol
1mol
x
=
56.1kJ/mol
How much heat is released when 2.500mol of NaOH(s) is dissolved in
water?
The molar heat of solution of NaOH(s) = -445.1kJ.
Thus 1 mol NaOH has a heat of solution of -445.1kJ
then 2.5mol of NaOH has a heat of solution = 2.5 x -445.1
= -1113kJ
11.6
HESS’S LAW
You are already aware that carbon can exist in two common allotropes - diamond and
graphite. Because graphite is more stable than diamond, you would expect diamond
to change into graphite. Although this does happen, it takes millions of years to do so
and obviously this reaction if far too slow for the enthalpy change to be measured.
However, Germain Hess calculated a law that assists in cases like these. This rule is
called Hess's Law of Heat Summation and states that if you add two or more
thermochemical equations to give a final equation, then you can also add the heat
changes to give the final heat changes.
Example: use Hess's Law to calculate the enthalpy changes for the conversion of
diamond to graphite.
a.
b.
C(s, graphite) +
C(s, diamond) +
O2(g) 
O2(g) 
CO2(g) H = -393.5kJ
CO2(g) H = -395.4kJ
Write the reverse of equation (a) to give:
c.
CO2(g) 
C(s, graphite) +
O2(g) H = +393.5kJ
(NB: When you write a reverse reaction, you must also change the sign of H)
If you now add equations (b) and (c), you get the equation for the conversion of
diamond to graphite. Remember, that as in algebra, the CO2(g) and O2(g) terms on
both sides of the equation cancel each other out.
ie:
C(s, diamond)  C(s, graphite)
Now does the same thing with the enthalpy values :ie:
H = -395.4kJ
+
H = +393.5kJ
H = -1.9kJ
Therefore, the conversion of diamond to graphite is an exothermic process.
Conversely, the change of graphite to diamond is an endothermic process.
11.7
STANDARD HEATS OF FORMATION
Sometimes it is hard to measure the heat change for a reaction. Materials may not be
available, or perhaps the reaction is too slow for an enthalpy change to be measured.
The chemicals involved may be too expensive to buy. In these cases you can
calculate the enthalpy change of the reaction from standard heats of formation.
The standard heat of formation (Hf) of a compound is the change in enthalpy that
accompanies the formation of one mole of the compound from its elements with all
substances in their standard states at 25C.
Many values of Hf have been measured and these can be seen in your text book.
For any of the free elements, eg: H2, F2, N2 or even C (as in graphite) the
Hf is = 0.0.
The H for a reaction is the difference between the standard heats of formation of all
the reactants and products.
i.e.:
H =
Hf (products)
Hf (reactants)
Example:
What is the standard heat of reaction for the reaction of carbon monoxide gas
with oxygen to form carbon dioxide gas?
The equation for this reaction is:
2CO(g)
+
O2(g) 
HfO2(g)
HfCO(g)
HfCO2(g)
=
2CO2(g)
=
0.0kJ/mol (free element)
=
-110.5kJ/mol
-393.5kJ/mol
Sum the Hf values of the reactants and then the products, taking into account the
number of moles of each.
Reactants:
221.0kJ
2molCO(g) x -110.5kJ
+
1molO2(g) x
1molCO(g)
Products:
2molCO2(g) x
-393.5kJ
0.0kJ
1molO2(g)
=
-787.0kJ
= -
1molCO2 (g)
But:
Then:
H
H
= Hf (products) - Hf (reactants)
= (-787.0kJ) - (-221.0kJ)
= -566.0kJ
H is negative, therefore the reaction is exothermic.