Why do energy changes accompany chemical changes?
Fuel-oxygen systems are the sources of energy because fuels cannot release any energy
until they chemically react with oxygen. When these chemical reactions take place it is the
breaking and making of chemical bonds that is associated with the overall enthalpy change
that accompanies a chemical reaction. This is why the enthalpy of combustion of hexane
(C6H14, ∆Hc = -4163 kJ mol-1) is more exothermic than that of hydrogen (H2, ∆Hc = -286 kJ
mol-1).
The overall enthalpy change of a chemical reaction depends on the:
•
number of chemical bonds broken and formed.
•
type of chemical bonds broken and formed.
For example, look at the two combustion reactions below:
More energy is given out when 1 mole of methane burns than when a mole of methanol
burns. The energy released during combustion comes from the making of bonds to oxygen.
Compare methane and methanol – methanol already has one O-H bond formed; it can be
thought of as methane that is partly oxidised.
As a general rule, the more oxygen a substance has in its molecules, the less energy it will
give out when 1 mole of it burns. However, fuels like ethanol (CH3CH2OH) are not poor fuels;
they are less polluting and often have a higher octane number.
When it comes to carrying fuels around, energy density is more important than enthalpy of
combustion. This is the amount of energy obtained by burning a kilogram of fuel (units are kJ
kg-1). Hexane and hydrogen are compared in the table below.
Fuel
Formula
Mr
∆Hc
-1
kJ mol
Energy
density
-1
kJ kg
Hexane
Hydrogen
C6H14(l)
H2(g)
86
2
-4163
-286
-48 400
-143 000
No. of
moles in 1
kg
No. of
molecules
in 1 kg
11.63
500
7 x 10
26
3 x 10
24
In a kilogram of hydrogen there are 43 times more molecules of hydrogen than of hexane.
Chemical Ideas 4.2 deals with energy changes involved in breaking and making chemical bonds.
Key terms in this chapter are:
Bond breaking.
Bond making.
Overall enthalpy change.
Molar Bond Enthalpy.
Enthalpy Cycle.
Bond Breaking, Bond Making, and Molar Bond Enthalpy
In this section you are going to learn:
•
•
•
•
What is happening in chemical reactions that results in their overall enthalpy change
being exothermic or endothermic.
How to define 'molar bond enthalpy' (bond dissociation enthalpy).
That bond breaking is an endothermic process.
That bond making is an exothermic process.
The answer to the question in the title involves the chemical bonds that hold atoms together
in molecules. In chemical reactions the rearrangement of atoms involves breaking chemical
bonds in reactant molecules and forming new bonds in product molecules. The atoms are
themselves not created or destroyed, but are simply rearranged. Breaking chemical bonds
requires energy; bond breaking is an endothermic process. Conversely, when chemical bonds
form, energy is given out; bond formation is an exothermic process.
•
If more energy is needed to break the chemical bonds in the reactant molecules than is
evolved when new bonds form in the product molecules then the overall enthalpy change
for the reaction will be endothermic.
•
If more energy is evolved when chemical bonds form in the product molecules than was
needed to break the bonds in the reactant molecules then the overall enthalpy change will
be exothermic.
The molar bond enthalpy is the energy required to break one mole of bonds between
pairs of atoms in the 'gaseous molecule'. These values are given in chemical data books.
-1
They are typically expressed in kJ mol .
Molar bond enthalpy is also called bond dissociation enthalpy, but there are other
terms too that refer to this.
Using Bond Enthalpy values to Calculate ∆H for a Reaction
In this section you are going to learn:
•
•
•
How to calculate ∆H for a reaction using molar bond enthalpies.
How to represent such calculations using enthalpy cycles.
Why ∆H values calculated using standard molar bond enthalpies differ slightly from
the ∆H values for the same reactions given in a chemical data book.
The overall enthalpy change, ∆H, for a reaction can be calculated from the appropriate bond
enthalpy values. It is important to remember that when ∆H values for chemical reactions are
calculated using bond enthalpies, all reactants and products are taken to be in the gaseous
state.
Note that even if standard bond enthalpy values are used, in these calculations the standard overall
enthalpy change of reaction calculated will differ slightly from that calculated by the method described in
the previous section. The reasons for this are discussed later.
This is what you do:
1. Write a balanced chemical equation for the reaction involved including state symbols (g), showing
all the chemical bonds in the reactant and product molecules.
2. Decide which chemical bonds are broken and how many (in moles) of each for each of the
reactant molecules.
3. Look up the appropriate bond enthalpies and take the correct number of moles of each.
4. Add all of these together to get a total endothermic value.
5. Do the same as above for the product molecules, but this time you will get a total exothermic
value.
6. Now add together the total endothermic and total exothermic values to obtain ∆H for the reaction.
Q1.
Calculate the enthalpy of combustion, ∆Hc, of ethane, C2H6(g). You need
to look up the required bond enthalpy values.
Total endothermic value .....................................................................
Total exothermic value........................................................................
∆Hc (CH4)............................................................................................
You can also represent the calculation of an enthalpy change for a reaction (∆H) using bond
enthalpies with a bond enthalpy cycle, but the calculation is the same. An example is shown
below for the enthalpy of combustion of propane:
Q2.
Draw an enthalpy cycle to find the overall molar enthalpy change for the
combustion of methanol, CH3OH(g), to form carbon dioxide and water
vapour. (Look up the required bond enthalpy values.)
Now compare your calculated value with that given in a chemical data
book.
∆Hc calculated from bond enthalpies ..................................................
∆Hc from data book.............................................................................
Why are the ∆Hc values different?
There are two reasons why the calculated enthalpy changes of combustion are different when
bond enthalpies are used:
1. Bond enthalpies are often averaged values.
2. Bond enthalpy values apply to molecules in the gaseous state.
You can be precise about the amount of energy required to break certain bonds, such as the FF bond in the fluorine (F2) molecule.
F2(g) →
2F(g)
∆H° = +158 kJ mol-1
However, for more complicated molecules a precise bond enthalpy for a particular chemical
bond depends to some extent on the environment in the molecule where the bond exists. That
is, a precise bond enthalpy depends on what other atoms are attached to the two atoms of the
bond to be broken. For this reason, tables of bond enthalpy values are averaged over those
found in a large number of different compounds.
Because bond enthalpy values are averaged and also apply to molecules in the gaseous state,
when they are used in calculations the answer will be only approximate. However, the
approximation is often very good and such calculations are useful in predicting the overall
enthalpy change, ∆H, for a chemical reaction. There is often close agreement with the
standard value for an enthalpy change given in a chemical data book.
In general, the higher the bond enthalpy, the shorter the bond.
Bond
C–C
C=C
C≡C
Average bond enthalpy
kJ mol-1
+ 347
+612
+838
Bond length
nm
0.154
0.134
0.120