Bond breaking and formation- Bond enthalpy
The bonds between atoms in molecules are a source of stored potential energy.
To break a chemical bond you
have to supply energy, that is bond breaking is an endothermic process.
However when chemical bonds are formed
energy is released, usually as heat.
The image opposite shows a molecule of carbon tetrachloride (CCl4).
This small molecule contains 4 C-Cl covalent
bonds. The bond enthalpy is the amount of energy
required to break chemical bonds. The stronger the bond the more
energy is required to break it or the more energy
will be released when it forms.
Bond enthalpy
The bond enthalpy for a diatomic molecule, also called
the bond dissociation enthalpy is the enthalpy change for the following
process:
A-B(g) → A(g) + B(g)
The energy required to break a particular covalent bond in an element such as hydrogen, oxygen or nitrogen is shown below.
H2(g) → 2H2(g) ΔH=+436kJmol -1
O2(g) → 2O2(g) ΔH=+497kJmol -1
N2(g) → 2N2(g) ΔH=+945kJmol-1
The bond dissociation energy is for a particular bond
such as a H-H or 0=0 bond. However many bonds such as C-C, C-H, C=O or O-H are
found in many different types of molecules. It is unlikely that say a O-H bond in a
water molecule would have the same bond enthalpy
as
a O-H bond in say an alcohol molecule, simply because the electron
distribution within
each of the O-H bonds is bound to be slightly different and this will lead to different bond strengths. So what
value do we use for say a O-H bond then? Well as a sort of work around we simply
gather data for O-H bond enthalpies from a large number of molecules
containing an O-H group and take an average bond enthalpy across all these molecules. However this will mean that if we
use average or mean bond enthalpies when calculating enthalpy changes
from a reaction the result may differ slightly from any
experimentally gathered result. However the differences are likely to be minor and using mean bond enthalpies
will give us a good idea
of the enthalpy changes taking place.
Mean bond enthalpies
The mean bond enthalpy of a covalent bond is defined as:
The mean bond enthalpy is the average of many values of the
bond dissociation enthalpy for a given bond found in a range of
different compounds.
As an example consider the following process which shows the breaking up of a molecule
of methane in the gas phase to form
individual atoms of carbon and hydrogen. To carry out this endothermic process
requires an input of 1652kJmol-1 of energy.
CH4(g) → C(g) + 4H(g) ΔH=+1652kJ/mol
Since this process represents the breaking of 4 C-H
bonds
we can calculate the mean bond enthalpy of a C-H bond in methane as 413kJ/mol (
1652kJ/4). It is unlikely that whatever route by which the C-H bonds in a methane molecule
are broken that all the C-H bonds
will require exactly
416kJ per mole to break them. It is unlikely once the C-H bond in a methane
molecule is first broken apart that a C-H bond in CH4 molecule
will have the same bond
dissociation enthalpy as a C-H bond in a CH3 or a CH2 molecule, but
using mean bond enthalpies will give a good approximation
of the enthalpy changes taking place.
| Bond |
Mean bond enthalpy/kJmol-1 |
| C-H |
+412 |
| H-H |
+436 |
| O=O |
+497 |
| O-H |
+463 |
| C-Cl |
336 |
| C=O |
+743 |
| C=C |
+612 |
Calculating the enthalpy change in a reaction using mean bond enthalpies
The table opposite gives some common mean bond enthalpy values,
these are mean values taken from a wide range
of compounds. We can use this data to calculate the
enthalpy changes for reactions taking place where
there is no change of state, this usually means reactions in the gaseous state. As a simple example consider the
combustion of methane gas to form carbon dioxide and water vapour:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
To calculate the enthalpy change for this reaction we simply follow
a series of simple steps:
- Calculate the total energy required to break all the bonds in the reactants
molecules.
- Calculate the total energy released
by bond formation in the products
- The enthalpy change for the reaction is simply:
- ΔH =Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)
If possible it is often very helpful if you can draw a quick sketch to show the molecular structure of the reactant and
product molecules,
this will help you see clearly all the bonds which are broken are
formed during the reaction. The image below shows this for the combustion of methane:
To calculate the enthalpy change for the reaction it is
simpy a case of adding up the individual mean bond enthalpies as shown in the table below:
| Bonds broken |
Mean bond enthalpy/kJmol-1 |
|
Bonds formed |
Mean bond enthalpy/kJmol-1 |
| C-H x 4 |
412 x 4=1648 |
|
C=O x 2 |
743 x2 = 1486 |
| O=O x 2 |
497 x 2=994 |
|
O-H x 4 |
463 x 4 = 1852 |
energy supplied to break all bonds in the reactants:
1648 + 994 = 2642. |
|
energy released by bond formation in products:
1486 + 1852= 3338 |
ΔH =Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)
=2642-3338
=-696kJmol-1 |
Since in this example more energy is released by
bond formation than is taken in by bond breaking
the reaction is exothermic with a negative enthalpy change
(ΔH=-ve).
Key points
- Mean bond enthalpies are averages of the
bond dissociation enthalpies averaged over many
molecules containing the particular bond and as such any
enthalpies change calculated using
mean bond enthalpies will only give an approximation to the actual enthalpy change of the reaction.
- The enthalpy change for a reaction is calculated from:
ΔH =Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)
- In an exothermic reaction more energy
is released by bond formation than is taken in by bond breaking. The opposite is true for an
endothermic reaction.
- When using mean bond enthalpies all reactants and products
must be in the gaseous state.
Practice questions
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