enthalpy change

Bond enthalpy

bond breaking is an endothermic process 
whereas bond formation is exothermic, it releases energy

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.

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 unliklely 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 and take an average bond enthalpy. 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.

dissociation of a moleule of methane

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 enothermic process requires an input of 1664kJmol-1 of energy.

CH4(g) C(g) + 4H(g)     ΔH=+1664kJ/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 416kJ/mol ( 1664kJ/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 will require 416kJ per mole to break them, since it is unlikely once the methane molecul is broken apart that a C-H bond in CH4 molecule will have the same bond dissociation enthalp as a C-H bond in say 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 reaction 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: 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:

model equation 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)
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

Practice questions

Check your understanding - Questions on mean bond enthalpies