When a chemical reaction has taken place it is usual that the products have either less stored energy than the reactants (overall exothermic) or more stored energy than the reactants (overall endothermic).
DH = SHproducts - SHreactantsEnthalpy level diagrams illustrate this:
When chemical reactions take place, chemical bonds are broken in the reactants and new chemical bonds are formed in the products. It is as a result of these processes that a reaction is overall exothermic (energy is given out to the surroundigs) or overall endothermic (energy is absorbed from the surroundings).
It is not quite so obvious that...
But, if a particular bond in a molecule is broken and then reformed the same amount of energy must be involved, because the First Law of Thermodynamics (also known as the 'law of conservation of energy') must apply.
F2(g) ® 2F(g) DH° = +158 kJ mol-1
2F(g) ® F2(g) DH° = -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, DH, for a chemical reaction. There is often close agreement with the standard value for an enthalpy change given in a chemical data book.
Bond enthalpies can be used to calculate an overall enthalpy change, DH, for a chemical reaction. It is an over-simplification, but bonds are broken in the reactants and new bonds are formed in the products.
This idea is illustrated below for the combustion of ethanol:
CH3CH2OH(g) + 3O2(g) ® 2CO2(g) + 3H2O(g)
Note that all reactants and products are gaseous.
This is quite straigtforward. First write out the balanced equation for the reaction showing full structural formulae. Now simply add together the molar bond enthalpies involved for the reactants to obtain a total endothermic value. Do the same for the products to obtain a total exothermic value. Finally, add the total endothermic and exothermic values together to find DH for the reaction.
|Here bond enthalpies are defined endothermically.|
Total endothermic value = (+347 x 1) + (+413 x 5) + (+358 x 1) + (+464 x 1) + (+498 x 3) = +4728 kJ
Total exothermic value = (-464 x 6) + (-805 x 4) = -6004 kJ
Sum total of bond breaking and bond making:
DHc = +4728 + - 6004 = -1276 kJ mol-1
The molar bond enthalpies and bond lengths in the table above show that in general...
For example, hydrogen chloride shows little tendencey to decompose into its constituent elements when heated. Strong heating of hydrogen bromide produces a brown colour of bromine vapour, while copious violet fumes of iodine form when a hot glass rod is plunged into a gas jar of hydrogen iodide.
HI(g) ® ½H2(g) + ½I2(g)
The stability of the hydrogen halides to thermal decomposition therefore decreases in the order:
HCl > HBr > HI
and this is due to the progressive decrease in the H ¾ X bond enthalpy.
With regard to the diatomic halogen molecules (X2), fluorine has an abnormally low molar bond enthalpy. This low value is in part explained by the repulsion between the non-bonding electrons on the fluorine atoms, the other halogen molecules having longer bond lengths making this repulsive force is less significant.