IB syllabus > energy (sl) > 6.4 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

6.4 - Bond enthalpies

6.4.1: Define the term average bond enthalpy. Bond enthalpies are quoted for the gaseous state and should be recognised as average values obtained form a number of similar compounds. Cross reference with 11.2.6.




6.4.2: Calculate the enthalpy change of a reaction using bond enthalpies.




Why is the enthalpy change of combustion more negative for propan-1-ol than propan-2-ol.

Fuels don't 'contain' energy and they don't all start out from the same energy level.

The energy is released by the formation of the bonds in the products - in this case the CO2 and the H2O. It is the strength of these bonds being created that releases the energy (bond formation is exothermic and releases energy)

Each combusting fuel has to break the atoms apart before forming the products and breaking these bonds in the propanol requires energy. The more stable a molecule is the more energy required to break it apart.

So propan-2-ol starts off from a position of greater stability (lower energy) and so the energy required to break its bonds is greater than for the propan-1-ol.

If more energy is required to break it apart but the same energy is released on forming the products bonds then the actual observed energy released will be less.

This can be expressed simply by a Hess's law diagram showing the propan-2-ol at a lower level than the propan-1-ol but the products at the same level (which in turn are at an even lower level in the diagram as the overall reaction is exothermic). The drop from propan-2-ol to the products is less than the drop from propan-1-ol to the products.

The OVERALL energy experienced by the experimenter is what is left after the endothermic breaking apart of the reactant molecules and the exothermic release of energy by the formation of the product molecules.

Another way to express this idea is by looking at the actual bond energies.

Even though the two molecules apparently have the same numbers and types of bonds, as explained in the previous post, the positive inductive effect (+I) of the two methyl groups makes the electron density more even in the case of the 2-isomer, hence the bonds are stronger and require more energy to break.

E(overall) = (sum of bond energies of the reactants) - (sum of bond energies of the products)

If the concept of electron induction by methyl groups is unfamiliar to you just accept the fact that alkyl groups have a tendency to 'push' electrons towards electronegative elements or positive charges in dipoles. This is called the +I (inductive) effect.

The OH creates a dipole between the oxygen and the neighbouring carbon in which the carbon carries a partial positive charge. It is this charge that destabilises the molecule in the 1- isomer with respect to the 2- isomer. n the 1- isomer there is only one alkyl group 'pushing ' electrons towards it whereas in the 2- isomer there are two alkyl groups performing the same function.






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