Advanced Higher ChemistryUnit 2

Applications of Hess’s Law 2

Mean Bond Enthalpies

The mean bond enthalpy is defined as the energy required to break a covalent bond or bonds in the gaseous state

H2(g) 2H(g) H = +435 kJmol-1

Or it can be the energy released when 1 mole of a particular type of covalent bond is formed

2H(g) H2(g) H= – 435 kJmol-1

Since the precise energy of a bond is subject to variation due to the effects of neighbouring atoms or groups, data books always quote an average or mean.

These are found on page 9 of Higher / Advanced Higher data book

Bond HH O=O CC CH CO C=O

mean B.E.(kJ mol-1)

432 497 346 414 365 798

The idea of measuring and calculating the bond enthalpies for various bonds allows another application of Hess’s law

It allows the enthalpy changes for a variety of chemical reactions or other bond enthalpies to be calculated that might otherwise be difficult.

The values for these are in the data book and can be used as follows to calculate enthalpy changes:-

1. The bond enthalpies for all of the bond breaking and bond making steps are calculated as a “profit and loss” account.

2. The total enthalpies for each side are added together

i.e. H = bond breaking steps + bond making steps

Note : it is helpful when doing these to draw any relevant molecules so as to count the number of bonds being broken and formed accurately.

e.g. 1 Calculate the enthalpy change for the reaction

H2(g) + Cl2(g) 2HCl(g)

Bonds Broken Bonds Made

H H 1 x 436 kJ

Cl Cl 1 x 243 kJ

Total + 679 kJ

2HCl 2 x 431 kJ

Total 862kJ

Hence the enthalpy change, H = 679 + ( 862) = 183kJ

Since two moles of product are formed H = 183 2 = 91.5kJmol-1

e.g. 2 Calculate the enthalpy change for the formation of ethene, C2H4

Bonds Broken Bonds Made

2C(s) + 2H2(g) C2H4(g)

2C(s) 2C(g) 2 x 715 kJ

2HH 2 x 436 kJ

Total + 2302 kJ

C=C 1 x 607 kJ

4CH 4 x 414 kJ

Total 2263 kJ

Hence the enthalpy of formation of ethene = 2302 + ( 2263) = + 39 kJmol-1