Transcript Bond Enthalpies - Miller, Jonathan

Applications of Hess’s Law
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
mean B.E.
(kJ mol-1)
HH
O=O
CC
CH
CO
C=O
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) +
Bonds Broken
HH

Cl2(g)
2HCl(g)
Bonds Made
1 x 436 kJ 2HCl 2 x  431 kJ
Cl  Cl 1 x 243 kJ
Total
+ 679 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
2C(s) +
2H2(g)
 C2H4(g)
Bonds Broken
Bonds Made
2C(s)  2C(g) 2 x 715 kJ C=C
1 x  607 kJ
2HH
2 x 436 kJ 4CH
4 x  414 kJ
Total
+ 2302 kJ Total
 2263 kJ
Hence the enthalpy of formation of ethene = 2302 + ( 2263)
= + 39 kJmol-1