Page 147 - Physical chemistry understanding our chemical world

114 ENERGY AND THE FIRST LAW OF THERMODYNAMICS

We defined the value of H c O during combustion as H (final) − H (initial) , so a neg-
ative sign for H O suggests the final enthalpy is more negative after combustion.
c
In other words, energy is given out during the reaction. Our Stone Age forebears
absorbed this energy by their fires in the night, which is another way of saying ‘they
warmed themselves’.

Why does butane burn with a hotter flame
than methane?

Bond enthalpies

Methane is easily bottled for transportation because it is a gas. It burns with a
clean flame, unlike coal or oil. It is a good fuel. The value of H O for methane is
c
−1
−1
−886 kJ mol , but H c O for n-butane is −2878 kJ mol . Burning butane is clearly
far more exothermic, explaining why it burns with a hotter flame. In other words,
butane is a better fuel.
The overall enthalpy change during combustion is H . An alternative way of
O
c
calculating an enthalpy change during reaction dispenses with enthalpies of forma-
tion H O and looks at the individual numbers of bonds formed and broken. We
f
saw in Chapter 2 how we always need energy to break a bond, and release energy
each time a bond forms. Its magnitude depends entirely on the enthalpy change
for breaking or making the bonds, and on the respective numbers of each. For
example, Equation (3.28) proceeds with six bonds cleaving (four C–H bonds and
two O=O bonds) at the same time as six bonds form (two C=O bonds and four H–O
bonds).
A quick glance at Worked Example 3.11 shows how the energy released during
combustion is associated with forming the CO 2 and H 2 O. If we
To simplify the cal- could generate more CO 2 and H 2 O, then the overall change in
culation, we pretend H would be greater, and hence the fuel would be superior. In
the reaction proceeds fact, many companies prefer butane to methane because it releases
with all bonds break- more energy per mole.
ing at once; then, an We can calculate an enthalpy of reaction with bond enthalpies by
instant later, different assuming the reaction consists of two steps: first, bonds break, and
bonds form, again all at then different bonds form. This approach can be simplified further
once. Such an idea is if we consider the reaction consists only of reactive fragments,
mechanistic nonsense and the products form from these fragments. The majority of the
but it simplifies the
calculation. molecule can remain completely unchanged, e.g. we only need to
consider the hydroxyl of the alcohol and the carboxyl of the acid
during a simple esterification reaction.

Worked Example 3.12 What fragments do we need to consider during the esterification
of 1-butanol with ethanoic acid?