|Standard Heats of Reaction
|Chemical scientists have agreed upon a standard reference set of conditions
for temperature and pressure. These conditions have of course been chosen
so that experiments can be done easily. Thus the standard reference temperature
is 25oC, which is just slightly above normal room temperature.
This is easily controlled in a water bath. The reference pressure is 1 atmosphere
or 101.325 kPa.
|When the enthalpy change of a reaction is determined with all reactants
and products at 1 atm and some given temperature, and when the scale of the
reaction is in the moles specified by the coefficients of the equation,
then H is the standard enthalpy change or the standard heat of reaction.
To show that a pressure of 1 atm is used, the symbol H is given a superscript,
o, to make the symbol Ho. Values of Ho usually
correspond to an initial and final temperature of 25oC, unless
|The units of Ho are normally kilojoules. For example, a
reaction between gaseous nitrogen and hydrogen produces gaseous ammonia according
to the equation
N2(g) + H2(g) ----> 2 NH3(g)
|When 1.00 mol of N2 react with 3.00 mol of H2 to
form 2.00 mol of NH3 at 25o and 1 atm, the reaction
releases 92.38 kJ. Hence for the reaction as written Ho = -92.38
|Often it is useful to make the enthalpy change of a reaction part
if its equation. When we do this we have to be very careful about the coefficients,
and we must indicate the physical states of all the reactants and products.
The reaction between gaseous nitrogen and hydrogen to form gaseous ammonia,
for example, releases 92.38 kJ is 2.00 mol of NH3 forms.
|But if we were to make twice as much, or 4.00 mol, of NH3 from
2.00 mol of N2 and 6.00 mol of H2, then twice as much
heat (184.8 kJ) would be released. On the other hand, if only 0.50 mol of
N2 and 1.50 mol of H2 were to react to form only 1.00
mol of NH3, then only half as much heat (46.19 kJ) would be released.
|An equation that includes its value of Ho is called a thermochemical
equation. The following three thermochemical equations for the formation
of ammonia, for example, give the quantitative data describe in the preceding
paragraph and correctly specify the physical states of all substances.
N2(g) + 3 H2(g) ----> 2 NH3(g) Ho
= -92.38 kJ
2 N2(g) + 6 H2(g) ----> 4 NH3(g) Ho
= -184.8 kJ
½ N2(g) + 1½ H2(g) ----> NH3(g)
Ho = -46.19 kJ
|When you read a thermochemical equation, always interpret the coefficient
as moles. This is why we must use fractional coefficients in such an
equation, where normally we try to avoid them (because we cannot have fractions
of molecules). In thermochemical equations, however, fractions are allowed,
because we can have fractions of moles.
|Thermochemical Equations for Experimentally Difficult
|Once we have the thermochemical equation for a particular reaction,
we automatically have all the information we need for the reverse reaction.
The thermochemical equation for the combustion of carbon to give carbon dioxide
C(s) + O2(g) -------> CO2(g) Ho = -393.5
|The reverse reaction would be experimentally impossible to preform,
for reasons that will become clear later. But from the above equation we can
write the reverse reaction.
CO2 (g) ------> C(s) + O2 (g) Ho = +393.5
|If we have the Ho for a given reaction, the Ho for
the reverse reaction has the same numerical value, but its algebraic sign