282 15 · DYNAMIC CHEMICAL EQUILIBRIA
●Operating at higher temperatures increases the reaction rate but reduces Kc(T)
enormously because the reaction is exothermic and the variation of equilibrium
constant with temperature is similar to that shown in Fig. 15.6. (For the Haber–
Bosch process, Kc= 60 mol^2 dm^6 at 227 °C and at 0.02 mol^2 dm^6 at 527 °C.)●Using higher partial pressures of hydrogen and nitrogen increases the equilibrium
concentration of ammonia. For example, using 75 atm of H 2 (g) and 25 atm of
N 2 (g) at 500 °C produces an equilibrium mixture in which 10% of all the
molecules present are ammonia. The use of high partial pressures in this way also
increases reaction rate, but (alone) is not enough to make the reaction fast enough.Haber realized that a catalyst was needed to force the reaction to achieve equilib-
rium faster. In the commercial process, hydrogen and nitrogen in a 3:1 ratio and at
250 atm pressure are reacted in the presence of iron catalyst at a ‘compromise’ tem-
perature of about 450 °C. Below 450 °C the reaction is too slow even with a catalyst.
The use of pressures above 250 atm involves the use of very powerful compressor
pumps which are too expensive to run.
Figure 15.7 shows the main stages of the commercial process. The reaction of
nitrogen and hydrogen takes place in the Fe catalyst bed. The equilibrium mixture
(containing ammonia and unused hydrogen and nitrogen) is passed into a heat
exchanger which cools down the mixture so that the ammonia (but not the reactants)
become liquefied. The reactants are then recirculated to the catalyst bed.Heterogeneous equilibria
Equilibria in which the substances are all in the same phase (e.g. all gases) are known
ashomogeneous equilibria. Equilibria which involves substances in different phases
are known as heterogeneous equilibria.
A simplification occurs with heterogeneous equilibria because the concentrations
of pure solids and pure liquids are constant at a fixed temperature (see page 137).
For example, the equilibrium expression for the production of carbon monoxide
from coke and carbon dioxide,C(s)CO 2 (g)\===\2CO(g)
is
[CO(g)]^2
Kc(T)
[CO 2 (g)][C(s)]15.6
Fig. 15.7The Haber–Bosch process.