Page 128 - Chemical equilibria Volume 4

104 Chemical Equilibria
and:
{ }
dH
d y = 2 [3.111b]
k
C
However, because of the stoichiometry of the reaction, the elementary
variations of the quantities of matter are not random, and necessarily we
have:
{ }
{ } =−
dCO d H 2 [3.112]
and therefore, by substituting back into relations [3.111], we obtain:

d y = k C [3.113]
d x k H

This is the equation of a straight line. The line passes through the point
representative of the initial mixture.

To demonstrate the use of the diagram, let us begin with a mixture
formed of 5 moles of carbon monoxide and 1 mole of water, and determine
the equilibrium compositions at 1600°C – a temperature where the constant
K () P is 0.322. The initial point of the mixture is at A according to Table 3.1.
27
The slope of the line representing the evolution is –5. The equilibrium at
1600°C is represented by point P, which is the point of intersection of the
straight line starting at A with the slope of –5 and the equilibrium isotherm
with the constant 0.322 (Figure 3.17). The coordinates of P are x = 0.87 and
y = 0.62. This gives us the composition at equilibrium:
– 5 × 0.87 = 4.35 moles of CO and therefore 0.65 moles of CO 2;

– 1 × 0.62 = 0.62 moles of H 2 and therefore 0.38 moles of H 2O.

In this chapter, we have defined and used a number of equilibrium
constants. The next chapter will be devoted to the evaluation of the
numerical values of those constants, either by experimentation or by
computation.