Chapter 16 | 803
EXAMPLE 16–4 Effect of Inert Gases on Equilibrium
Composition
A mixture of 3 kmol of CO, 2.5 kmol of O 2 , and 8 kmol of N 2 is heated to
2600 K at a pressure of 5 atm. Determine the equilibrium composition of
the mixture (Fig. 16–13).
Solution A gas mixture is heated to a high temperature. The equilibrium
composition at the specified temperature is to be determined.
Assumptions 1 The equilibrium composition consists of CO 2 , CO, O 2 , and
N 2. 2 The constituents of the mixture are ideal gases.
Analysis This problem is similar to Example 16–3, except that it involves
an inert gas N 2. At 2600 K, some possible reactions are O 2 ∆2O (ln KP
7.521), N 2 ∆2N (ln KP28.304), –^12 O 2 ^1 – 2 N 2 ∆NO (ln KP
2.671), and CO –^12 O 2 ∆CO 2 (ln KP2.801 or KP16.461). Based
on these KPvalues, we conclude that the O 2 and N 2 will not dissociate to
any appreciable degree, but a small amount will combine to form some
oxides of nitrogen. (We disregard the oxides of nitrogen in this example, but
they should be considered in a more refined analysis.) We also conclude that
most of the CO will combine with O 2 to form CO 2. Notice that despite the
changes in pressure, the number of moles of CO and O 2 and the presence of
an inert gas, the KPvalue of the reaction is the same as that used in Exam-
ple 16–3.
The stoichiometric and actual reactions in this case are
Stoichiometric:
Actual:
products reactants inert
(leftover)
C balance:
O balance:
Total number of moles:
Assuming ideal-gas behavior for all components, the equilibrium constant
relation (Eq. 16–15) becomes
Substituting, we get
Solving for xyields
x2.754
16.461
x
13 x 21 2.5x> 221 >^2
a
5
13.5x> 2
b
1 > 2
KP
NCOnCO 22
NCOnCONOnO 22
a
P
Ntotal
b
nCO 2 nCOnO 2
Ntotalxyz 8 13.5
x
2
8 2 xy 2 z¬or z2.5
x
2
3 xy¬ or y 3 x
3CO2.5O 2 8N 2 ¡xCO 2 yCOzO 2 8N 2
CO^12 O 2 ∆CO 2 1 thus nCO 2 1, nCO1, and nO 2 ^122
3 kmol CO
Initial
composition
x CO 2
Equilibrium
composition at
2600 K, 5 atm
y CO
8 kmol N 2 z O 2
2.5 kmol O 2
8 N 2
FIGURE 16–13
Schematic for Example 16–4.
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