Modern Engineering Thermodynamics

H 2 O !ð1 − yÞH 2 O + yðv H2 ÞH 2 + yðv O2 ÞO 2
(Continued )
15.12 Chemical Equilibrium and Dissociation 633
Degree of dissociation, y
FIGURE 15.10
Example 15.15, Solution, part a.
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
CO 2 → (1 − y)CO 2 + (y)CO + (y/2)O 2
p T = p 0 = 0.100 MPa
0.0
0 1000 2000 3000 4000 5000 6000
Equilibrium temperature, T(K)
0.8
Degree of dissociation, y
FIGURE 15.11
Example 15.15, Solution, part b.
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.1
CO 2 → (1 − y)CO 2 + (y)CO + (y/2)O 2
1.0
10.0
Total mixture pressure, pm (MPa)
p 0 = 0.100 MPa
T = 3000. K
100.0
Exercises
43. Use the solution given in Example 15.15 to determine the dissociation equation for CO 2 at 3000. K and 5.000 MPa.
Answer: CO 2 → 0.8472(CO 2 ) + 0.1528(CO) + 0.0764(O 2 ).
44. Use the solution given in Example 15.15 to determine the dissociation equation for CO 2 at 3000. K and 0.01000 MPa.
Answer: CO 2 → 0.3194(CO 2 ) + 0.6806(CO) + 0.3403(O 2 ).
45. Use the solution given in Example 15.15 to determine the total mixture pressure that produces the following
dissociation reaction for CO 2 at 3000 K: CO 2 → 0.50(CO 2 ) + 0.50(CO) + 0.25(O 2 ). Answer: p m = 0.94 Pa.
An equivalent and perhaps more straightforward way of approaching a dissociation equilibrium reaction is illustrated
in the next example.
EXAMPLE 15.16
Determine the amount of H 2 produced as a function of temperature in the thermal dissociation of water vapor at the SRS
pressure (i.e., p m = p°).
Solution
The overall irreversible dissociation reaction equation for water vapor is