Modern Engineering Thermodynamics

634 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.16 (Continued)
Table 15.9 Example 15.16, Typical Values
y log K e T(K)
0.001 −4.650 1710
0.010 −3.147 2020
0.100 −1.615 2850
0.500 −0.349 3950
subject to the following reversible equilibrium dissociation reaction equation:
H 2 O ⇆ H 2 + 1 2 O 2
Thus, v H2 = 1:0 and v O2 = 1 2
, and the overall reaction equation becomes
H 2 O !ð1 − yÞH 2 O + yH 2 + ðy/2ÞO 2
Thus, v H2O = 1, v H2 = 1, and v O2 = 1 2
, then Eq. (15.38) gives
K e = χ ð1 + 1/2 − 1Þ
H2 χ1/2 O2
p m
χ H2O p°
where
χ H2O =
2ð1 − yÞ
2 + y
χ H2
= 2y
2 + y
and χ O2
= y
2 + y
so that
K e =
y 1/2
y
ð1Þ 1/2 = y 1/2
y
1 − y 2 + y
1 − y 2 + y
Arbitrarily choosing values for y and solving for K e from the previous equation, then looking up the corresponding temperature
in Table C.17 in Thermodynamic Tables to accompany Modern Engineering Thermodynamics gives the desired relation between the
amount of H 2 present (y) and the system temperature (T). Table 15.9 illustrates some typical values.
Exercises
46. Determine the value of the equilibrium constant and the reaction temperature in Example 15.16 when the degree of
dissociation is y = 0.250. Answer: K e = 1/9 = 0.1111, then T = 3280 K.
47. If the equilibrium constant in Example 15.16 is 1.0, determine the degree of dissociation y for this reaction.
Answer: y = 2/3.
48. If the degree of dissociation in Example 15.16 is 0.8384, determine the base-10 logarithm of the equilibrium constant
and the temperature of the dissociation reaction. Answer: log 10 (K e ) = 0.450 and T = 5000. K.
15.13 RULES FOR CHEMICAL EQUILIBRIUM CONSTANTS
Tables of equilibrium constants are usually limited in size. This limitation can often be overcome by combining
reactions available in a table to produce a desired reaction. For example, many tables include equilibrium constants
for the water dissociation reaction H 2 O ⇆ H 2 + ð1/2ÞO 2 . But what do you do if you need the reverse reaction
for the formation of water H 2 + ð1/2ÞO 2 ⇆ H 2 O? Are the equilibrium constants for these two reactions the
same? The following three chemical equilibrium constant rules can be used to determine the equilibrium constant
for a reaction equation that is not listed in any table but can be constructed from reactions with known
equilibrium constants that are listed in a table.
EQUILIBRIUM CONSTANT RULE 1
Let K e1 be the equilibrium constant for the reaction ν A A + ν B B ⇆ ν C C + ν D D and let K e2 be the equilibrium constant for the
reverse reaction ν C C + ν D D ⇆ ν A A + ν B B: Then, these two equilibrium constants are related as follows: K e2 = 1/K e1 .