Modern Engineering Thermodynamics

630 CHAPTER 15: Chemical Thermodynamics
EXAMPLE 15.14 (Continued)
b. At 2000: K = 3600: R and 0.100 MPa, Eq. (15.34) with Tables 15.7 and C.16c in Thermodynamic Tables to accompany
Modern Engineering Thermodynamics give
and
g • H2O = ðg f °Þ H2O + ½ðhð2000: KÞ − hð298 KÞŠ H2O − 2000:½ðsð2000: KÞŠ H2O + 298½ðs°ð298 KÞŠ H2O
g • H2
= −228,583 + ð35,540:1 − 4258:3Þð2:3258Þ − 2000:ð63:221Þð4:1865Þ + 298ð188:833Þ
= −628900 kJ/kbmole
= 0 + ð26,398:5 − 3640:3Þð2:3258Þ − 2000:ð44:978Þð4:1865Þ + 298ð130:684Þ
= −285,000 kJ/kgmole
g • O2
= 0 + ð29,173:9 − 3725:1Þð2:3258Þ
− 2000:ð64:168Þð4:1865Þ + 298ð205:138Þ = −417,000 kJ/kgmole
Note: The multipliers 2.3258 and 4.1865 in these equations are necessary to convert the Btu/lbmole and Btu/(lbmole·R)
values in Table C.16c into kJ/kgmole and kJ/(kgmole ·K), respectively.
Then,
RT ln K e
= −629,000 − ð−285,000Þ − ð1/2Þð−417,000Þ
= −136,000 kJ/kgmole
so
and
ln K e =
−136,000
8:3143ð2000:Þ = −8:18
K e = expð−8:18Þ = 2:81 × 10 −4
Exercises
40. Find the equilibrium constant for part b of Example 15.14 from the Table C.17 in Thermodynamic Tables to accompany Modern
Engineering Thermodynamics. Note that the numbers in this table are log 10 (K e ), and log 10 (K e ) = 2.30528 × log 10 (K e ) =
2.30528 × ln(K e ). Answer: From Table C.17, we find that K e =10 −3.531 = 0.000294.
41. Using the technique of Example 15.14, determine the equilibrium constant for the dissociation of N 2 ≈ 2 N at 2000. K.
Compare your answer with that obtained in Table C.17. Answer: log 10 (K e ) = 12.02.
42. Using the technique of Example 15.14, determine the equilibrium constant for the dissociation reaction of water
given in Example 15.14 at a temperature of 5300. R ≈ 3000. K. Compare your result with that given in Table C.17.
Answer: K e = 0.0476.
The magnitude of K e is a good indicator of the degree to which a reaction goes to completion. Generally, if K e is
less than about 0.01 (or ln K e < −4.6), then the reaction does not occur to any significant degree. However, if
K e is greater than about 100 (or ln K e > 4.6), then the reaction essentially goes to completion.
If the components of the reaction are ideal gases that obey the Gibbs-Dalton ideal gas mixture law, then the
partial pressures can be expressed in terms of the mole fraction χ i and the total mixture pressure p m as
(see Eq. (12.23))
Then, Eq. (15.36) reduces to
p i = χ i p m
K e =
∏
P′
∏
R′
ðχ i Þ v i
∑
p i − ∑
m P′v
R′
p°
ð Þ v i
χ i
v i
(15.37)
Note that the v i and the repeated multiplication ranges (P’ and R’) in Eqs. (15.36) and (15.37) come from the
equilibrium reaction equation, but the p i and χ i in these equations come from only the products of the irreversible
reaction equation. For example, consider an equilibrium reaction equation for a simple dissociation of the form
v A A ⇆ v B B + v C C