Modern Engineering Thermodynamics

15.12 Chemical Equilibrium and Dissociation 629
WHAT DOES THE SYMBOL ðÞ MEAN?
In these equations, we use ∏ as the symbol for repeated multiplication, just as we use ∑ as the repeated summation symbol;
that is,
and
∑ N
α 1
i=1
ð Þ = α 1 + α 2 + α 3 + … + α N ,
∏ N
= ðα 1 Þ× ðα 2 Þ× ðα 3 Þ× ð…Þ × ðα N Þ
i=1
K e is the equilibrium constant for the reaction, defined as
2
K e = exp6
4
∑
R’
v i g • i
−∑
P’
RT
3
v i g • i
7
5 (15.35)
and, from the previous equation, we also have
The equilibrium constant:
K e =
∏
P′
∏
R′
ðp i /p°
ðp i /p°
Þ v i
Þ v i
ð
= p C/p° Þ vC ðp D /p° Þ vD ð…Þ
ðp A /p° Þ vA ðp B /p° Þ vB ð…Þ
(15.36)
Equation (15.36) indicates that the equilibrium constant is a measure of how much product has been generated
by the reaction. Equation (15.36) (or (15.37) later) is normally used to find the actual concentrations v i if the
equilibrium constant K e is known, whereas Eq. (15.35) is normally used to find K e if the v i are known.
EXAMPLE 15.14
Calculate the equilibrium constant for the reversible equilibrium water vapor dissociation reaction equation H 2 O ⇆ H 2 + 1 2 O 2
at 0.1 MPa and at the following temperatures: (a) 298 K and (b) 2000. K.
Solution
Here, we do not have an irreversible reaction equation to contend with, so we can find K e directly from Eq. (15.35). Then,
RT ln ðK e Þ = ∑
R′
where the g • i are determined from Eq. (15.34).
a. At T = T° =298 K, Eq. (15.34) reduces to g • i = ðg f °Þ i , then
v i g • i −∑ v i g • i = g • H2O − g• − 1 H2 2 g• O2
P′
RT ln ðK e Þ = ðg f °Þ H2O − ðg f °Þ H2 − 1 2 ðg f °Þ O2
and, since H 2 and O 2 are elements, their molar specific Gibbs function of formation is zero. Then, from Table 15.7, we
have for water vapor H 2 O(g)
ln K e =
ðg f °Þ H2O /RT =
− 228:583
0:0083143ð298Þ = −92:3
and
K e = expð−92:3Þ = 8:22 × 10 −41 (Continued )