Modern Engineering Thermodynamics

15.14 The van’t Hoff Equation 635
EQUILIBRIUM CONSTANT RULE 2
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
reaction αðν A A + ν B BÞ ⇆ αðν C C + ν D DÞ, where α is any constant. Then, these two equilibrium constants are related as follows:
K e2 = ðK e1 Þ α .
EQUILIBRIUM CONSTANT RULE 3
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 a second
reaction: ν E E + ν F F ⇆ ν G G + ν H H. Then, the equilibrium constant for a third reaction, formed by multiplying the first
reaction by a constant α and adding it to the second reaction multiplied by a constant β,
is
αðν A A + ν B BÞ + βðν E E + ν F FÞ ⇆ αðν C C + ν D DÞ + βðν G G + ν H HÞ
K e3 = ðK e1 Þ α ðK e2 Þ β :
EXAMPLE 15.17
Determine the equilibrium constants for the following reactions at 5000. K.
a. H 2 + ð1/2ÞO 2 ⇆ H 2 O
b. O 2 + N 2 ⇆ 2NO
c. O 2 + 3:76 N 2 ⇆ 2O + 7:52 N
Solution
a. Table C.17 gives equilibrium constant values for the spontaneous dissociation of water into hydrogen and water at
5000. K (H 2 O ⇆ H 2 + ð1/2ÞO 2 )asK e1 = 10 0.450 = 2.82. The formation of water from hydrogen and oxygen gases is simply
the reverse reaction, so rule 1 can be used to determine its equilibrium constant as K e2 = 1/K e1 = 1/2.82 = 0.355.
b. The reaction O 2 + N 2 ⇆ 2NO is the same as the reaction 2½ð1/2ÞO 2 + ð1/2ÞN 2 Š ⇆ 2 NO, which can be obtained from the
reaction ð1/2ÞO 2 + ð1/2ÞN 2 ⇆ NO that appears in Table C.17 by multiplying it by α = 2. From Table C.17 at 5000. K, we
find that the equilibrium constant for the reaction ð1/2ÞO 2 + ð1/2ÞN 2 ⇆ NO is K e1 = 10 −0.298 = 0.504. Then, using rule 2,
we can calculate the equilibrium constant for the reaction O 2 + N 2 ⇆ 2NO as K e2 = (K e1 ) α .
c. In Table C.17, we find the reactions O 2 ⇆ 2O and N 2 ⇆ 2N. If we multiply the second reaction by 3.76 and add it to the
first reaction, we get the given reaction O 2 þ 3:76N 2 ⇆ 2O þ 7:52N and we can use rule 3 with α = 1 and β = 3.76.
At 5000. K, the equilibrium constant for the reaction O 2 ⇆ 2O is K e1 = 10 −1.719 = 52.4 and the equilibrium constant for
the reaction N 2 ⇆ 2N is K e2 = 10 −0.570 = 0.269, then rule 3 gives the equilibrium constant for the combined reaction as
K e3 = (K e1 ) 1 (K e2 ) 3.76 = (52.4) 1 (0.296) 3.76 = 0.376.
Exercises
49. Determine the equilibrium constant for the reaction H 2 + ð1/2ÞO 2 ⇆ H 2 O at 2000. K. Answer: K e = 3396.
50. Determine the equilibrium constant for the reaction at 5000. K. Answer: K e = 1/0.2535 = 3.945.
51. Determine the equilibrium constant for the reaction O 2 + N 2 ⇆ NO + O + N at 5000. K. Answer: K e = 1.890.
15.14 THE VAN’T HOFF EQUATION
Both the equilibrium constant K e and the molar Gibbs function g • i
depend on the mixture temperature T. To
investigate the temperature dependence of K e , we differentiate Eq. (15.35) with respect to temperature to obtain
!
!
dK e
dT = − 1
v i g • i − T d
v i g • i K e
RT 2
∑
R ′ v i g • i −∑
P ′
dT ∑ R ′ v i g • i −∑
P ′