Page 99 - Air pollution and greenhouse gases from basic concepts to engineering applications for air emission control

3.3 Chemical Kinetics and Chemical Equilibrium 73

This equation describes the conversion between K P and K C as

cþd a b Dn
K P ¼ K c RTð Þ ¼ K c RTÞ ð3:33Þ
ð
where Dn is a general expression of the number of coefficient difference and Dn ¼
ð c þ dÞ ða þ bÞ for the equilibrium reaction aA þ bB $ cC þ dD.
Example 3.7: SOx chemical equilibrium
Consider a mixture of 64 kg of SO 2 and 32 kg of O 2 mixed with 56 kg of inert N 2 in
a sealed reactor. After ignition the system eventually reaches chemical equilibrium
at temperature T = 1,000 K and pressure P ¼ 2 atm. Assume nitrogen does not
participate in the chemical reactions. If the equilibrium constant (K P ) at 1,000 K is
1.8, determine the corresponding chemical mole fractions in the equilibrium
products.
Solution
The mole amount of the mixture before reaction can be determined as follows:

64,000 g
¼ ¼ 1,000 mole
n SO 2
64 g=mole
32,000 g
¼ ¼ 1,000 mole
n O 2
32 g=mole
56,000 g
¼ ¼ 2,000 mole
n N 2
28 g=mole
After ignition the chemical reaction most likely proceed as follows:

1
SO 2 þ O 2 $ SO 3
2

Since N 2 does not participate in the reaction as an inert gas, it works only as a
dilution gas throughout the process. When the final products reach equilibrium, we
assume there is s mole of SO 3 . Then we can determine the mole amount of other
gases before combustion and after reaching equilibrium by mass balance as follows.
Mole amount n SO 2 n O 2 n SO 3 n N 2 Total n
Prior to combustion 1,000 1,000 0 2,000 4,000
At equilibrium after 1,000 − s 1,000 − 0.5 s s 2,000 4,000 − 0.5 s
reaction

The partial pressure-based equilibrium constant can be calculated using
Eq. (3.31)