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72 3 Basics of Gas Combustion

CO þ 3H 2 $ CH 4 þ H 2 O ðR1Þ

What is the equilibrium constant expression and value for the following reaction?

CH 4 þ H 2 O $ O þ 3H 2 ðR2Þ

Solution
According to Eq. (3.28), the equilibrium constant for Reaction R 1 is

½ CH 4 Š½H 2 OŠ
K C1 ¼ ¼ 4
3
½ COŠ½H 2 Š
Similarly, for R 2
3
½ COŠ½H 2 Š 1 1
K C2 ¼ ¼ ¼ ¼ 0:25
½ CH 4 Š½H 2 OŠ K C1 4

Chemical equilibrium can also be described in terms of partial pressures of the
gases when all the products are considered as ideal gases. The partial pressure-
based equilibrium constant equation is described as

c
P P d
C D
K P ¼ a b ð3:29Þ
P P
A B
According to Dalton’s Law, Eq. (2.40), P i ¼ y i P, then Eq. (3.29) becomes
c d
y y cþd a b
C D
K P ¼ a b P ðÞ ð3:30Þ
y y
A B
With mole fractions y i ¼ n i =n into this equation becomes
c
n n d cþd a b
P
C D
K P ¼ a b ð3:31Þ
n n n
A B
where n i is the mole amount of the ith gas and n is the total mole amount of all the
gases in the reaction system. And n n A þ n B þ n C þ n D Þ as there may be other
ð
gases present in the system.
Considering the ideal gas law, PV ¼ nRT,
c d c d
ð n C =VÞ n D =VÞ cþd a b ½CŠ ½DŠ cþd a b
ð
K P ¼ ð RTÞ ¼ ð RTÞ ð3:32Þ
a b a b
ð
ð n A =VÞ n B =VÞ ½AŠ ½BŠ