Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and `H_(2)`. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
`CO(g) + H_(2)O (g) hArr CO_(2)(g) + H_(2)(g)`
If a reaction vessel at `400^(@)C` is charged with an equimolar mixture of CO and steam such that `p_(co) = p_(H_(2)O) = 4.0` bar, what will be the partial pressure of `H_(2)` at equilibrium? `K_(p) = 10.1 at 400 ^(@)C`

Explanation

Given: `P_(CO)=P_(H2O=4.0` Bar, `K_(P)=10.1`
Asked: Partial pressure of `H_(2)` at equilibrium
Formulae: `K_(P)= ((P_(CO_(2)))(P_(H_(2))))/((P_(CO))(P_(H_(2)O))`
Substitution & Calculation:
Suppose the partial pressure of `H_(2)` at equilbrium=P` bar
CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g)`
Initial pressure `4.0 "bar" 4.0` bar
At eqm. `(4-P) (4-P) P P`
`K_(P)=(P^(2))/((4-P)^(2))=0.1` (Given)
`therefore (P)/(4-P)=sqrt0.1=0.316`
`P= 1.264-0.316P or 1.316P=1.264or P=0.96` bar