Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of the two `2` 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)=0.1` 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