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

Suppose the partial pressure of `H_2` at equilibrium =p bar
`{:(,CO(g) + , H_2O(g) hArr, CO_2(g)+,H_2(g)),("Initial pressure","4.0 bar","4.0 bar",0,0),("At equi","(4-p)bar","(4-p)bar","p bar","p bar"):}`
`K_p=(p_(CO_2)xxp_(H_2))/(p_(CO)xxp_(H_2O))`
`=(pxxp)/((4-p)xx(4-p))`=0.1
`p/(4-p)=sqrt0.1`=0.316
p=1.264-0.316 p
or 1.316 p=1.264
`therefore` p=0.96 bar