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_2O_((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_2O)` = 4.0 bar, what will be the partial pressure of `H_2` at equilibrium ? `K_p` = 10.1 at `400^@`C.

Explanation

Suppose in reaction , x bar of CO react with x bar of `H_2O` and So, x bar of `CO_2` and `H_2` is create. (Partial pressure (x) bar `prop` concentration mol `L^(-1)`)
`{:("Equilibrium reaction :",CO_((g))+,H_2O_((g)) hArr , CO_(2(g)) + , H_(2(g))),("Initial partial pressure bar :",4.0,4.0,0,0),("Change in pressure bar",-x,-x,+x,+x),("Partial pressure at equilibrium :",(4-x),(4-x),x,x):}`
In this gas reaction `K_p` is as under.
`K_p=((p_(CO_2))(p_(H_2)))/((p_(CO))(p_(H_2O)))`
`10.1 =((x)(x))/((4-x)(4-x))`
`therefore 10.1 = x^2/(4-x)^2` (take square root both sides)
`therefore 3.178-x/(4-x)`
`therefore` 12.712-3.178x =x
`therefore` 12.712=4.178x
`therefore x=12.712/4.178`=3.0426 `approx` 3.04 bar
`therefore p_(H_2) = p_(CO_2)`= x = 3.04 bar
`therefore p_(H_2) = p_(CO_2)`=x=3.04 bar
`p_(CO)= p_(H_2O)` = (4-x)=(4-3.04)=0.96 bar