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,
$CO_{(g)} + H_2O_{(g)} \longleftrightarrow CO_{2(g)} + H_{2(g)}$
If a reaction vessel at $400^{\circ}C$ 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 $400^{\circ}C$

(D) Let the partial pressure of both carbon dioxide and hydrogen gas be $p$. The given reaction is:
$CO_{(g)} + H_2O_{(g)} \longleftrightarrow CO_{2(g)} + H_{2(g)}$
Initial: $4.0 \ bar \quad 4.0 \ bar \quad 0 \quad 0$
At equilibrium: $(4.0 - p) \ bar \quad (4.0 - p) \ bar \quad p \ bar \quad p \ bar$
It is given that $K_P = 10.1$
Now,
$\frac{P_{CO_2} \times P_{H_2}}{P_{CO} \times P_{H_2O}} = K_P$
$\frac{p \times p}{(4.0 - p)(4.0 - p)} = 10.1$
$\frac{p}{4.0 - p} = \sqrt{10.1} = 3.178$
$p = 12.712 - 3.178p$
$4.178p = 12.712$
$p = 3.04 \ bar$
Hence,at equilibrium,the partial pressure of $H_2$ will be $3.04 \ bar$.