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.

Dihydrogen gas used in Haber's process is produced by reacting methane

1.	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.
Answer» <P> Solution :Let the PARTIAL pressure of `H_(2)` at equilibrium be 'p' bar. So, at equilibrium the partial pressures of the species involved in equilibrium will be as follows: `{:(,CO(g),+,H_(2)O(g),HARR,CO_(2)(g),+,H_(2)(g)),("Initial pressure (bar)",4.0,,4.0,,,,),("Equilibrium pressure (bar)",4-p,,4-p,,p,,p):}` For this reaction, `K_(p)=(p_(CO_(2))xxp_(H_(2)))/(p_(CO)xxp_(H_(2)O))=(pxxp)/((4-p)(4-p))=10.1` or, `(p)/(4-p)=3.178 or, p=(4xx3.178)/(4.178)=3.04` `therefore` The partial pressure of `H_(2)` at equilibrium=3.04 bar.