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 H2. In the second stage, CO formed in the first stage is reacted with more steam in water gas shift reaction,

CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g)


If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that CO H2O p = p = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C

The given reaction is CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g)

Given: Temperature = 400° C


Partial pressure of CO (PCO) = Partial pressure of H2O (PH2O) = 4.0 bar


Kp = 10.1 bar


Formula: Kp =


Let the partial pressure of H2 (PH2) at equilibrium = P bar



By applying the formula given below:


Kp =


Where Px and PY are the partial pressures of x and y(products)


PA and PB are the partial pressures of a and b (reactants)


Hence, Kp =


As Kp = 10.1 bar (given)


∴ 10.1 =


⇒10.1 =


By taking square roots of both the sides, we get


⇒√10.1 =


⇒0.3178 =


⇒12.712– 0.3178 P = P


⇒4.178 P = 12.712


⇒P = 3.042 bar


Thus, the partial pressure of H2 (PH2) at equilibrium = P bar = 3.042 bar.


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