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? K_p= 10.1 at 400°C

Question

Philoid · Accepted Answer

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

Given: Temperature = 400° C

Partial pressure of CO (P_CO) = Partial pressure of H₂O (P_H2O) = 4.0 bar

K_p = 10.1 bar

Formula: K_p =

Let the partial pressure of H₂ (P_H2) at equilibrium = P bar

By applying the formula given below:

K_p =

Where P_x and P_Y are the partial pressures of x and y(products)

P_A and P_B are the partial pressures of a and b (reactants)

Hence, K_p =

As K_p = 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 (P_H2) at equilibrium = P bar = 3.042 bar.