Question

For the reaction CO(g) + H2O(g) <=> CO2(g) + H2(g), Kp = 0.64 at 900 K. If the reaction is initiated with just H2(g) and CO2(g), each at a partial pressure of 0.540 atm, what is the partial pressure of CO(g) at equilibrium?

Answer 1

For the reaction CO(g) + H2O(g) <=> CO2(g) + H2(g), Kp = 0.64 at 900 K. If the reaction is initiated with just H2(g) and CO2(g), each at a partial pressure of 0.540 atm, what is the partial pressure of CO(g) at equilibrium?

Given chemical transformation is,

CO(g) + H₂O(g) <----------------> CO₂(g) + H₂(g) Kp = 0.64

Kp = (P_CO2 x P_H2 )/ (P_CO x P_H2O)

Let partial pressure at equilibrium be P for CO and H2O i.e.P_CO =P_H2O= P

And P_CO2 = P_H2 = 0.54

Let us put all these values,

0.64 = [(0.54) x (0.54)]/(P x P)]

0.64 = (0.54)² / P²

Taking square roots of both sides,

0.8 = 0.54/P

P = 0.54/0.8

P = 0.675 atm