Get Answers to all your Questions

header-bg qa

7.31     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 second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

              CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}+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 H2 at equilibrium? Kp= 10.1 at 400°C  

Answers (1)


We have,
The partial pressure of CO and H_2O is 4 bar and the K_p = 10.1
Let p be the partial pressure of  CO_2 and H_2 at equilibrium. The given reaction is-

                                    CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}+H_{2}(g)
Initial concentration        4 bar               4 bar             0              0
At equilibrium                4 - p                 4 - p             p          p

Therefore, we can write,
10.1= \frac{p_{CO_2}.p_{H_2}}{p_{CO}.p_{H_2O}}
\\10.1= \frac{p^2}{(4-p)^2}\\ 3.178 = \frac{p}{(4-p)}\\ p=(4-p)\times 3.178
By solving the above equation we get,  p = 3.04

Hence, the partial pressure of dihydrogen at equilibrium is 3.04 bar

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support