One mole of H 2 O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H 2 O (g) + CO (g) ⇌ H 2 (g) + CO 2 (g) Calculate the equilibrium constant for the reaction.

Question

Philoid · Accepted Answer

Initially there was 1 mole of H₂O and CO in a 10 L flask.

At equilibrium, 40% of H₂O has reacted thus amount of H₂O remaining is 0.6 mole.

The amount of CO reacted is also 0.4 mole.

Amount of H₂ and CO₂ formed is also 0.4 mole

H₂O (g)+ CO (g)⇌H₂ (g) + CO₂ (g)

H₂O (g) + CO (g) ⇌H₂ (g) + CO₂ (g)

Equilibrium constant for the reaction is,

K_c =

=

= 4/9

One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,H2O (g) + CO (g) ⇌H2 (g) + CO2 (g)Calculate the equilibrium constant for the reaction.

One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
H₂O (g) + CO (g) ⇌H₂ (g) + CO₂ (g)

Calculate the equilibrium constant for the reaction.