Question

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,
\(H_2O (g) + CO (g) ⇋ H_2 (g) + CO_2 (g)\)
Calculate the equilibrium constant for the reaction.

Answer 1

Given:

Initial moles of H₂O = 1 mol
Initial moles of CO = 1 mol
Volume of vessel = 10 L
Temperature = 725 K
Reaction:
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)

At equilibrium, 40% of water reacts.

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Step 1: Calculate moles reacted

40% of 1 mol H₂O reacts:

Moles of H₂O reacted = 0.40 × 1 = 0.40 mol

From the stoichiometry of the reaction:
H₂O : CO : H₂ : CO₂ = 1 : 1 : 1 : 1

So,
CO consumed = 0.40 mol
H₂ formed = 0.40 mol
CO₂ formed = 0.40 mol

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Step 2: Write equilibrium moles

H₂O = 1 − 0.40 = 0.60 mol
CO = 1 − 0.40 = 0.60 mol
H₂ = 0.40 mol
CO₂ = 0.40 mol

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Step 3: Convert moles into molar concentrations

Volume of vessel = 10 L

[H₂O] = 0.60 / 10 = 0.06 M
[CO] = 0.60 / 10 = 0.06 M
[H₂] = 0.40 / 10 = 0.04 M
[CO₂] = 0.40 / 10 = 0.04 M

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Step 4: Write equilibrium constant expression

K_c = ([H₂][CO₂]) / ([H₂O][CO])

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Step 5: Substitute values

K_c = (0.04 × 0.04) / (0.06 × 0.06)

K_c = 0.0016 / 0.0036

K_c = 0.44

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Final Answer:

The equilibrium constant for the reaction at 725 K is
K_c = 0.44