Question:
The reaction,

CO(g) + 3H₂(g) ↔ CH₄(g) + H₂O(g)

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.

Answer:

Let the concentration of methane at equilibrium be x.

CO(g) + 3H₂(g) ↔ CH₄(g) + H₂O(g)

At equilibrium 0.3/1 M 0.1/1 M x 0.02/1 M

It is given that K_c= 3.90.

Therefore,

Kc = [CH₄(g)] [ H₂O(g)] / [ CO(g) ] [H₂(g)]³

⇒ [x] [0.02] / (0.3) (0.1)³ = 3.90

⇒ x = (3.90) (0.3) (0.1)³/ [0.02]

⇒ x = 0.00117 / 0.02

= 0.0585 M

= 5.85 x 10^-2

Hence, the concentration of CH₄ at equilibrium is 5.85 × 10^-2 M.

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