SELECT * FROM question_mgmt as q WHERE id=3068 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

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 H2 and 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, Kc 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 Kc= 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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