SELECT * FROM question_mgmt as q WHERE id=3047 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:

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass

C (s) + CO2 (g) ↔ 2CO (g)

Calculate Kc for this reaction at the above temperature.

Answer:

Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g

And, mass of CO2 = (100 - 90.55) = 9.45 g

Now, number of moles of CO, nco = 90.55/28 = 3.234 mol

Number of moles of CO2,  nco2 = 9.45/44 = 0.215 mol

Partial pressure of CO,

For the given reaction,

Δn = 2 - 1 = 1

we know that,

Kp  =  Kc  (RT)Δn

⇒ 14.19 = Kc  (0.082 x 1127)1

⇒ Kc = 14.19 / 0.082 x 1127

= 0.154 (arrpox.)


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