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at 1127 k and 1 atm pressure a gaseous mixture of...

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 K_c 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, n_co = 90.55/28 = 3.234 mol

Number of moles of CO₂, n_co2 = 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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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 K_c for this reaction at the above temperature.

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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.

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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 K_c for this reaction at the above temperature.