At 1000°C and a pressure of 16 atm, the equilibrium constant of the reaction:
CO₂(g) + C(s) ⇌ 2CO(g)
is such that for every nine moles of CO, there is one mole of CO₂. For what pressure of the m

Question

At 1000°C and a pressure of 16 atm, the equilibrium constant of the reaction:
CO₂(g) + C(s) ⇌ 2CO(g)
is such that for every nine moles of CO, there is one mole of CO₂. For what pressure of the mixture, is the ratio CO:CO₂ = 4:1? The temperature remains 1000°C.

Prashant Kotian · Accepted Answer

Correct answer: A. At P=16 atm, CO:CO₂ = 9:1, so P_CO = 14.4 atm, P_CO₂ = 1.6 atm. K_p = (14.4)² / 1.6 = 129.6. At the new condition, CO:CO₂ = 4:1. Let P_CO₂ = x, P_CO = 4x. K_p = (4x)² / x = 16x = 129.6 => x = 8.1 atm. Total new pressure = 5x = 5(8.1) = 40.5 atm. Therefore, correct answer is A.

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