At 675 K, H2(g) and CO2(g) react to form CO(g) and H2O(g) with Kp = 0.16 for H2(g) + CO2(g) CO(g) + H2O(g). If 0.25 mole H2 and 0.25 mole CO2 are heated together at 675 K, what is the mole percent of CO(g) in the equilibrium mixture?

Question

At 675 K, H2(g) and CO2(g) react to form CO(g) and H2O(g) with Kp = 0.16 for H2(g) + CO2(g) <=> CO(g) + H2O(g). If 0.25 mole H2 and 0.25 mole CO2 are heated together at 675 K, what is the mole percent of CO(g) in the equilibrium mixture?

Accepted Answer

14.28. Taking x as moles reacted, Kp = x²/(0.25-x)², so x/(0.25-x) = sqrt(0.16) = 0.4, giving x = 1/14 mole. Mole % CO = (x/0.5)*100 = (1/14)/0.5 *100 = 14.28%.

At 675 K, H2(g) and CO2(g) react to form CO(g) and H2O(g) with Kp = 0.16 for H2(g) + CO2(g) <=> CO(g) + H2O(g). If 0.25 mole H2 and 0.25 mole CO2 are heated together at 675 K, what is the mole percent of CO(g) in the equilibrium mixture?

Solution

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