At 823 °C, Kp = 490 for the equilibrium reaction CoO(s) + CO(g) ⟷ Co(s) + CO2(g) What is the value of Kc at the same temperature for the equilibrium below? Co(s) + CO2(g) ⟷ CoO(s) + CO(g)

490

Explanation:

823°C = (823 + 273)K = 1096K

Kp = Kc (RT)∇ⁿ

∇ⁿ = change in mole in product - change in mole in reactant.

Equation of reaction;

Co(s) + CO2(g) ⟷ CoO(s) + CO(g)

∇ⁿ = 2 - 2 = 0

Kp = Kc * (RT)∇ⁿ

Kp = Kc when ∇ⁿ = 0

Since anything or number raised to power of 0 = 1

Kp = 490 = Kc

Kc = 490.

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