The dissociation of calcium carbonate has an equilibrium constant K p = 1.16 at 800 C. CaCO3(s)
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Question:
The dissociation of calcium carbonate has an equilibrium constant K p = 1.16 at 800 °C.
CaCO3(s) ↔ CaO(s) + CO2(g)
a. What is Kc for the reaction?
b. If you place 22.5 g of CaCO3 in a 9.56 L container at 800°C, what is the pressure of CO2 in the container?
c. What percentage of the original 22.5 g sample of CaCO3 remains undecomposed at equilibrium?
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