The dissociation of calcium carbonate has an equilibrium constant K p = 1.16 at 800 C. CaCO3(s)

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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?

Related Book For answer-question

General Chemistry

ISBN: 978-1439043998

9th edition

Authors: Darrell Ebbing, Steven D. Gammon

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Posted Date: Apr 15, 2020 08:39 AM

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The dissociation of calcium carbonate has an equilibrium constant K p = 1.16 at 800 C. CaCO3(s)

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General Chemistry

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