At 900 degrees C, K p = 1.04 for the reaction CaCO 3 (s) --> <-- CaO (s) + CO 2

Q: At 900 degrees C, K p = 1.04 for the reaction CaCO 3 (s) -->

CaCO3 (s) CaO (s) + CO2 (g) Kp = pCO2 = 1.04 given system has pCO2 = 2.55 atm Qp = pCO2 initilal = 2.55 hence Qp > Kp which means some amount of CO2 gets reacted back with CaO to form CaCO3 so that we get less value of CO2 i.e until we get pCO2 = 1.04 Hence equilibrium shifts left side

ID: 1070699 • Letter: A

Question

At 900 degrees C, Kp = 1.04 for the reaction CaCO3 (s) --><-- CaO (s) + CO2 (g). At a low temperature, dry ice (solid CO2), calcium oxide, and calcium carbonate are introduced into a 50 L reaction chamber. The temperature is raised to 900 degrees C, resulting in the dry ice converting to gaseous CO2. For the following mixture, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at 900 degrees C?

655 g CaCO3, 95 g CaO, Pco2 = 2.55 atm

Explanation / Answer

CaCO3 (s) <---> CaO (s) + CO2 (g)

Kp = pCO2 = 1.04

given system has pCO2 = 2.55 atm

Qp = pCO2 initilal = 2.55

hence Qp > Kp which means some amount of CO2 gets reacted back with CaO to form CaCO3 so that we get less value of CO2 i.e until we get pCO2 = 1.04

Hence equilibrium shifts left side

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At 900 degrees C, K p = 1.04 for the reaction CaCO 3 (s) --> <-- CaO (s) + CO 2

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