equilibrium and pressure problem
« on: April 01, 2008, 08:29:23 PM »

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Once again, I came across an equilibrium problem and I don’t know where to start…can someone guide me through this one?

At 900 C, Kp= 1.04 for the reaction:

CaCO3(s) (—->equilibrium sign<—-) CaO(s) + CO2 (g)

At a low temperature, dry ice (solid CO2), calcium oxide and calcium carbonate are introduced into a 50.0 L reaction chamber. The temperature is raised to 900 Celcius, resulting in the dry ice converting to gaseous CO2. For the mixture below, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at 900 celcius?

655 g CaCO3 95.0 g CaO PCO2= 2.55 atm

this is how I think I should start it

covert the CaCO3 and the CaO into moles
divide it by 50.0 L (to get molar concentration)

and that’s all I know..I’m still not sure what we’re looking for. It ofcourse has to do with the initial amount of calcium oxide..but what does the amount represent? is it concentration? Kp? Please help!

Re: equilibrium and pressure problem
« Reply #1 on: April 01, 2008, 10:22:02 PM »

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Actually, it’s much simpler than that.

Kp = pCO2 (the other product and reactant do not appear in the equilibrium expression because they are solids).

At equilibrium at 900 C, Kp has a value of 1.04 , which means the pCO2 at equilibrium must be equal to 1.04 atm.

However, the actual pCO2 is given as 2.55 atm, which is much larger than it should be if the system were at equilibrium at 900C.

Therefore, the system will shift to the left, increasing the amount of solid CaCO3, and reducing the amount of CaO and CO2. Assuming that the temperature remains constant this process will continue until the pCO2 falls to 1.04 atm, at which point the system will be at equilibrium.

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