Chemistry Equilibrium

At 900degrees C, Kp = 1.04 for the reaction
CaCO3 (s) --> CaO(s) + CO2(g)
At a low temperature, dryi ice (solid CO2), calcium oxide and calcium carbonate are introduced into a 50L reaction chamber. The temperature is raised to 900dgrees C, resulting in the dry ice converting to gaseous CO2. For the following mixtures, determine whether the system is at equilibrium. If the system is not at equilibrium determine the amount of each species when equilibrium is established
a) 655g CaCO3, 95.0g CaO, P(CO2) = 2.55atm
b) 780g CaCO3, 31.00 g CaO, P(CO2)= 0.104

Begin: The expression for Kp does not include solids, so it is simply.....Kp = Pco2 = 1.04.

Because solids are involved in the questions, converting Kp to Kc would help fro questions (a) and (b):

Kp = Kc (RT)^∆n where n = (moles of gas products – moles of gas reactants) and T = Kelvin

1.04 = Kc (0.0821)(1173 K)^1

Kc = 0.0108

This can now be interpreted that Kc = [CO2]…. and at equilibrium, [CO2] = 0.0108 M

1a) Now, in the information for (a), since the P(CO2) = 2.55, we can either compare this value to the Kp = 1.04 and see that it is not at equilibrium, or I can interpret the Kp = 2.55 as 2.55 atm, then find moles from PV=nRT and then find Molarity:

2.55 atm x 50 L = n x 0.0821 x 1173 K

n = 1.32 moles….molarity = 1.32 moles/50L = 0.0264 M….. which is greater than the 0.0108 M that we got earlier when we converted the Kp to Kc.

Either way, the Kc or Kp is smaller than the given atm or Molarity of CO2. It is not at equilibrium and will need to shift to the left.

2a) OK, now if the reaction shifts to the left, this will affect the amount of grams of CaO and CaCO3 present. We can use an ICE chart to determine the amounts of everything at equilibrium. (The molarity of CO2 will be changed to moles because the amount of solids we are going to calculate have to be moles, not molarity)

note: 0.0264 M CO2 in 50 L = 1.32 moles CO2