[sparkly]

On this one, since the Kc is very low logically the reactant is going to be very much favored.  I changed the grams of each compound to moles then divided by ten because its a 10 liter vessel. Then I plugged those concentrations into the Keq formula.  But I don't know what to do with the answers I get. How do they compare? Of course I have no idea if that was what I was supposed to do....


#9 At 900oC, Kc = 0.0108 for the reaction:

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

A mixture of CaCO3, CaO, and CO2 is placed in a 10.0 Liter vessel at 900oC. For the following mixtures, will the amount of CaCO3 increase, decrease, or remain the same as the system approaches equilibrium?

(a) 15.0 grams CaCO3, 15.0 grams CaO, and 4.25 grams CO2;

(b) 2.50 grams CaCO3, 25.0 grams CaO, and 5.66 grams CO2;

(c) 30.5 grams CaCO3, 25.5 grams CaO, and 6.48 grams CO2.

[Vidya]

Now take out Q (Quotient Of concentrations) for each part .
Q has the same formula as Kc .Kc is at equilibrium but Q is at any concentrations.
Now you need to compare  Q with Kc
If Q>KC  It means products are more than reactants so it will shift to left.
Similarly try to interpret if
Q< KC and Q=KC