[kandie_kane]

The question is:
Consider the equilibrium reaction
CaC03 (s) --> CaO (s) + CO2 (g)
Write the approprate expression for the thermodynamic constant, assuming that the gaseous compounds obey the ideal-gas equation of state.  Evaluate Kp at 298.15K.

I have tried this problem and have done the following

deltaG = RT ln Kp
-394.395 Kj/mol /(0.0083145 kj/mol k X 298.15 K ) = ln Kp
ln Kp= -159.08
kp =

I am unsure if I am supposed to leave out the solids in this equation or if they should be considered.  Also I unsure about the thermodynamic constant is and how to write an expression for it.

[Albert]

deltaG CO2 (g) = -394.4 kJ/mol
deltaG CaO (s) = -604.2 kJ/mol
deltaG CaCO3 (s) = -1128.8 kJ/mol

deltaG reaction = + 130.2 kJ/mol

deltaG reaction / (RT) = ln Keq
130.2 kJ/mol / (8.314E-3 kJ/K*mol * 298.15) = 52.525

Keq = 6.476 E+22