Question

calculate the partial pressure of CO2 over calcium carbonate at 1000K

Answer 1

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

K_p = p_CO2 * [CaO]/[CaCO₃] = p_CO2 ....( it is conventional to take the concentration (activity) of pure solid phases                                                                              (e.g., CaCO3, CaO) to be equal to 1)

To do the calculations we need the thermodynamic values(dH, dG, dS etc) of the reactant and the products. You have not given the values. In that case we can use the standard table of thermodynamic values (at 25^oC) and calculate the K_P value at 25^oC and derive the same at 1000K using the following equation:

ln (K₂/K₁) = dH⁰ [T₂-T₁/T₁T₂].......(where dH⁰ is assumed to be constant over the temperature range)

From the standard thermodynamic table we get,

dH° for the reaction = 177.8 kJ/mol

dS° for the reaction = 160.5 J/Kmol

dG^o for the reaction = 130.431 kJ/mol

WE know, dG^o = -RT ln K_P

At 25^oC, 130.431*10³ J/mol = - (8.314 J/K.mol)(298K) ln K_P

K_P = 1.37*10^-23

Multiply this with standard P= 1 atm

P_CO2 at 298 K = 1.37*10^-23 atm

ln (K₂/K₁) = (dH⁰/R) [T₂-T₁/T₁T₂]

ln(K₂/1.37*10^-23) = [(177.8 *10³ J/mol)/(8.314J/mol/K)] [ 1000-298/1000*298]

K₂ = 0.1

P_CO2 at 1000 K = 0.1 atm