Chemistry Thermodynamic Math Question 5

Question

Compute the temperature at which calcite is in equilibrium with CaO(s) and CO₂(g) at 0.8 bar.

SubstanceΔfH0	kJ/molS0	J/mol K
CaCO3(s) (calcite)	-1207.6	91.7
CaO(s)	-634.9	38.1
CO2(g)	-393.5	213.8

Answer 1

Calcite is CaCO₃ so the reaction of interest would be CaCO₃(s) <==> CaO(s) + CO₂(g)

For the reaction to be at equilibrium, ∆G_rxn should be equal to 0. So using ∆G = ∆H-T∆S = 0 we solve for T

But before we can do that, we need to find the ∆H and ∆S for the reaction, and that is ∑products - ∑reactants

∆H_rxn = (-634.9 kJ/mol + -303.5 kJ/mol) - (-1207.6 kJ/mol) = 269.2 kJ/mole

∆S_rxn = (0.0381 kJ/mol k + 0.2138 kJ/mol K) - 0.0917 kJ/mol K) = 0.1602 kJ/mol K

∆G = ∆H - T∆S = 0

269.2 - T(0.1602) = 0

T = 1680 K