The thermodynamic values from part A will be useful as you work through part B:

ΔH∘rxn 178.5kJ/mol

ΔS∘rxn 161.0J / (mol⋅K)

Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C:

CaCO3 (s) →CaO (s) + CO2 (g)

1.33 * 10⁻²³

Explanation:

Let's consider the following equation.

CaCO₃ (s) → CaO (s) + CO₂ (g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = ΔH°rxn - T * ΔS°rxn

where,

ΔH°rxn: standard enthalpy of the reaction T: absolute temperature (25 °C + 273.15 = 298 K) ΔS°rxn: standard entropy of the reaction

ΔG°rxn = 178.5 kJ/mol - 298 K * 0.1610 kJ/mol. K

ΔG°rxn = 130.5 kJ/mol

Then, we can calculate the equilibrium constant (K) using the following expression.

ΔG°rxn = - R * T * ln K

where,

R: ideal gas constant

ΔG°rxn = - R * T * ln K

ln K = - ΔG°rxn / R * T

ln K = - (130.5 * 10³ J/mol) / (8.314 J/mol. K) * 298 K

K = 1.33 * 10⁻²³