Question

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)

Answer 1

1.33 × 10⁻²³

Step-by-step 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