Question

The decomposition of calcium carbonate, CaCO3(s) Right arrow. CaO(s) + CO2(g), has the following values for free energy and enthalpy at 25.0°C.

Delta.G = 130.5 kJ/mol
Delta.H = 178.3 kJ/mol

What is the entropy of the reaction? Use Delta.G = Delta.H – TDelta.S.
mc022-7.jpg160.3 J/(molmc022-8.jpgK)
mc022-9.jpg47.8 J/(molmc022-10.jpgK)
160.3 J/(molmc022-11.jpgK)
1,912 J/(molmc022-12.jpgK)

Answer 1

ΔS = (ΔH − ΔG)/T = (178.3 − 130.5) kJ·mol⁻¹ / 298.15 K
= 47.8 kJ·mol⁻¹ / 298.15 K = 0.1605 kJ·mol⁻¹·K⁻¹ = 160.5 J·mol⁻¹·K⁻¹

So ≈ 160.3–160.5 J/(mol·K).