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

G = 130.5 kJ/mol

H = 178.3 kJ/mol

What is the entropy of the reaction? Use G = H - TS.

A. - 160.3 J / (mol. K)

B. - 47.8 J / (mol. K)

C. 160.3 J / (mol. K)

D. 1,912 J / (mol. K)

The decomposition of calcium carbonate is shown below:

CaCO3 (s) - - > CaO (s) + CO2 (g)

at 25 degrees Celcius:

G = 130.5 kJ/mol

H = 178.3 kJ/mol

We need to find the Entropy of the reaction using the formula:

G = H - TS

where G = Gibbs Free Energy

H = Enthalpy

T = Temperature

S = Entropy

130.5 kJ/mol = 178.3 kJ/mol - (298 K*S)

S = 0.1604 kJ/mol K