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PLEASE PLEASE, I AM CURRENTLY DOING THE TEST AND I NEED HELP.


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)

Respuesta :

dbeardown013
The answer is B because you subtract H-G = TS and ts = -47.8 J/(mol.K)

Hope this helps for future problems