Question

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/molh = 178.3 kj/molwhat is the entropy of the reaction? use g = h � ts.160.3 j/(molk)47.8 j/(molk)160.3 j/(molk)1,912 j/(molk)

Answer 1

47.8 J/(mol·K)

Step-by-step explanation:

To calculate the entropy change (ΔS) for the decomposition of calcium carbonate, we can use the equation:

ΔG = ΔH - TΔS

where ΔG is the change in free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

Given:

ΔG = 130.5 kJ/mol

ΔH = 178.3 kJ/mol

T = 25.0°C = 298.15 K

We can rearrange the equation to solve for ΔS:

ΔS = (ΔH - ΔG)/T

Substituting the given values, we get:

ΔS = (178.3 - 130.5) kJ/mol / 298.15 K

ΔS = 47.8 J/(mol·K)

Therefore, the entropy change for the decomposition of calcium carbonate is 47.8 J/(mol·K).