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)

160.3kJ/(mol.K)

Explanation:

Given parameters:

ΔG° = 130.5kJ/mol

ΔH° = 178.3kJ/mol

Temperature of reaction = 25°C to kelvin, 25+273 = 298K

Unknown:

ΔS  = ?

Solution:

Using the free energy change formula, we can express the standard entropy change as the subject of the formula:

               ΔG° = ΔH° - TΔS

                ΔS = -

where ΔS is the standard entropy change of the reaction

            ΔG° is the free energy change of the reaction

             ΔH° is the standard enthalpy change of the reaction

ΔS = - = 0.1603kJ/(mol.K) = 160.3kJ/(mol.K)

Answer

C.160.3J/mol

Explanation:

We are given that

G=

T=

We have to find the entropy of the reaction.

Substitute the values in the formula

 (1KJ=1000J)

Hence, the entropy of the reaction=S=160.3J/mol

law of conservation of matter