Question

Consider the following reaction:CaCO3(s)→CaO(s)+CO2(g).Estimate ΔG∘ for this reaction at each of the following temperatures. (Assume that ΔH∘ and ΔS∘ do not change too much within the given temperature range.)A. 285 KΔG∘ =B.1025 KΔG∘=C.1475 KΔG∘ =

Answer 1

Answer:

a) ΔG° = 132.5 kJ

b) ΔG° = 13.68 kJ

c) ΔG° = -58.58 kJ

Explanation:

The values of the enthalpy of formation(H°f) and the entropy (S°) for the substances in the reaction are:

CaCO₃(s): H°f = -1206.9 kJ/mol   S° = 92.9 J/K.mol

CaO(s): H°f = -635.09 kJ/mol  S° = 39.75 J/K.mol

CO₂(g): H°f = -393.51 kJ/mol   S° = 213.74 J/K.mol

ΔH° = ∑n*H°f,products - ∑n*H°f,reactants (where n is the number of moles)

ΔH° = (-393.51 + (-635.09)) - (-1206.9) = 178.3 kJ

ΔS° = ∑n*S°,products - ∑n*S°,reactants

ΔS° = (213.74 + 39.75) - 92.9 = 160.59 J/K = 0.1606 kJ/K

ΔG° = ΔH° - TΔS°

a) ΔG° = 178.3 - 285*0.1606 = 132.5 kJ

b) ΔG° = 178.3 - 1025*0.1606 = 13.68 kJ

c) ΔG° = 178.3 - 1475*0.1606 = -58.58 kJ

Answer 2

The values for ΔG° at the given temperatures are 132.5 kJ, 13.68 kJ and -58.58 kJ respectively.

What is the relationship between change standard free energy ΔG°, enthalpy, ΔH°, entropy, ΔS°, and temperature, T?

The change in standard free energy ΔG°, enthalpy, ΔH°, entropy, ΔS°, and temperature, T are related by the formula:

• ΔG° = ΔH° - TΔS°

Equation of reaction and reaction data information

The equation of reaction is given as:

• CaCO₃(s) ----> CaO(s) + CO₂(g)

From the reaction data, the values of the enthalpy of formation(H°f) and the entropy (S°) for the substances in the reaction are as follows:

• CaCO₃(s): H°f = -1206.9 kJ/mol;   S° = 92.9 J/K.mol
• CaO(s): H°f = -635.09 kJ/mol;  S° = 39.75 J/K.mol
• CO₂(g): H°f = -393.51 kJ/mol;   S° = 213.74 J/K.mol

Calculating ΔH° and ΔS° of the reaction

ΔH° = ∑n * H°f of products - ∑n * H°f of reactants

where n is the number of moles

n = 1 for both products and reactants

ΔH° = (-393.51 + (-635.09)) - (-1206.9)

ΔH° = 178.3 kJ

ΔS° = ∑n*S°,products - ∑n*S°,reactants

ΔS° = (213.74 + 39.75) - 92.9 = 160.59 J/K

ΔS° = 0.1606 kJ/K

Calculating ΔG° for the reaction at the various temperatures

Using ΔG° = ΔH° - TΔS°

And assuming that  ΔH° and ΔS° do not change much:

a) At 285 K

ΔG° = 178.3 - 285 * 0.1606

ΔG° = 132.5 kJ

b) At 1025 K

ΔG° = 178.3 - 1025 * 0.1606

ΔG° = 13.68 kJ

c) At 1475 K

ΔG° = 178.3 - 1475 * 0.1606

ΔG° = -58.58 kJ

Therefore, the values for ΔG° at the given temperatures are 132.5 kJ, 13.68 kJ and -58.58 kJ respectively.

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