Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)

Estimate ΔGo for this reaction at each temperature and predict whether or not the reaction is spontaneous. (Assume that DH and DS do not change too much within the given temperature range.)

Standard Thermodynamic Quantities for Selected Substances at 25 ∘C

Substance ΔH∘f(kJ/mol) S∘(J/mol⋅K)
CaCO3(s) −1207.6 91.7
CaO(s) −634.9 38.1
CO2(g) −393.5 213.8

Express your answer using one decimal place. ΔG∘ =

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

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the child is heterozygous

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