Question

The Romans used calcium oxide, CaO, to produce a strong mortar to build stone structures. Calcium oxide was mixed with water to give calcium hydroxide, which reacted slowly with carbon dioxide in the air to give calcium carbonate. Ca(OH)2(s) + CO2 → CaCO3(s) + H2O(g) Calculate the standard enthalpy change for this reaction. kJ/mol-rxn What is the change in enthalpy if 3.50 kg of calcium hydroxide reacts with a stoichiometric amount of carbon dioxide? kJ

Answer 1

The standard enthalpy change for the reaction of calcium hydroxide with carbon dioxide to form calcium carbonate and water is -177.8 kJ/mol-rxn. If 3.50 kg of calcium hydroxide reacts, the total enthalpy change is approximately -8395.094 kJ.

Step-by-step explanation:

To calculate the standard enthalpy change for the reaction of calcium hydroxide forming calcium carbonate and water from carbon dioxide, we use the reaction Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(g). The reverse reaction is CaCO3(s) → CaO(s) + CO2(g) with ΔH = 177.8 kJ/mol. Since reversing the reaction changes the sign of ΔH, the enthalpy change for the forward reaction is -177.8 kJ/mol-rxn.

For 3.50 kg of calcium hydroxide, first convert the mass to moles (using the molar mass of Ca(OH)2 which is approximately 74.09 g/mol). 3.50 kg is 3500 g, so the number of moles is 3500 g ÷ 74.09 g/mol ≈ 47.23 mol. To find the total change in enthalpy, multiply the moles by the enthalpy change per mole: 47.23 mol × (-177.8 kJ/mol) ≈ -8395.094 kJ.

Answer 2

ΔH°rxn = -68.7 kJ/mol

ΔH = -3249.5 kJ

Step-by-step explanation:

The standard enthalpy change of the reaction (ΔH°rxn) can be calculated by the standard enthalpy of formation of the compounds (H°f), which can be found in thermodynamics tables.

H°f, Ca(OH)₂(s) = -986.6 kJ/mol

H°f, CO₂(g) = -393.5 kJ/mol

H°f, CaCO₃(s) =-1207.0 kJ/mol

H°f, H₂O(g) = -241.8 kJ/mol

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

ΔH°rxn = (-241.8 - 1207.0) - (-393.5 - 986.6)

ΔH°rxn = -68.7 kJ/mol

The molar mass of calcium hydroxide is 74 g/mol (Ca = 40 g/mol, O = 16 g/mol, and H = 1 g/mol), so the number of moles, is the mass (3500g) divided by the molar mass:

n = 3500/74 = 47.3 moles

The change in enthalpy is the standard enthalpy of the reaction multiplied by the number of moles of the limiting reactant. Because stoichiometry is 1:1, and they have the same amount of moles, there's no reactant in excess:

ΔH = -68.7 * 47.3

ΔH = -3249.5 kJ