What is the enthalpy of formation of calcium sulfate from the reaction of calcium carbonate and sulfuric acid?

The reaction between calcium carbonate (CaCO₃) and sulfuric acid (H₂SO₄) can be summarized by the following chemical equation:

CaCO₃(s) + H₂SO₄(aq) → CaSO₄(s) + H₂O(l) + CO₂(g)

To determine the enthalpy of formation (ΔH_f) of calcium sulfate (CaSO₄), we will apply Hess’s law and the given values. We know the standard enthalpy of formation (ΔH_f) of calcium carbonate is 1206 kJ/mol, and the reaction enthalpy (ΔH_rxn) is given as -49.8 kJ/mol.

The enthalpy change for the reaction can be expressed as:

ΔH_rxn = ΔH_f(CaSO₄) + ΔH_f(H₂O) + ΔH_f(CO₂) – ΔH_f(CaCO₃) – ΔH_f(H₂SO₄)

We will need the standard enthalpies of formation for water (H₂O) and carbon dioxide (CO₂):

• ΔH_f(H₂O) = -285.8 kJ/mol
• ΔH_f(CO₂) = -393.5 kJ/mol

For sulfuric acid (H₂SO₄), the standard enthalpy of formation is approximately -814 kJ/mol. Plugging all these values into our equation, we can solve for ΔH_f(CaSO₄):

-49.8 = ΔH_f(CaSO₄) + (-285.8) + (-393.5) – 1206 – (-814)

This simplifies to:

-49.8 = ΔH_f(CaSO₄) – 285.8 – 393.5 – 1206 + 814

Combining constants:

-49.8 = ΔH_f(CaSO₄) – 1071.3

Finally, solving for ΔH_f(CaSO₄):

ΔH_f(CaSO₄) = -49.8 + 1071.3 = 1021.5 kJ/mol

Thus, the enthalpy of formation of calcium sulfate is approximately 1021.5 kJ/mol.