The standard heat of formation, in kcalmol of Ba2+ is[Given standard heat of formation of 2 SO4− ion(aq)= –216 kcalmol, standard heat of crystallizationof BaSO4(s) = –4.5 kcalmol, standard heat offormation of BaSO4(s) = –349 kcalmol]

Answer

Application of Hess’s Law to Calculate Enthalpy of Formation

Consider the precipitation reaction:

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

This reaction forms solid barium sulfate from its aqueous ions. The associated enthalpy change is the enthalpy of crystallization:

ΔH° = ΔH°_crys. = −4.5 kcal mol⁻¹

Using Hess’s Law

We want to find the standard enthalpy of formation for Ba²⁺(aq):

ΔH°_f[BaSO₄(s)] = ΔH°_f[Ba²⁺(aq)] + ΔH°_f[SO₄²⁻(aq)] + ΔH°_crys.

Substituting the known values:

• ΔH°_f[BaSO₄(s)] = −349 kcal mol⁻¹
• ΔH°_f[SO₄²⁻(aq)] = −216 kcal mol⁻¹
• ΔH°_crys. = −4.5 kcal mol⁻¹

−349 = ΔH°_f[Ba²⁺(aq)] − 220.5

⇒ ΔH°_f[Ba²⁺(aq)] = −128.5 kcal mol⁻¹

Conclusion

The standard enthalpy of formation of Ba²⁺(aq) is:

ΔH°_f[Ba²⁺(aq)] = −128.5 kcal mol⁻¹

This example illustrates how Hess’s Law is applied to calculate unknown thermodynamic values using known enthalpy data.