Consider the information given below
ΔH⁰/ kJmol-1
Enthalpy of lattice formation for CaF2 -2611
Enthalpy of hydration for Ca2+ ions -1650
Enthalpy of hydration for F- ions -506
Using the information given above, calculate the enthalpy of solution for CaF2
Reverse eqn 1(change sign of dH), add to eqn 2 and 2 x eqn 3.
To calculate the enthalpy of solution for CaF2, you need to consider the enthalpy of lattice formation and the enthalpies of hydration for both Ca2+ ions and F- ions.
The enthalpy of solution (ΔHsol) is given by the equation:
ΔHsol = ΔHlattice + ΔHhydration
Where:
ΔHsol is the enthalpy of solution
ΔHlattice is the enthalpy of lattice formation
ΔHhydration is the sum of the enthalpies of hydration for the ions involved
Given that:
ΔHlattice = -2611 kJ/mol
ΔHhydration (Ca2+) = -1650 kJ/mol
ΔHhydration (F-) = -506 kJ/mol
Substituting these values into the equation:
ΔHsol = -2611 + (-1650) + (-506)
ΔHsol = -2611 - 1650 - 506
ΔHsol = -4767 kJ/mol
Therefore, the enthalpy of solution for CaF2 is -4767 kJ/mol.