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The reaction is

2NaCl + CaF2 --> 2NaF + CaCl2

and the enthalpies of formation (kJ/Mol) of the solids at STP are

2NaCl = (-411)*2

CaF2 = -1220

Total = -2042

2NaF = (-469)*2

CaCl2 = -796

Total = -1734

The calculations show that the reactants are more stable than the products by some 307kJ/mol, so that the reaction would not proceed. (For more advanced readers, the value for the Gibbs free energy of formation will not be very much different from this, as the entropy terms will be relatively small in comparison with the enthalpy terms.) A major factor in this is the relatively high lattice enthalpy of CaF2.

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