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Topic 15 – of and hydration www.msjchem.com

Enthalpy of solution and hydration

Ɵ The enthalpy change of solution (ΔH sol) is the enthalpy change when one of an ionic substance dissolves in water to give a solution of infinite dilution.

Example: NaCl(s)  NaCl(aq)

Ɵ The enthalpy of hydration (ΔH hyd) is the enthalpy change when one mole of gaseous ions dissolve in water to give a solution of infinite dilution.

+ + Example: Na (g)  Na (aq)

A solution of infinite dilution is one where there is a sufficiently large excess of water that adding any more does not cause any further heat to be absorbed or evolved.

Ɵ 1) Construct an enthalpy cycle and calculate the enthalpy change of solution (ΔH sol) for calcium fluoride (CaF2).

Ɵ 2+ -1 ΔH hyd Ca (kJmol ) -1616 Ɵ - -1 ΔH hyd F (kJmol ) -504 Ɵ -1 ΔH lat (CaF2) (kJmol ) +2651

Ɵ 2) Construct an enthalpy cycle and calculate the lattice enthalpy (ΔH lat) for NaOH

Ɵ + -1 ΔH hyd Na (kJmol ) -424 Ɵ - -1 ΔH hyd OH (kJmol ) -519 Ɵ -1 ΔH sol (NaOH) (kJmol ) -44.5

Ɵ + + 3) Explain why the enthalpy change of hydration (ΔH hyd ) of Na is greater than K

Ɵ 2+ + 4) Explain why the enthalpy change of hydration (ΔH hyd ) of Mg is greater than Li

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Topic 15 – Enthalpy of solution and hydration www.msjchem.com

1) Enthalpy cycle for CaF2

Ɵ ΔH sol CaF2(s) CaF2(aq)

Ɵ Ɵ ΔH lat ΔH hyd

2+ - Ca (g) + 2F (g)

Ɵ Ɵ Ɵ ΔH sol = ΔH lat + ΔH hyd

Ɵ ΔH sol = +2651 + (-1616 + 2 x -504)

Ɵ -1 ΔH sol = +27 kJmol

2) Enthalpy cycle for NaOH

Ɵ ΔH sol NaOH(s) NaOH(aq

) Ɵ Ɵ ΔH lat ΔH hyd

+ - Na (g) + OH (g)

Ɵ Ɵ Ɵ ΔH lat = ΔH sol - ΔH hyd

Ɵ ΔH lat (NaOH) = -44.5 – (-424 + -519)

Ɵ -1 ΔH lat (NaOH) = +898 kJmol

+ + Ɵ 3) Na has a smaller ionic radius than K , therefore a greater ΔH hyd (stronger electrostatic + attraction between the Na ion and water molecule).

2+ Ɵ 4) Mg has a higher charge on the ion, therefore a greater ΔH hyd (same reason as above).

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