Topic 15 – Enthalpy of solution and hydration www.msjchem.com Enthalpy of solution and hydration Ɵ The enthalpy change of solution (ΔH sol) is the enthalpy change when one mole 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 MSJChem - Video tutorials for IB chemistry www.msjchem.com Topic 15 – Enthalpy of solution and hydration www.msjchem.com Answers: 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). MSJChem - Video tutorials for IB chemistry www.msjchem.com .