Mr Cole Chemistry

Enthalpy of Solution Questions

Enthalpy of Solution & Hydration

Thermodynamics Worksheet

Definitions & Concepts

1. Define the following terms.
a) Enthalpy of Solution

The enthalpy change when 1 mole of a solid ionic substance fully dissolves in water to form an infinitely dilute solution.

b) Enthalpy of Hydration

The enthalpy change when one mole of gaseous ions fully dissolve in water to form an infinitely dilute solution.

2. Which of the following pairs have the highest (most exothermic) enthalpy of hydration? Explain your answer.
a) Na⁺ and Li⁺

Li+. Lithium ions are smaller than sodium ions but have the same charge. This means they have a larger charge density, resulting in stronger electrostatic attraction to the partially negative oxygen in water molecules.

b) Na⁺ and Mg²⁺

Mg2+. Magnesium ions have a greater charge (+2 vs +1) and a smaller ionic radius than sodium ions. This gives them a significantly higher charge density, leading to much stronger attraction to water molecules.

c) Cl⁻ and Br⁻

Cl–. Chloride ions have a smaller radius than bromide ions but the same charge. The higher charge density of Cl– results in stronger attraction to the partially positive hydrogen atoms in water molecules.

d) O²⁻ and F⁻

O2-. Oxide ions have a double negative charge compared to the single charge of fluoride, while having a similar radius. This higher charge density creates stronger electrostatic attraction to water molecules.

3. Explain why the enthalpy of hydration of H+ is significantly larger than other ions with a charge of +1.

H+ is simply a proton with an extremely small radius and incredibly high charge density. Unlike other ions which form ion-dipole attractions, H+ effectively forms a dative covalent bond with water to form the hydronium ion (H3O+), which releases a significant amount of energy.

Calculations

4. The following questions all require the use of the information in the tables below.
Enthalpy of Hydration / kJmol-1
Li+-520
I–-293
Na+-406
Mg2+-1926
Cl–-770
Enthalpy of Solution / kJmol-1
NaBr-0.6
MgCl2-343
BaCl2-20.6
Lattice Enthalpy (Formation) / kJmol-1
LiI-759
NaBr-742
BaCl2-2056
a) Draw a Born-Haber cycle and calculate the enthalpy of solution of LiI using the data in the table above.
Born Haber Cycle for LiI Solution
Working Out: Formula: ΔHsol = ΣΔHhyd – ΔHLE(formation) ΔHsol = [ΔHhyd(Li+) + ΔHhyd(I–)] – ΔHLE ΔHsol = [-520 + (-293)] – (-759) ΔHsol = -813 + 759 ΔHsol = -54 kJ mol-1
b) Draw a Born-Haber cycle and calculate the enthalpy of hydration of Bromide ions using the data in the table above.
Born Haber Cycle for Bromide Hydration
Working Out (for NaBr): ΔHsol = ΔHhyd(Na+) + ΔHhyd(Br–) – ΔHLE(formation) -0.6 = -406 + ΔHhyd(Br–) – (-742) -0.6 = -406 + 742 + ΔHhyd(Br–) -0.6 = 336 + ΔHhyd(Br–) ΔHhyd(Br–) = -0.6 – 336 ΔHhyd(Br–) = -336.6 kJ mol-1

Comment: This value fits the trend (between Cl– at -770 and I– at -293, though much closer to Iodide than Chloride, likely due to size effects).

c) Draw a Born-Haber cycle and calculate the lattice enthalpy of formation of MgCl₂ using the data in the table above.
Born Haber Cycle for MgCl2 Lattice Enthalpy
Working Out: ΔHsol = ΔHhyd(Mg2+) + 2ΔHhyd(Cl–) – ΔHLE(formation) -343 = [-1926 + 2(-770)] – ΔHLE -343 = [-1926 – 1540] – ΔHLE -343 = -3466 – ΔHLE ΔHLE = -3466 + 343 ΔHLE = -3123 kJ mol-1

Comment: This is larger (more exothermic) than BaCl2 (-2056 kJ mol-1), which is expected as Mg2+ is smaller than Ba2+, leading to stronger ionic bonding.

d) Draw a Born-Haber cycle and calculate the enthalpy of hydration of barium using the data in the table above.
Born Haber Cycle for Barium Hydration
Working Out (for BaCl2): ΔHsol = ΔHhyd(Ba2+) + 2ΔHhyd(Cl–) – ΔHLE(formation) -20.6 = ΔHhyd(Ba2+) + 2(-770) – (-2056) -20.6 = ΔHhyd(Ba2+) – 1540 + 2056 -20.6 = ΔHhyd(Ba2+) + 516 ΔHhyd(Ba2+) = -20.6 – 516 ΔHhyd(Ba2+) = -536.6 kJ mol-1

Comment: This is significantly less exothermic than Mg2+ (-1926 kJ mol-1), which is expected as the larger Ba2+ ion has a lower charge density and attracts water molecules less strongly.

e) 50.0g of NaCl was dissolved in 1.00dm³ of water. The temperature change that occurred was -2.00°C. Calculate the lattice enthalpy of formation of NaCl.
Calculation for NaCl
Step 1: Calculate Enthalpy of Solution Q = mcΔT m = 1000 g (mass of 1 dm3 water), c = 4.18, ΔT = 2.00 Q = 1000 × 4.18 × 2.00 = 8360 J = 8.36 kJ Moles NaCl = Mass / Mr = 50.0 / (23 + 35.5) = 50.0 / 58.5 = 0.8547 mol ΔHsol = +Q / mol (positive because temp decreased/endothermic) ΔHsol = +8.36 / 0.8547 = +9.78 kJ mol-1
Step 2: Calculate Lattice Enthalpy ΔHLE = ΣΔHhyd – ΔHsol ΔHLE = [ΔHhyd(Na+) + ΔHhyd(Cl–)] – ΔHsol ΔHLE = [-406 + (-770)] – (+9.78) ΔHLE = -1176 – 9.78 ΔHLE = -1185.8 kJ mol-1
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