 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/10/b/i/solution

Use the balanced aqueous-acid reaction

Ln (s) + 3 H^{+} (aq) ⟶ L n^{3 +} (aq) + \frac{3}{2} H_{2} (g) .

A Hess cycle first atomizes the metal, then ionizes its gas-phase atoms, then hydrates the triply charged gas ions:

Ln (s) Ln (g) L n^{3 +} (g) ⟶ Ln (g) ⟶ L n^{3 +} (g) + 3 e^{-} (g) ⟶ L n^{3 +} (aq) Δ_{at} H^{\circ} > 0, I_{1} + I_{2} + I_{3} > 0, Δ_{hyd} H^{\circ} < 0.

Close the reference leg by combining the three electrons with

3 H^{+} (aq)

to make

\frac{3}{2} H_{2} (g)

. Calling that element-independent contribution

C_{H}

gives

Δ H_{acid}^{\circ} = Δ_{at} H^{\circ} + I_{1} + I_{2} + I_{3} + Δ_{hyd} H^{\circ} + C_{H} .

A single-ion standard enthalpy of formation uses the conventional hydrogen-ion reference; it must not be treated as a balanced creation of net charge from neutral metal without an electron/reference term. The constant hydrogen leg does not affect comparisons along the lanthanide series.

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