 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/5/a/solution

Δ_{r} S^{\circ} = 282.15 - 217.94 - 202.79 = - 138.58 J K^{- 1} mo l^{- 1} .

Using

Δ_{r} G^{\circ} = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ}

Δ_{r} H^{\circ} = - 73.01 + 298 (- 0.13858) = - 114.30684 kJ mo l^{- 1} .

The formation enthalpy of elemental fluorine gas is zero in its standard state. Thus

Δ_{f} H^{\circ} (Cl F_{3}) = Δ_{r} H^{\circ} + Δ_{f} H^{\circ} (ClF) = - 164.60 kJ mo l^{- 1} .

The large negative entropy change is consistent with converting two gas molecules to one; it is already included in the Gibbs-energy relation.

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