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Question #44243

Difficulty: 8/10 | 🟢 Free

The Question

A 1.0 molal1.0\text{ molal} aqueous solution of a binary electrolyte \ce{AB} is prepared at 300 K300\text{ K}. The freezing point of this solution is −3.0∘C-3.0^\circ\text{C}. (Given: KfK_f of water =2.0 K kg mol−1= 2.0\text{ K kg mol}^{-1}, assume the Van't Hoff factor ii is independent of temperature).

For the dissolution of \ce{AB(s)}, the standard enthalpy of solution at infinite dilution is ΔHsol∞=+15.0 kJ mol−1\Delta H_{sol}^{\infty} = +15.0\text{ kJ mol}^{-1}. At 1.0 m1.0\text{ m} concentration, ion-pairing and non-ideal interactions contribute an excess enthalpy term ΔHexcess=+4.5 kJ mol−1\Delta H_{excess} = +4.5\text{ kJ mol}^{-1}, such that the actual enthalpy of solution is ΔHsol=ΔHsol∞+ΔHexcess\Delta H_{sol} = \Delta H_{sol}^{\infty} + \Delta H_{excess}.

The entropy of dissolution for a hypothetical, fully dissociated ideal solution (i=2i=2) is +80 J K−1 mol−1+80\text{ J K}^{-1}\text{ mol}^{-1}. Given that the actual entropy of dissolution scales linearly with the Van't Hoff factor (ii), determine the Gibbs free energy of dissolution (ΔGsol\Delta G_{sol}) for 1 mole1\text{ mole} of \ce{AB} at 300 K300\text{ K}.

Options

  • A.

    +1.5 kJ mol−1+1.5\text{ kJ mol}^{-1}

  • B.

    −3.0 kJ mol−1-3.0\text{ kJ mol}^{-1}

  • C.

    −4.5 kJ mol−1-4.5\text{ kJ mol}^{-1}

  • D.

    −9.0 kJ mol−1-9.0\text{ kJ mol}^{-1}

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Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Spontaneity Criteria|Van’t Hoff Factor (i) Application|Degree of Dissociation/Association

Why This Question Is Tricky

Common Mistakes

The problem tests the intersection of colligative properties and solution thermodynamics in non-ideal regimes.

Students often memorize that dissolution of salts is 'entropy-driven' and assume ideal behavior (ΔHsol=ΔHsol∞\Delta H_{sol} = \Delta H_{sol}^{\infty} and i=2i=2).

The trap lies in failing to account for the reduced entropy due to ion-pairing (i<2i < 2) and the positive excess enthalpy (ΔHexcess\Delta H_{excess}) arising from solute-solute interactions at higher concentrations.

Ignoring either leads to incorrect, overly negative (spontaneous) ΔG\Delta G values.

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Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptSpontaneity Criteria
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree