JEE Main · 2020 · Shift-IIeasyTHERMO-028

Lattice enthalpy and enthalpy of solution of NaCl are 788 and 4\,kJ mol-1 respectively. The hydration enthalpy of NaCl…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Lattice enthalpy and enthalpy of solution of NaCl are 788788 and 4 kJ mol−14\,\text{kJ mol}^{-1} respectively. The hydration enthalpy of NaCl is:

Options
  1. a

    −780 kJ mol−1-780\,\text{kJ mol}^{-1}

  2. b

    780 kJ mol−1780\,\text{kJ mol}^{-1}

  3. c

    −784 kJ mol−1-784\,\text{kJ mol}^{-1}

    ✓
  4. d

    784 kJ mol−1784\,\text{kJ mol}^{-1}

Correct Answerc

−784 kJ mol−1-784\,\text{kJ mol}^{-1}

Detailed Solution

🧠 For dissolution: ΔHsol=ΔHlattice+ΔHhydration\Delta H_{\text{sol}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}} When an ionic solid dissolves, it first breaks apart (costs lattice energy) then the ions get hydrated (releases hydration energy).

🗺️ Calculation ΔHsol=ΔHlattice (dissociation)+ΔHhydration\Delta H_{\text{sol}} = \Delta H_{\text{lattice (dissociation)}} + \Delta H_{\text{hydration}} +4=+788+ΔHhydration+4 = +788 + \Delta H_{\text{hydration}} ΔHhydration=4−788=−784 kJ mol−1\Delta H_{\text{hydration}} = 4 - 788 = -784\text{ kJ mol}^{-1}

Negative sign confirms hydration is exothermic (energy is released as water molecules stabilise the ions).

⚠️ Trap: Make sure you use the dissociation form of lattice enthalpy (+788), not the formation form (−788). If you use the wrong sign, you get +784 instead of −784.

Answer: (c)\boxed{\text{Answer: (c)}}

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