Skip to content
Example · Example 9

Q.The lattice enthalpy of NaCl(s)\text{NaCl}(s) is +788 kJ mol−1+788\ \text{kJ mol}^{-1} and the hydration enthalpies of Na+(g)\text{Na}^+(g) and Cl−(g)\text{Cl}^-(g) are −406 kJ mol−1-406\ \text{kJ mol}^{-1} and −378 kJ mol−1-378\ \text{kJ mol}^{-1} respectively. Calculate the enthalpy of solution of NaCl\text{NaCl} in water.

West Bengal WbchseTextbookSubjectiveImportance★★★★★est
30% · 9/30 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For an ionic solid dissolving in water, ΔsolH∘=ΔlatticeH∘+ΔhydH∘(cation)+ΔhydH∘(anion)\Delta_{sol}H^\circ = \Delta_{lattice}H^\circ + \Delta_{hyd}H^\circ(\text{cation}) + \Delta_{hyd}H^\circ(\text{anion}), where the lattice enthalpy here is taken as the (positive) energy needed to separate the solid into gaseous ions, and the hydration enthalpies are the (negative) energies released when each gaseous ion is surrounded and stabilized by water molecules. Substituting the given values: ΔsolH∘=(+788)+(−406)+(−378)=788−784=+4 kJ mol−1\Delta_{sol}H^\circ = (+788) + (-406) + (-378) = 788 - 784 = +4\ \text{kJ mol}^{-1}. The small positive (endothermic) value shows that the energy released hydrating the ions very nearly, but not quite, compensates for the energy required to break apart the ionic la …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.