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NCERT Exemplar · Q60

Q.The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCl(s).

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Lattice enthalpy cannot be measured directly because we cannot start with gaseous NaCl(s) and break it into gaseous ions. Instead, we use the Born–Haber cycle, a thermochemical loop that combines measurable enthalpy changes (sublimation, ionization, dissociation, electron affinity, and formation) to calculate lattice enthalpy indirectly via Hess's law.

Why we need an indirect method

Lattice enthalpy is defined as the energy required for the process:

NaCl(s)→Na+(g)+Cl−(g)\text{NaCl}(s) \to \text{Na}^+(g) + \text{Cl}^-(g)

We cannot perform this reaction in a laboratory. Solid NaCl does not simply vaporize into separated ions when heated—it melts, then vaporizes as ion pairs, and eventually dissociates at extremely high temperatures where other decomposition pathways interfere. No calorimeter can isolate this single step cleanly.

The solution lies in recognizing that enthalpy is a state function: the total enthalpy change around any closed cycle is zero. If we construct a cycle connecting the same initial and final states through steps we can measure, we can solve for the one unknown step.

The Born–Haber cycle for NaCl

The cycle connects the formation of solid NaCl from its elements to the formation of gaseous ions through two different routes.

Route 1 (direct): Formation of NaCl(s) from elements in their standard states:

Na(s)+12Cl2(g)→NaCl(s)ΔHf∘\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \to \text{NaCl}(s) \qquad \Delta H_f^\circ

This is the standard enthalpy of formation, measured directly by calorimetry.

Route 2 (stepwise through gaseous ions): Break this into measurable steps:

  1. Sublimation of sodium:

Na(s)→Na(g)ΔHsub\text{Na}(s) \to \text{Na}(g) \qquad \Delta H_\text{sub}

Energy needed to convert solid sodium metal to gaseous atoms.

  1. Ionization of sodium:

Na(g)→Na+(g)+e−ΔHIE\text{Na}(g) \to \text{Na}^+(g) + e^- \qquad \Delta H_\text{IE}

First ionization energy of sodium.

  1. Dissociation of chlorine:

12Cl2(g)→Cl(g)12ΔHdiss\frac{1}{2}\text{Cl}_2(g) \to \text{Cl}(g) \qquad \frac{1}{2}\Delta H_\text{diss}

Half the bond dissociation energy of Cl2\text{Cl}_2.

  1. Electron affinity of chlorine:

Cl(g)+e−→Cl−(g)ΔHEA\text{Cl}(g) + e^- \to \text{Cl}^-(g) \qquad \Delta H_\text{EA}

Energy released when chlorine gains an electron (usually negative).

  1. Lattice formation (reverse of lattice enthalpy):

Na+(g)+Cl−(g)→NaCl(s)−ΔHlattice\text{Na}^+(g) + \text{Cl}^-(g) \to \text{NaCl}(s) \qquad -\Delta H_\text{lattice}

This is the negative of what we want to find.

Applying Hess's law

Since enthalpy is a state function, the sum of enthalpies around the cycle equals zero:

ΔHf∘=ΔHsub+ΔHIE+12ΔHdiss+ΔHEA−ΔHlattice\Delta H_f^\circ = \Delta H_\text{sub} + \Delta H_\text{IE} + \frac{1}{2}\Delta H_\text{diss} + \Delta H_\text{EA} - \Delta H_\text{lattice}

Rearranging for lattice enthalpy:

ΔHlattice=ΔHsub+ΔHIE+12ΔHdiss+ΔHEA−ΔHf∘\Delta H_\text{lattice} = \Delta H_\text{sub} + \Delta H_\text{IE} + \frac{1}{2}\Delta H_\text{diss} + \Delta H_\text{EA} - \Delta H_f^\circ

Every term on the right can be measured experimentally:

  • ΔHsub\Delta H_\text{sub}: heat solid sodium and measure energy absorbed.
  • ΔHIE\Delta H_\text{IE}: spectroscopic measurement of ionization energy.
  • ΔHdiss\Delta H_\text{diss}: calorimetry or spectroscopy on Cl2\text{Cl}_2 molecules. …

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