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If the enthalpy of sublimation of Li is $155 \text{ kJ mol}^{-1}$, enthalpy of dissociation of $\text{F}_2$ is $150 \text{ kJ mol}^{-1}$, ionization enthalpy of Li is $520 \text{ kJ mol}^{-1}$, electron gain enthalpy of F is $-313 \text{ kJ mol}^{-1}$, standard enthalpy of formation of LiF is $-594 \text{ kJ mol}^{-1}$. The magnitude of lattice enthalpy of LiF is _________ $\text{kJ mol}^{-1}$. (Nearest Integer)

Born-Haber Cycle for LiF Formation

The Born-Haber cycle relates the standard enthalpy of formation ($\Delta H_f^\circ$) of an ionic compound to various thermodynamic quantities, including lattice enthalpy ($\Delta H_{\text{lattice}}^\circ$). The cycle allows us to calculate the lattice enthalpy using the following relationship:

$ \Delta H_f^\circ(\text{LiF}) = \Delta H_{\text{sub}}^\circ(\text{Li}) + \Delta H_{IE}^\circ(\text{Li}) + \frac{1}{2} \Delta H_{\text{diss}}^\circ(\text{F}_2) + \Delta H_{\text{EGE}}^\circ(\text{F}) + \Delta H_{\text{lattice}}^\circ(\text{LiF}) $

Calculating Lattice Enthalpy

We need to find the lattice enthalpy ($\Delta H_{\text{lattice}}^\circ(\text{LiF})$). Rearranging the Born-Haber equation:

$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = \Delta H_f^\circ(\text{LiF}) - \Delta H_{\text{sub}}^\circ(\text{Li}) - \Delta H_{IE}^\circ(\text{Li}) - \frac{1}{2} \Delta H_{\text{diss}}^\circ(\text{F}_2) - \Delta H_{\text{EGE}}^\circ(\text{F}) $

Substituting Given Values

Substitute the provided enthalpy values into the rearranged equation:

  • Standard enthalpy of formation of LiF ($\Delta H_f^\circ(\text{LiF})$) = $-594 \text{ kJ mol}^{-1}$
  • Enthalpy of sublimation of Li ($\Delta H_{\text{sub}}^\circ(\text{Li})$) = $155 \text{ kJ mol}^{-1}$
  • Ionization enthalpy of Li ($\Delta H_{IE}^\circ(\text{Li})$) = $520 \text{ kJ mol}^{-1}$
  • Enthalpy of dissociation of F₂ ($\Delta H_{\text{diss}}^\circ(\text{F}_2)$) = $150 \text{ kJ mol}^{-1}$, so $\frac{1}{2} \Delta H_{\text{diss}}^\circ(\text{F}_2) = 75 \text{ kJ mol}^{-1}$
  • Electron gain enthalpy of F ($\Delta H_{\text{EGE}}^\circ(\text{F})$) = $-313 \text{ kJ mol}^{-1}$

Calculation:

$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = (-594) - (155) - (520) - (75) - (-313) $

$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = -594 - 155 - 520 - 75 + 313 $

$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = -1344 + 313 $

$ \Delta H_{\text{lattice}}^\circ(\text{LiF}) = -1031 \text{ kJ mol}^{-1} $

Magnitude of Lattice Enthalpy

The question asks for the magnitude of the lattice enthalpy.

Magnitude = $ |-1031 \text{ kJ mol}^{-1}| = 1031 \text{ kJ mol}^{-1} $

The magnitude of the lattice enthalpy of LiF is 1031 kJ mol⁻¹.

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    In the light of the above statements, choose the correct answer from the options given below :
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