Question

How do use the following five values to calculate a lattice energy (in kilojoules per mol) for sodium hydride, ${\text{NaH}}$?
${{\text{E}}_{{\text{ea}}}}$ for ${\text{H}}$ $ = - 72.8{\text{ kJ/mol}}$
${{\text{E}}_{{\text{i1}}}}$ for ${\text{Na}}$ $ = + 495.8{\text{ kJ/mol}}$
Heat of sublimation for ${\text{Na}}$ $ = + 107.3{\text{ kJ/mol}}$
Bond dissociation energy for ${{\text{H}}_{\text{2}}}$ $ = 435.9{\text{ kJ/mol}}$
Net energy change for the formation of ${\text{NaH}}$ from its elements $ = - 60{\text{ kJ/mol}}$

Answer 1

Hint: To solve this we use the equation of the Born Haber cycle. The Born Haber cycle is related to the formation of an ionic compound by the reaction of a metal with a halogen or other non-metallic elements like oxygen. The metal which reacts is a metal that belongs to group 1 or group 2 of the periodic table.

Formula Used:
$\Delta {H_{{\text{formation}}}} = \Delta {H_{{\text{sublimation}}}} + \Delta {H_{{\text{ionisation}}}} + \Delta {H_{{\text{bond dissociation}}}} + \Delta {H_{{\text{electron gain}}}} + \Delta {H_{{\text{lattice}}}}$

Complete step-by-step answer:The reaction for the formation of ${\text{NaH}}$ is as follows:
\[{\text{Na}}\left( {\text{s}} \right) + \dfrac{1}{2}{{\text{H}}_2}\left( {\text{g}} \right) \to {\text{NaH}}\left( {\text{s}} \right){\text{ }}\Delta {{\text{H}}_{f\left( {{\text{overall}}} \right)}} = - 60{\text{ kJ/mol}}\]
We know that the amount of energy required to break a chemical bond is known as bond dissociation energy..
From the reaction, we can see that half mole of hydrogen gas reacts. Thus,
Bond dissociation energy $ = \dfrac{1}{2} \times 435.9{\text{ kJ/mol}}$
Bond dissociation energy $ = 217.95{\text{ kJ/mol}}$
We are given that the heat of sublimation for ${\text{Na}}$ is $ + 107.3{\text{ kJ/mol}}$. Electrons gain energy for ${\text{H}}$ is $ - 72.8{\text{ kJ/mol}}$. Ionisation energy for ${\text{Na}}$ is $ + 495.8{\text{ kJ/mol}}$.
Using the Born Haber cycle,
$\Delta {H_{{\text{formation}}}} = \Delta {H_{{\text{sublimation}}}} + \Delta {H_{{\text{ionisation}}}} + \Delta {H_{{\text{bond dissociation}}}} + \Delta {H_{{\text{electron gain}}}} + \Delta {H_{{\text{lattice}}}}$
Rearrange the equation for the lattice enthalpy as follows:
$\Delta {H_{{\text{lattice}}}} = \Delta {H_{{\text{formation}}}} - \Delta {H_{{\text{sublimation}}}} - \Delta {H_{{\text{ionisation}}}} - \Delta {H_{{\text{bond dissociation}}}} - \Delta {H_{{\text{electron gain}}}}$
We are given that the energy of formation is $ - 60{\text{ kJ/mol}}$. Heat of sublimation for ${\text{Na}}$ is $ + 107.3{\text{ kJ/mol}}$
. Ionisation energy for ${\text{Na}}$ is $ + 495.8{\text{ kJ/mol}}$. Bond dissociation energy is $217.95{\text{ kJ/mol}}$. Electrons gain energy for ${\text{H}}$ is $ - 72.8{\text{ kJ/mol}}$.
Thus,
$\Delta {H_{{\text{lattice}}}} = \left( { - 60{\text{ kJ/mol}}} \right) - \left( { + 107.3{\text{ kJ/mol}}} \right) - \left( { + 495.8{\text{ kJ/mol}}} \right) - \left( {217.95{\text{ kJ/mol}}} \right) - \left( { - 72.8{\text{ kJ/mol}}} \right)$
$\Delta {H_{{\text{lattice}}}} = - 808.25{\text{ kJ/mol}}$
$\Delta {H_{{\text{lattice}}}} = - 2810{\text{ kJ mo}}{{\text{l}}^{ - 1}}$
Thus, the lattice energy for sodium hydride, ${\text{NaH}}$ is $ - 808.25{\text{ kJ/mol}}$.

Note: The energy required to convert one mole of an ionic solid to its gaseous ionic constituents is known as the lattice enthalpy. The lattice enthalpy measures the strength of the forces between the ions in an ionic solid. Higher the lattice enthalpy, stronger are the forces.