Question

The ionization enthalpy of ${\mathrm{Na}}^{+}$formation from $\mathrm{Na}\left(\mathrm{g}\right)$ is $495.8\mathrm{kJ}{\mathrm{mol}}^{-1}$, while the electron gain enthalpy of $\mathrm{Br}$ is$-325.0\mathrm{kJ}{\mathrm{mol}}^{-1}$. Given the lattice enthalpy of $\mathrm{NaBr}$ is$-728.4\mathrm{kJ}{\mathrm{mol}}^{-1}$. The energy for the formation of $\mathrm{NaBr}$ ionic solid is$(-)$_______ $\times {10}^{-1}\mathrm{kJ}{\mathrm{mol}}^{-1}$

Answer 1

The energy for the formation of $\mathrm{NaBr}$ an ionic solid is $5576\mathrm{kJ}$

• Explanation: Cations and anions are held together by electrostatic forces to form ionic solids. Ionic solids are hard, brittle, and have high melting points due to the strength of these interactions.
• A covalent bond is formed when two atoms share electrons.
• Given data is$\mathrm{Na}\left(\mathrm{s}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{g}\right)∆\mathrm{H}=495.8$; $\mathrm{Br}\left(\mathrm{g}\right)\to {\mathrm{Br}}^{-}\left(\mathrm{g}\right)$=$-325.0\mathrm{kJ}{\mathrm{mol}}^{-1}$; $\mathrm{NaBr}$=$-728.4\mathrm{kJ}{\mathrm{mol}}^{-1}$

$∆{\mathrm{H}}_{\mathrm{f}}\left(\mathrm{NaBr}\right)=∆{\mathrm{H}}_{\mathrm{I}.\mathrm{E}}\left(\mathrm{Na}\right)+∆{\mathrm{H}}_{\mathrm{E}.\mathrm{G}.\mathrm{E}}\left(\mathrm{Br}\right)+∆{\mathrm{H}}_{\mathrm{L}.\mathrm{E}}$

• $∆\mathrm{H}=495.8-325-728.4$
• $-557.6\mathrm{kJ}=-5576\times {10}^{-1}\mathrm{kJ}$=$5576\mathrm{kJ}$

The energy for the formation of $\mathrm{NaBr}$ an ionic solid is$5576\mathrm{kJ}$