Question:
The ionization enthalpy of Na$^+$ formation is 495.8, electron gain enthalpy of Br is -325.0, and lattice enthalpy of NaBr is -728.4 kJ mol$^{-1}$. The energy for the formation of NaBr ionic solid is (-) __________ $\times 10^{-1}$ kJ mol$^{-1}$.
The ionization enthalpy of Na$^+$ formation is 495.8, electron gain enthalpy of Br is -325.0, and lattice enthalpy of NaBr is -728.4 kJ mol$^{-1}$. The energy for the formation of NaBr ionic solid is (-) __________ $\times 10^{-1}$ kJ mol$^{-1}$.
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This is a simplified Born-Haber cycle calculation. The lattice enthalpy released is the dominant factor that makes the formation of ionic solids exothermic.
Updated On: Feb 3, 2026
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Correct Answer: 5576
Solution and Explanation
Step 1: Total Energy ($\Delta H_f$) = $\text{Ionization Energy} + \text{Electron Gain Enthalpy} + \text{Lattice Enthalpy}$.
Step 2: $\Delta H_f = 495.8 + (-325.0) + (-728.4)$.
Step 3: $\Delta H_f = 495.8 - 1053.4 = -557.6$ kJ/mol.
Step 4: To express in form $(-) x \times 10^{-1}$, we get $5576 \times 10^{-1}$.
Step 2: $\Delta H_f = 495.8 + (-325.0) + (-728.4)$.
Step 3: $\Delta H_f = 495.8 - 1053.4 = -557.6$ kJ/mol.
Step 4: To express in form $(-) x \times 10^{-1}$, we get $5576 \times 10^{-1}$.
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