COMEDK · Chemistry · 4. Chemical Bonding and Molecular Structure
For one mole of \(\mathrm{NaCl}(s)\) the lattice enthalpy is

- A \(-788 \mathrm{~kJ} / \mathrm{mol}\)
- B \(+878 \mathrm{~kJ} / \mathrm{mol}\)
- C \(+788 \mathrm{~kJ} / \mathrm{mol}\)
- D \(-878 \mathrm{~kJ} / \mathrm{mol}\)
Answer & Solution
Correct Answer
(A) \(-788 \mathrm{~kJ} / \mathrm{mol}\)
Step-by-step Solution
Detailed explanation
According to Hess law, \[ \begin{gathered} \Delta_{f} H^{\circ}=S+D+I \mathrm{E}+\mathrm{E} A+U \\ -411.2=108.4+121+495.6-348.6+U \\ U=-787.6 \mathrm{~kJ} / \mathrm{mol}=-788 \mathrm{~kJ} / \mathrm{mul} \end{gathered} \]
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