Question

For one mole of $NaC{l}_{(s)}$ the lattice enthalpy is

• A. $-788kJ/mol$
• B. $+878kJ/mol$
• C. $+788kJ/mol$
• D. $-878kJ/mol$

Answer 1

Answer: (A)

$\Delta /H°=S+D+J.E.+E.A.+U$
$-411.2=108.4+121+495.6-348.6+U$
$U=-787.6kJ/mol$
$=-788kJ/mol$