ThermochemistryHard

Question

The following are various ΔH values (kJ per mol), ΔfH (NaCl) = – 411.2; ΔfH (Na, g) = 107.3; ΔfH (Cl, g) = 121.7; ΔiH (Na, g) = 495.4; ΔegH (Cl, g) = – 348.5. The lattice enthalpy of NaCl(s) is

Options

A.495.4 kJ
B.107.3 kJ
C.411.2 kJ
D.787.1 kJ

Solution

$- 411.2 = 107.3 + 121.7 + 495.4 - 348.5 - \left( \text{L.E.} \right)$

$\therefore\text{L.E. = 787.1 kJ}$

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