\[\Delta_fH^\circ(\mathrm{MX}) = \Delta_\text{sub}H(\mathrm M) + I(\mathrm M) + \tfrac12D(\mathrm X_2) - E_\text{ea}(\mathrm X) + U_L\]
含义Meaning
用盖斯定律把离子晶体的生成焓分解为升华、电离、解离、电子亲和、晶格形成各步，从而由实验数据求出无法直接测量的晶格能（1919）。
A Hess’s-law cycle splitting the formation enthalpy of an ionic solid into sublimation, ionisation, dissociation, electron attachment and lattice formation, giving the otherwise unmeasurable lattice energy (1919).
符号说明Symbols
| 符号Symbol | 含义 Meaning | 单位 Unit |
|---|---|---|
| \(\Delta_\text{sub}H\) | 金属升华焓sublimation enthalpy of the metal | kJ·mol⁻¹ |
| \(I\) | 电离能ionisation energy | kJ·mol⁻¹ |
| \(D\) | X₂ 键解离能bond energy of X₂ | kJ·mol⁻¹ |
| \(E_\text{ea}\) | 电子亲和能（放出为正）electron affinity (positive if released) | kJ·mol⁻¹ |
| \(U_L\) | 晶格焓（形成晶体，负值）lattice enthalpy of formation (negative) | kJ·mol⁻¹ |
推导Derivation
路径 1：M(s) + ½X₂(g) → MX(s)，焓变 ΔfH°。
Route 1: direct formation, ΔfH°.
路径 2：M(s)→M(g)→M⁺(g)；½X₂→X(g)→X⁻(g)；M⁺ + X⁻ → MX(s)。
Route 2: gaseous ions then lattice.
两路径焓变相等，解出 \(U_L\)。
Equate the two and solve for \(U_L\).
适用条件与局限Assumptions & limitations
各步数据误差会累积到 U_L；与玻恩–朗德计算值比较可判断键的共价性。
Errors accumulate in U_L; comparison with Born–Landé reveals covalency.
例题Worked example
NaCl：−411 = 107 + 496 + 122 − 349 + U_L → U_L = −787 kJ/mol。
NaCl: −411 = 107 + 496 + 122 − 349 + U_L → U_L = −787 kJ/mol.