The enthalpy of sublimation has commonly been predicted using the equipartition theorem. If the lattice energy is assumed to be approximately half the packing energy, then the following thermodynamic corrections can be applied to predict the enthalpy of sublimation. Assuming a 1 molar ideal gas gives a correction for the thermodynamic environment (pressure and volume) in which pV = RT, hence a correction of 1RT. Additional corrections for the vibrations, rotations and translation then need to be applied. From the equipartition theorem gaseous rotation and translation contribute 1.5RT each to the final state, therefore a +3RT correction. Crystalline vibrations and rotations contribute 3RT each to the initial state, hence −6RT. Summing the RT corrections; −6RT + 3RT + RT = −2RT. This leads to the following approximate sublimation enthalpy. A similar approximation can be found for the entropy term if rigid bodies are assumed.