Geological conditions are sometimes so extreme that quantities neglected in normal laboratory experi — Thermodynamics and Thermochemistry Chemistry Question
Question
Geological conditions are sometimes so extreme that quantities neglected in normal laboratory experiments take on an overriding importance. For example, consider the formation of diamond under geophysically typical conditions. The density of graphite is 2.4 g/cm3 and that of diamond is 3.6 g/cm3 at a certain temperature and 500 kbar. By how much does δ U_trans differs from δ H_trans for the graphite to diamond transition?
💡 Solution & Explanation
For C(graphite) -> C(diamond): δ H = δ U + P * δ V => δ U - δ H = -P * δ V. Molar mass of Carbon = 12 g/mol. Molar volume of graphite: V_g = 12 / 2.4 = 5 cm3/mol. Molar volume of diamond: V_d = 12 / 3.6 = 3.33 cm3/mol. δ V = V_d - V_g = -1.67 cm3/mol = -1.67 * 10^-6 m3/mol. P = 500 kbar = 5 * 10^10 Pa. Thus: P * δ V = 5 * 10^10 Pa * (-1.67 * 10^-6 m3/mol) = -83.33 kJ/mol. Thus, δ U and δ H differ by 83.33 kJ/mol.