CHEM 1062 Lecture Notes - Lecture 26: Joule, Thermodynamics
Document Summary
Chem 1062: lecture #26 standard states and the 3 rd law of thermodynamics. Liquid or solid: most stable form often at 1 atm, but varies depending on chemical composition. For a perfect crystal at 0ok, the enthalpy is 0. Only one way to achieve perfect order. A(0k, perfect crystal) a (standard state) S = sa o, sa: standard state , is standard entropy value for a. Can calculate go with so and ho to determine spontaneity: - sa, are both values found in the table that also contains ho for compounds. 2so2 (g) + o2 (g) 2so3 (g) 248 j 205 j 257 j: = (-396 kj/mole * 2 mole) - (-297 kj/mole * 2 mole) - (0, = -198 kj. So = (257 j/mole*k * 2 mole) - (205 j/mole*k * 1 mole) - (248 j/mole*k * 2 mole) So = -0. 187 kj at 25oc (298 k)
