Laws of Thermodynamics - Chemical Principles - Lecture Slides, Slides of Chemical Principles

Download Laws of Thermodynamics - Chemical Principles - Lecture Slides and more Slides Chemical Principles in PDF only on Docsity! The Second and Third Laws of Thermodynamics • Outline – Definition of the Second Law – Determining S – Definition of the Third Law docsity.com The Second Law • The Second Law: In any spontaneous process, there is always an increase in the entropy of the universe. • From our definitions of system and surroundings: Suniverse = Ssystem + Ssurroundings docsity.com Example • For the following reaction at 298 K: Sb4O6(s) + 6C(s) 4Sb(s) + 6CO2(g) H = 778 kJ What is Ssurr? Ssurr = -H/T = -778 kJ/298K = -2.6 kJ/K docsity.com The Third Law • Recall, in determining enthalpies we had standard state values to use. Does the same thing exist for entropy? • The third law: The entropy of a perfect crystal at 0K is zero. • The third law provides the reference state for use in calculating absolute entropies. docsity.com What is a Perfect Crystal? Perfect crystal at 0 K Crystal deforms at T > 0 K docsity.com Big Example • Is the following reaction spontaneous at 298 K? 2Fe(s) + 3H2O(g) Fe2O3(s) + 3H2(g) S°rxn = S°system = -141.5 J/K S°surr = -H°sys/T = -H°rxn/T (Is S°univ > 0?) H°rxn= H°f(Fe2O3(s)) + 3H°f(H2(g)) - 2H°f(Fe (s)) - 3 H°f(H2O(g)) docsity.com Big Example (cont.) S°surr = -H°sys/T = 348 J/K H°rxn= -100 kJ S°univ = S°sys + S°surr = -141.5 J/K + 348 J/K = 207.5 J/K S°univ > 0 ; therefore, reaction is spontaneous docsity.com Entropy and Phase Changes • Phase Change: Reaction in which a substance goes from one phase of state to another. • Example: H2O(l) H2O(g) @ 373 K • Phase changes are equilibrium processes such that: Suniv = 0 docsity.com