Bài giảng Chapter 17 Free Energy and Thermodynamics

Chemistry, Julia Burdge, 2st Ed McGraw Hill Chapter 17 Free Energy and Thermodynamics Mr Truong Minh Chien ; losedtales@yahoo.com http://tailieu.vn/losedtales http://mba-programming.blogspot.com 2011, NKMB Co., Ltd Don’t get ahead of the game • There is a lot of theory in this chapter • Keep terms separate (5), and equations Universe, System, Surrounds, Enthalpy, and Entropy are all different, and you must know each • Memorize the Laws of Thermodynamics • You can break a California State Law but you can NOT break a Thermodynamic Law Chemistry, Julia Burdge, 2nd e., McGraw Hill First Law of Thermodynamics • you can’t win! • First Law of Thermodynamics: Energy cannot be Created or Destroyed the total energy of the universe cannot change though you can transfer it from one place to another • ∆Euniverse = = ∆Esystem + ∆Εsurroundings • Think of a match burning, crumbling paper, etc Chemistry, Julia Burdge, 2nd e., McGraw Hill First Law of Thermodynamics • Conservation of Energy • For an exothermic reaction, “lost” heat from the system • • • goes into the surroundings two ways energy “lost” from a system, converted to heat, q used to work, w Energy conservation requires that the energy change in the system equal the heat released + work done  ∆E = q + w  ∆E = ∆H + P∆V ∆E is a state function internal energy change independent of how done Chemistry, Julia Burdge, 2nd e., McGraw Hill Energy Tax • you can’t break even! • to recharge a battery with 100 kJ of • useful energy will require more than 100 kJ every energy transition results in a “loss” of energy  conversion of energy to heat which is “lost” by heating up the surroundings ∆E = q + w ∆E = ∆H + P∆V Tro, Chemistry: A Molecular Approach Heat Tax fewer steps generally results in a lower total heat tax Tro, Chemistry: A Molecular Approach Thermodynamics and Spontaneity • thermodynamics predicts whether a process will proceed under the given conditions spontaneous process nonspontaneous processes require energy input to go • spontaneity is determined by comparing the free energy of the system before the reaction with the free energy of the system after reaction if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable • spontaneity ≠ fast or slow Tro, Chemistry: A Molecular Approach Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end Tro, Chemistry: A Molecular Approach Reversibility of Process • any spontaneous process is irreversible  it will proceed in only one direction • a reversible process will proceed back and forth between the two end conditions  equilibrium  results in no change in free energy • if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction Tro, Chemistry: A Molecular Approach Thermodynamics vs Kinetics Tro, Chemistry: A Molecular Approach 10 Relative Standard Entropies Allotropes • the less constrained the structure of an allotrope is, the larger its entropy Tro, Chemistry: A Molecular Approach 41 Relative Standard Entropies Molecular Complexity • larger, more complex molecules generally have larger entropy • more available energy states, allowing more dispersal of energy through the states Tro, Chemistry: A Molecular Approach Molar S°, Substance Mass (J/mol∙K) Ar (g) 39.948 154.8 NO (g) 30.006 210.8 42 Relative Standard Entropies Dissolution • dissolved solids generally have larger entropy • distributing particles throughout the mixture Tro, Chemistry: A Molecular Approach Substance S°, (J/mol∙K) KClO3(s) 143.1 KClO3(aq) 265.7 43 Substance NH3(g) S°, J/mol⋅K 192.8 O2(g) 205.2 NO(g) 210.8 H2O(g) standard entropies from Appendix IIB 188.8 Ex 17.4 –Calculate ∆S° for the reaction NH3(g) + O2(g) → NO(g) + H2O(g) Given: Find: ∆S, J/K Concept Plan: Relationships: Solution: S°NH3, S°O2, S°NO, S°H2O, ( ( ) ( ∆S = Σn pSproducts − Σnr Sreactants ) ( ∆S = Σn pSproducts − ΣnrSreactants ) ) ∆S = [ 4(S NO( g ) ) + 6(SH O( g ) )] − [4(S NH ( g ) ) + 5(SO ( g ) )] J J J J = [ 4(210.8 K ) + 6(188.8 K )] − [4(192.8 K ) + 5(205.2 K )] J = 178.8 K Check: ∆S is +, as you would expect for a reaction with more gas product molecules than reactant molecules Calculating ∆G° • at 25°C: ∆Goreaction = ΣnGof(products) - ΣnGof(reactants) • at temperatures other than 25°C: assuming the change in ∆Horeaction and ∆Soreaction is negligible ∆G° reaction = ∆H° reaction – T∆S° reaction Tro, Chemistry: A Molecular Approach 45 ... determined by comparing the free energy of the system before the reaction with the free energy of the system after reaction if the system after reaction has less free energy than before the reaction,... Energy Tax • you can’t break even! • to recharge a battery with 100 kJ of • useful energy will require more than 100 kJ every energy transition results in a “loss” of energy  conversion of energy. .. theory in this chapter • Keep terms separate (5), and equations Universe, System, Surrounds, Enthalpy, and Entropy are all different, and you must know each • Memorize the Laws of Thermodynamics