tailieunhanh - Lecture Organic chemistry - Chapter 2: Alkanes, thermodynamics, and kinetics

Chapter 2 - Alkanes, thermodynamics, and kinetics. In this chapter, students will be able to understand: All reactions are equilibria, chemical thermodynamics and kinetics, equilibria, equilibria and free energy, potential energy diagrams, rate determining transition state,.and other contents. | Combustion How warm, how fast? Petroleum!! Chapter 2: Alkanes, Thermodynamics, And Kinetics All Reactions Are Equilibria “Barrier” kcal/mol Exothermicity CH4 + O2 CO2 + 2H2O What governs these equilibria? ~20 high -213 kcal mol-1 Equilibrium lies very much to the right. kcal mol-1 CH3Cl + Na+ -OH CH3OH + Na+ Cl- or Chemical Thermodynamics: Energy changes during reaction, extent of “completion of equilibration,” “to the left/right,” “driving force.” 2. Chemical Kinetics: How fast is equilibrium established; rates of disappearance of starting materials or appearance of products Chemical Thermodynamics And Kinetics The two principles may or may not go in tandem [ ] = concentration in mol L-1 Equilibria Two typical cases [A] [reactants] [B] [products] K = equilibrium constant K = [C][D] [A][B] If K large: reaction “complete,” “to the right,” “downhill.” How do we quantify? Gibbs free energy, ∆G° A + B C + D A B K K = K = 1. 2. Gibbs Free Energy ∆G° The “° ” refers to molecules in . | Combustion How warm, how fast? Petroleum!! Chapter 2: Alkanes, Thermodynamics, And Kinetics All Reactions Are Equilibria “Barrier” kcal/mol Exothermicity CH4 + O2 CO2 + 2H2O What governs these equilibria? ~20 high -213 kcal mol-1 Equilibrium lies very much to the right. kcal mol-1 CH3Cl + Na+ -OH CH3OH + Na+ Cl- or Chemical Thermodynamics: Energy changes during reaction, extent of “completion of equilibration,” “to the left/right,” “driving force.” 2. Chemical Kinetics: How fast is equilibrium established; rates of disappearance of starting materials or appearance of products Chemical Thermodynamics And Kinetics The two principles may or may not go in tandem [ ] = concentration in mol L-1 Equilibria Two typical cases [A] [reactants] [B] [products] K = equilibrium constant K = [C][D] [A][B] If K large: reaction “complete,” “to the right,” “downhill.” How do we quantify? Gibbs free energy, ∆G° A + B C + D A B K K = K = 1. 2. Gibbs Free Energy ∆G° The “° ” refers to molecules in their standard states, 1 atm, 25 °C (298 kelvin), 1 M (concentration). ∆G° = -RT lnK = logK T in kelvin (0 kelvin = -273 °C) R = gas constant ~ 2cal deg-1 mol-1 Large K : Large negative ∆G° : downhill Josiah Willard Gibbs (1839–1903) When K = 1 (A/B = 50/50), then ΔGº = 0 kcal mol-1 K = 10, then ΔGº = kcal mol-1 K = 100, then ΔGº = kcal mol-1 K = , then ΔGº = + kcal mol-1 At 25ºC (298 kelvin): ΔGº = logK Equilibria And Free Energy A B K Therefore, for ΔGº = kcal mol-1 CH3Cl + Na+ -OH CH3OH + Na+ Cl- What is order of magnitude of K roughly? ~ 17, . ~1017 huge! Enthalpy ∆H° And Entropy ∆S° ∆G° = ∆H° - T ∆S° kcal mol-1 Enthalpy ∆H° = heat of the reaction, arising mainly from changes in combined bond strengths: ∆H° = (sum of strength of bonds broken) – (sum of strengths of bonds made) cal kelvin-1 mol-1 or entropy units (.) temperature in kelvin 7/29/2015 © Univesity of California 7 ∆H° negative: called “exothermic” positive: called .

• Hoá học
• Organic chemistry
• Lecture Organic chemistry
• Hóa học hữu cơ
• Bài giảng Hóa học hữu cơ
• Chemical thermodynamics
• Potential energy diagrams