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Transcript
Chapter 4 The Study of Chemical Reactions 1 4.1 Introduction If we want to understand a reaction, we must also know the mechanism, the step-by-step pathway from reactants to products. • To know how well the reaction goes to products, we study its thermodynamics, the energetics of the reaction at equilibrium. • To use a reaction in a realistic time period (and to keep the reaction from becoming violent), we study its kinetics, the variation of reaction rates with different conditions and concentrations of reagents Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 2 4.2 Chlorination of Methane 4.3 The Free-Radical Chain Reaction The chain reaction consists of three kinds of steps: 1. The initiation step, which generates a reactive intermediate. 2. Propagation steps, in which the reactive intermediate reacts with a stable molecule to form a product and another reactive intermediate 3. Termination steps, side reactions that destroy reactive intermediates and tend to slow or stop the reaction. Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 3 Example: Chlorination of methane (substitution reaction) 4.31 The Initiation Step: Generation of Radicals The unpaired electron is called the odd electron or the radical electron. Species with unpaired electrons are called radicals or free radicals. Radicals are electron-deficient because they lack an octet Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 4 Examples of free radicals 4-32 Propagation Steps Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 5 Problem 1: Draw Lewis structures for the following free radicals. (a)The ethyl radical, (b)The tert-butyl radical, (c) The isopropyl radical (2-propyl radical) (d) The iodine atom Problem 2: Write the propagation steps leading to the formation of dichloromethane CH2Cl2 from chloromethane. Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 6 4.33 Termination Reactions Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 7 4.4 Equilibrium Constants and Free Energy Thermodynamics is the branch of chemistry that deals with the energy changes accompanying chemical and physical transformations. These energy changes are most useful for describing the properties of systems at equilibrium. The equilibrium concentrations of reactants and products are governed by the equilibrium constant of the reaction. Example : The chlorination of methane 8 Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition From the value of Keq we can calculate the change in free energy (sometimes called Gibbs free energy) The relationship between Keq and is given by where R = 8.314 J kelvin-mol (1.987 cal kelvin-mol), the gas constant T = absolute temperature, in kelvins. e = 2.718 the base of natural logarithms Notice: Absolute temperatures (in kelvins) are correctly given without a degree sign, as in the equation 25 °C = 298 K Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 9 Problem : Calculate the value of f or the chlorination of methane Solution : Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 10 4.5 Enthalpy and Entropy Two factors contribute to the change in free energy: the change in enthalpy and the change in entropy multiplied by the temperature. 4.51 Enthalpy The change in enthalpy is the heat of reaction Exothermic (negative value of ) Endothermic (positive value of ) 4.52 Entropy Is often described as randomness, disorder, or freedom of motion. Reactions tend to favour products with the greatest entropy Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 11 Notice: The negative sign in the entropy term of the free-energy expression. The positive value of the entropy change indicating that the products have more freedom of motion than the reactants, makes a favorable (negative) contribution to Problem : Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 12 4.6 Bond-Dissociation Enthalpies The bond-dissociation enthalpy (BDE, also called bond-dissociation energy) is the amount of enthalpy required to break a particular bond homolytically—that is, in such a way that each bonded atom retains one of the bond’s two electrons. In contrast, when a bond is broken heterolytically, one of the atoms retains both electrons. Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 13 4.7 Enthalpy Changes in Chlorination Problem : a) Using bond-dissociation enthalpies from Table 4-2 (page 143), calculate the heat of reaction for each step in the free-radical bromination of methane. b) Calculate the overall heat of reaction Reference: Organic Chemistry", L.G. Wade, Printice Hall, 8th Edition 14