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UK CHE 230 - Equilibria

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1 EQUILIBRIA 8.1. Definitions. By convention, we write and read chemical equations left to right. Thus, starting materials are written to the left of reaction products. Let’s look for example at a reaction coordinate diagram for a SN2 reaction (see, Jones Figure 8.1). A reaction coordinate diagram describes energy changes during the course of a reaction. Thus for a SN2 reaction, we see that starting materials and products lie at energy minima and are separated an energy barrier, the maximum of which is called the transition state. If a transition state is too high in energy, it will not be accessible under normal conditions and no reaction will occur. If there are several accessible transition states of comparable energy then a mixture of products will be formed and specificity (formation of a single product) will not be possible in the reaction. Therefore, the study of how reactions occur depends on an analysis of the starting materials, the products, and the transition states separating them. To understand chemical reactivity, we need to know: (i) which are more stable, starting materials or products? And (ii) how high the transition state separating starting materials and products is. In a SN2 reaction, we see that two nucleophiles (Nu:- and L:-) compete for an electrophilic carbon center. The reaction can potentially proceed in either direction and hence is an equilibrium. In fact, all chemical reactions can be regarded as equilibria. Multistep reactions are simply a series of equilibria in which species occupying energy minima are separated by transition states. For example, three equilibria are present in the SN1 solvolysis reaction of tert-butyl iodide (see, Jones Figure 8.2). The first equilibrium is highly unfavorable due to formation of a cation-anion pair that is less stable than the covalent iodide. The second equilibrium depicts exothermic formation of the protonated alcohol (an oxonium ion) while the third equilibrium gives the product (t-butyl alcohol) via deprotonation of the oxonium ion. 8.2. Equilibrium While all reactions can be viewed as equilibria, it is important to understand that thermodynamics (the free energy difference between starting materials and products) may dictate that one side of an equilibrium will be overwhelmingly favored over the other. Thus, whether a chemical reaction at equilibrium lies towards starting materials or products is determined by thermodynamics. For a general chemical reaction at equilibrium: aA + bB cC + dD The concentration of starting materials and products are related to one another by the equilibrium constant, K, which is given by:2 [C]c[D]d[A]a[B]bK = And the free energy difference between the starting materials and products at equilibrium is given by: ∆G˚ = -RT ln K or ∆G˚ = -2.3RT log K R = gas constant, 1.99 x 10-3 kcal/˚mol; T = temperature in Kelvins; ∆G˚ = Gibbs free energy change - the free energy difference between starting materials and products in their standard states. When K > 1, ∆G˚ is negative hence the products are favored (the total free energy of the products is less than the total free energy of the starting materials). The reaction is said to be exergonic. When K < 1, ∆G˚ is positive hence the starting materials are favored. The reaction is said to be endergonic. In a multistep reaction, it usually does not matter if an early step involves an unfavorable equilibrium, as long as ∆G˚ for the overall reaction is negative and starting materials have sufficient energy to overcome all of the energy barriers Indeed, only a small difference in energy between starting materials and products is needed to obtain a large excess of one over the other at equilibrium. We can rewrite the relationship between K and ∆G as: K = e-∆G˚/RT or K = 10-∆G˚/2.3RT Using either expression, we find that a reaction at 25 ˚C that gives products that are more stable than starting materials by 1kcal/mol has an equilibrium constant (K) = 5.4, which translates into ~85% of product at equilibrium (see Jones Table 8.1). So, is it only possible to obtain desired products that are favored at equilibrium? Of course not, what we have to do is run the reaction under conditions that shift the equilibrium and drive the reaction in the direction we want! Remember Le Chatelier’s principle states that a system at equilibrium responds to stress in such a way as to relieve that stress. Thus, for the general reaction aA + bB cC + dD if the equilibrium lies towards starting materials, we can drive the reaction towards products by increasing the concentration (using an3 excess) of one of the starting materials. Alternatively, we can remove one of the products from solution (by choosing conditions, such as solvent and/or temperature, that lead to precipitation of one the products). Both of these approaches serve to drive the reaction towards product in order to maintain equilibrium! 8.3. Gibbs Standard Free Energy ∆G˚ is composed of enthalpy (∆H˚) and entropy (∆S˚) terms. Thus, ∆G˚ = ∆H˚ - T∆S˚ ∆H˚ refers to heat of reaction, which is a measure of the change in bond strengths in the reaction. ∆S˚ is a measure of the change in freedom of motion in the reaction. The more ordered the transition state, the more negative the entropy change and hence the less negative ∆G˚. A reaction is exothermic when the difference in enthalpy between products and starting materials is negative, and usually K > 1 and the reaction proceeds towards products. However, sometimes the enthalpy change (∆H˚) may not be in the same direction as the free energy change (∆G˚), which also involves entropy (∆S˚). Entropy becomes increasingly more important to determining the value of ∆G˚ for a reaction as the reaction temperature is increased because the entropy term is temperature-dependent. For example, formation of cyclohexene from 1,3-butadiene and ethylene is favored at low temperature (∆G˚ < 0) because of


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UK CHE 230 - Equilibria

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