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University of Florida
CHM 2045


 • Reversible Reactions: Rate Determining Step o A. Rate determining step: If a complex reaction is separated into elementary steps, the rate of slowest step determines the rate of the overall reaction o B. If the slow step is the first step, the rate law can be derived directly from this step and no other o C. If the slow step is other than the first step, the slow step is still the rate-determining step, but steps prior to the slow step will contribute to the rate law. o D. Steps after the slow step make contribution to the rate law • Reversible Reaction Rates: First Step Rate-Determining Example o 1. NO (g2 + NO (g)2 NO (g) + 3O(g) slow step o 2. NO (3) + CO(g)  NO (g) + 2O2(g) fast step o 3. Rate = k [NO ] 2 1 2 • Reversible Reaction Rates: Second Step Rate-Determining Example o 1. 2NO(g) + Br (g)  NOBr fast step 2 2 o 2. NOBr (g2 + NO(g)  2NOBr(g) slow step o 3. Rate = k 2NOBr ][2O] o 4. [NOBr ] 2epends upon the first step; if we assume it reaches equilibrium very quickly, [NOBr ] 2an be written in terms of the equilibrium constant K fcr step 1, [NOBr ] 2 K [NO][Br ] c 2 2 o 5. Rate = k [cO][Br ]k2[N2] = k k [c 2 [Br ] 2 o 6. Or we can use Equilibrium Approximation • Reversible Reaction Rates: Equilibrium Approximation o A. Equilibrium approximation assumes all steps prior to the rate-limiting step are in equilibrium o B. It requires that the slow step be significantly slower than the fast steps o C. Example  1. Since the first step is considered to be in equilibrium, set the forward reaction rate equal to the reverse reaction rate:1k [NO][Br2] = k1 [NOBr 2, and then solve for [NOBr ] 2  2. [NOBr ]2= (k 1k- 1[NO][Br ]2  3. Rate law: rate = 1 /k1 [NO][Br2]k2[NO] = k k2 1- 1NO] [Br ] 2  4. k2 1/k1 = kobserved o D. If there is not a step that is significantly slower than the others, use Steady State Approximation • Reversible Reaction Rates: Steady State Approximation o A. Steady State Approximation: the concentration of the intermediate is considered to be small and hardly changing; it leads to the same result as equilibrium concentration • Catalysis o A. Catalyst: substance that increases the rate of reaction, and creates an intermediate, without being consumed or permanently altered, by:  1. Enhance product selectivities – increase the steric factor, p  2. Reduce energy consumption – lower the activation energy, E a o B. o C. Catalysts work by providing an alternative reaction mechanism that competes with the uncatalyzed mechanism 
 o D. Cannot alter the equilibrium constant, Kc, of a reaction or the composition of the mixture at equilibrium, thus it increases the rate of the forward and reverse reactions o E. Types: heterogeneous and homogeneous o F. Since they alter reaction mechanisms, they require separate rate constants o G. Turnover Number: the number of reactions occurring at one active site on one enzyme • Catalysis: Heterogeneous Catalysts o A. Heterogeneous Catalyst: in a different phase than the reactants and products o B. Usually solids while the reactants and products are liquids and gases o C. Reactants adsorb (binding of molecules to the surface), not absorb (uptake of molecules into the interior) to the solid catalyst  1. Physically adsorb via Van der Waals forces (minor)  2. Chemically adsorb via covalent bonds (major) o D. Reactants bind to the surface because the catalyst molecules there have unfilled valence electrons and this process is always exothermic o E. Binding strengths determine the rate of catalysis  1. Too weak: not enough adsorption, doesn’t speed up enough  2. Too strong: requires too much energy to remove, cancels out effects o F. Reaction rate can be enhanced by increasing the surface area of the catalyst • Catalysis: Homogeneous Catalysts o A. Homogeneous Catalysts: the same phase as the reactants and products, usually a gas or liquid, e.g. aqueous acid or base solutions are homogenous catalysts o B. Autocatalysis: when reactions generate a catalyst as a product o C. Concentration: usually small compared to [reactant] and [product]; increasing [catalyst] will increase the rate of reaction o D. Concentration: if large compared to [product] and [reactant], changing it won’t do much to the rate of reaction • Effects of Solvent on Rate o A. Liquid molecules make ~100 times more collisions per second that gas molecules because of the closer proximity; however, most are between solvent molecules and thus don’t result in a reaction o B. Rate constant, k, in liquid is affected by both:  1. Solvent • a. Solvent-reactant bonds • b. Electrical insulation
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