Class Notes (806,696)
United States (312,146)
Chemistry (160)
CHM 2045 (77)
chu (15)
Lecture

# 7____

3 Pages
47 Views

School
University of Florida
Department
Chemistry
Course
CHM 2045
Professor
chu
Semester
Spring

Description
13  • 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   13    14  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
More Less

Related notes for CHM 2045

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.