Chemical Kinetics AP Chemistry Lecture Outline Name: Factors that govern rates of reactions. Generally... (1)...as the concentration of reactants increases, rate (2)...as temperature increases, rate (3)...with a catalyst, rate (4)...as reactant surface area increases, rate Reaction rates are usually measured in terms of concentration. units are usually M/s rate x t [ x ] (M) Conc. of Reactant x v. Time time (s) instantaneous rate: the reaction rate at any given time it is equal to the slope of the [ ] time curve at any point [ x ] (M) Conc. of Reactant x v. Time For point P, one finds the instantaneous rate by: time (s) 1
Coefficients in the balanced equation must be considered when comparing rates for substances in that reaction. e.g., H 2 (g) + Cl 2 (g) 2 HCl(g) In general, for the reaction aa + bb cc + dd EX. 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) If the instantaneous rate of disappearance of N 2 O 5 is 4.2 x 10 7 M/s at a given time, find the instantaneous rates for NO 2 and O 2. Generally, reaction rates are proportional to the [ ] of reactants. rate law: contains a rate constant, k For most reactions, a rate law has the form rate = k [reac. 1] m [reac. 2] n... where m and n are... m and n are determined experimentally Usually, reaction orders are 0, 1, or 2, but some are fractions or are ( ) EX. H 2 (g) + Cl 2 (g) 2 HCl(g) From experiment, the rate = k [ H 2 ] [ Cl 2 ]. Find... (1) reaction order of H 2 (2) overall reaction order (3) units of rate constant 2
If a reaction is zero order in a particular reactant, changing its concentration... 1 st order: 2 nd order: The rate constant k is NOT affected by concentration, but IS affected by... Rate laws can be converted into equations that give the concentrations of substances at any time during the course of a reaction. First-Order Reactions these depend on the [ ] of a single reactant (A) raised to the 1 st power rate A t By integrating both sides, we get... k A EX. (CH 3 ) 2 O(g) CH 4 (g) + H 2 (g) + CO(g) Assume that the rate of this reaction is a first-order process, with k = 6.8 x 10 4 s 1. If the initial pressure of (CH 3 ) 2 O is 135 torr, what is its partial pressure after 1420 s? Half-life of a reaction, t 1/2 : the time required for a reactant s [ ] to drop to ½ of its orig. value [A] t 1/2 = ½ [A] o From the basic equation ln [A] t = ln [A] o kt Put ½ [A] o here and t 1/2 here, and we ll get... For 1 st order reactions... t 1/2 is independent of initial concentration the concentration of reactants is cut in half... every half-life 3
Second-Order Reactions (that are 2 nd order in just one reactant) 1 A t 1 A o kt and t 1 / 2 k 1 A o ** Above equations do NOT apply to 2 nd -order reactions that are 1 st -order in two reactants. e.g., rate = k [A] [B] As temperature increases, rate increases because... collision model: explains how rate is affected by [ ] and by temperature greater [ ]: higher temperature: activated complex: E a E a of reverse reaction Energy E Time About E... it is for endothermic reactions it is for exothermic reactions The fraction of molecules with an energy greater than the activation energy E a is given by: E a in J/mol R = 8.314 J/mol-K T = absolute temperature (i.e., in K) 4
For the reaction to occur, collisions must take place with molecules oriented in a certain way. e.g., Cl + ONCl Cl 2 + NO The Arrhenius equation relates the rate constant k to the activation energy E a and the reaction temperature: k = rate constant at a particular temperature A = the frequency factor, which is related to collision frequency and the odds that the molecules have a favorable orientation If we know E a, we can relate the rate constants at two different temperatures without having to bother with the frequency factor A. EX. Using the following information, find... (1)...E a (2)...k at 430.0 K T ( o C) k (s 1 ) 189.7 2.52 x 10 5 251.2 3.16 x 10 3 5
Reaction Mechanisms the processes by which reactions occur sometimes, mechanisms are dependent on temperature elementary steps: molecularity: defined by the number of molecules that participate as reactants in an elementary step unimolecular bimolecular termolecular multistep mechanisms: sequences of elementary steps are needed to go from R to P e.g., For the reaction NO 2 + CO NO + CO 2 Elementary Step 1: Elementary Step 2: Above, NO 3 is an intermediate. All multistep mechanisms have them. The rate law and relative speed of each elementary step (no matter how many there are) determine the overall rate law for the reaction; that is, the mechanism gives us the overall rate law. Rate laws for the elementary steps are found as follows... Molecularity uni- Elementary Step A > P Rate Law for that Elementary Step bibiterterter- A + A > P A + B > P A + A + A > P A + A + B > P A + B + C > P These rate laws are for elementary steps, not necessarily for the R 6 P reaction.
EX. Assuming that H 2 (g) + Br 2 (g) 2 HBr(g) occurs as a single elementary step, predict the rate law. Most reactions have multiple elementary steps. The slowest of these is the... EX. 2 NO(g) + Br 2 (g) 2 NOBr(g) has the following elementary steps: Elementary Step 1: Elementary Step 2: Write the correct rate law. EX. An elementary step in a multistep mechanism is Br 2 (g) 2 Br(g) What expression relates the concentration of Br(g) to that of Br 2 (g)? catalysis: the process by which a catalyst changes the rate and mechanism of a chemical reaction a catalyst is NOT consumed during the course of a reaction Homogeneous catalysts are present in the same phase as the reacting molecules. Recall the Arrhenius equation: k = Ae Ea/RT a catalyst can alter the frequency factor A and/or the activation energy E a ** Catalysts provide... 7
Heterogeneous catalysts exist in a different phase than the reacting molecules. The first step in heterogeneous catalysis is of the reactant molecules onto the metal surface specifically, onto the, which are the locations at which the reactants attach to the metal catalyst. then, the products detach from the catalyst enzymes: large protein molecules usually very specific names end in... substrates: substances (i.e., reactants) that react at the active sites of enzymes The lock-and-key model explains how an enzyme affects a substrate molecule and changes it into a new substance. Lock-and-Key Model turnover number: Enzyme inhibitors bind to the active site or alter the unique shape of an enzyme molecule, destroying enzyme s activity e.g., 8