Review Unit 4 (Chp 14): Chemical Kinetics (rates) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Review Unit 4 (Chp 14): Chemical Kinetics (rates) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Chemical Kinetics Chemical Kinetics is the study of: The rate at which reactants are consumed or products are produced during a chemical rxn. The factors that affect the rate of a reaction according to Collision Theory (temperature, concentration, surface area, & catalyst). The mechanism, or sequence of steps, for how the reaction actually occurs. The rate laws (equations) used to calculate rates, rate constants, concentrations, & time. (M∙s–1) k (?∙s–1) [A] (mol∙L–1) t (s)

minimum E required to start reaction The Collision Model reactant bonds break, then product bonds form. Reaction rates depend on collisions between reactant particles by: collision frequency enough energy proper orientation activation energy: minimum E required to start reaction (Eact ) unsuccessful successful reactants COLLISION products

Potential Energy Diagram demo transition state activated complex …aka… Energy Profile Eact Potential Energy  Demo: Transition State Ball Toss ∆Hrxn Reaction progress 

4 Factors That Affect Reaction Rates Concentration Temperature Reactant Surface Area (particle size) Catalyst (↑ collisions) (↑ collisions and ↑ energy) (lowers Ea by changing mech.) Boltzmann Distribution more particles over the Eact old foamy (elephant toothpaste)

Rate Ratios = ? aA + bB cC + dD Rate 2 HI(g)  H2(g) + I2(g) 1 a [A] t = b [B] c [C] d [D] = − − 2 HI(g)  H2(g) + I2(g) Initial rate of production of H2 is 0.050 M∙s–1. What is the rate of consumption of HI? = ? 2 mol HI 0.050 M∙s–1 H2 x = 0.10 M∙s–1 HI 1 mol H2 mol L∙s –0.10 M∙s–1 HI mol ratio

Rate Laws rate = k[A]x[B]y[C]z recall… Rate equations (or rate laws) have the form: rate = k[A]x[B]y[C]z rate constant order with respect to reactants A, B, & C …or… number of each particle involved in collision that affects the rate overall order of reaction = x + y +… Example: overall order = ___ order (4 particles in collision) rate = k[BrO3–][Br–][H+]2 4th

Concentration and Rate Data Initial Concentrations Rate in M per unit time Mixture [BrO3–] / M [Br–] / M [H+] / M A 0.0050 0.25 0.30 10 B 0.010 20 C 0.50 40 D 0.60 160 Mathematically or conceptually show how to determine the orders and rate law. Rate = k [BrO3–] [Br–] [H+]2

Orders in Rate Laws …must be found experimentally (from data). …do NOT come from the coefficients of reactants of an overall reaction. …represent the number of reactant particles (coefficients) in the RDS of the mechanism. …zero order reactants have no effect on rate b/c they do not appear in the RDS of the mechanism (coefficient of 0 in RDS). …are typically 0, 1, 2, but be any # or fraction.

Integrated Rate Laws & Linear Graphs zero-order m = rate(M/s) [A] [A]0 = initial conc. at t = 0 [A]t = conc. at any time, t 1st or 2nd order t first-order second-order m = k m = –k 1 [A] ln [A] t t ln [A]t = –kt + ln [A]0 1 [A]t = kt + [A]0 on equation sheet (kinda)

Half-Life Half-life (t1/2): time at which half of initial amount is reacted. [A]t = 0.5 [A]0 Half-life (t1/2) is constant for 1st order only. = t1/2 0.693 k given on exam 11

Mechanisms Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO  2 NOBr (slow) RDS From RDS Step 2: rate = k2 [NOBr2] [NO] b/c step 1 is in equilibrium From Step 1: rateforward = ratereverse k1 [NO] [Br2] = k−1 [NOBr2] solve for [NOBr2] substitute for [NOBr2] k1 k−1 [NO] [Br2] = [NOBr2] 12

Mechanisms k2k1 k−1 rate = [NO] [Br2] [NO] rate = k [NO]2 [Br2] Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO  2 NOBr (slow) RDS rate = k2 [NOBr2] [NO] k2k1 k−1 rate = [NO] [Br2] [NO] rate = k [NO]2 [Br2] substitute for [NOBr2] k1 k−1 [NO] [Br2] = [NOBr2] 13