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 transition state activated complex …aka… Energy Profile Eact Potential Energy  ∆Hrxn Reaction progress 

4 Factors that Affect Rates Concentration Temperature Exposed Surface Area (size) Catalyst (↑ collisions) (↑ collisions and ↑ energy) (↑ collisions) (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

Rate Law & Orders from Mechanisms Step 1: NO + Br2 NOBr2 (slow) (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] 8

Rate Law & Orders from Mechanisms 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] 9

Rate Law & Orders from Exp. 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 rate = k [BrO3–]x[Br–]y[H+]z rate = k [H+]z then… x = 1 if… 2 = [2]x 4 = [2]z z = 2 Rate = k [BrO3–]1 [Br–]1 [H+]2

Orders in Rate Laws …only 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). …typically 0, 1, 2, but can be any # or fraction

Units of k (rate constant) HW p. 619 #28 Units of k give info about order. Rate is usually (M∙s–1) or (M∙min–1) (next slide) Order Rate Law k Units 0th 1st 2nd 3rd rate = k[A]0 rate = k[A] rate = k[A]2 rate = k[A]2[B] M = ? s M = ?∙M s M = ?∙M2 s M = ?∙M3 s M s 1 s 1 M∙s 1 M2∙s M∙s–1 s–1 M–1∙s–1 M–2∙s–1

Model for Solving Orders HW p. 619 #28 Model for Solving Orders Determine the rate law for the reaction (from experimental data). general rate law rate = k [ClO2]x [OH–]y (a) rate1 rate2 (0.0248) k (0.060)x(0.030)y = = (0.00276) k (0.020)x(0.030)y (0.0248) 0.060 x 0.020 0.030 y 0.030 = (0.00276) 9 = (3)x (1)y 9 = (3)x x = 2 (2nd order)

rate = k [ClO2]2 [OH–]y rate = k [ClO2]2 [OH–]1 rate3 rate2 (0.00828) HW p. 619 #28 (cont.) Determine the rate law for the reaction (from experimental data). rate = k [ClO2]2 [OH–]y (a) rate3 rate2 (0.00828) k (0.020)2(0.090)y = = (0.00276) k (0.020)2(0.030)y 3 = (3)y y = 1 (1st order) rate = k [ClO2]2 [OH–]1 OR rate = k [ClO2]2 [OH–]

rate = k [ClO2]2 [OH–] Exp 1: (0.0248) = k (0.060)2(0.030) (0.0248) HW p. 619 #28 (cont.) Calculate the rate constant (with units). rate = k [ClO2]2 [OH–] (b) Exp 1: (0.0248) = k (0.060)2(0.030) (0.0248) k = (0.060)2(0.030) k = 230 M–2∙s–1 M = ?∙M2∙M s

rate = k [ClO2]2 [OH–] rate = (230)(0.010)2(0.025) rate = 0.00058 HW p. 619 #28 (cont.) Calculate the rate when [ClO2] = 0.010 M and [OH–] = 0.025 M. rate = k [ClO2]2 [OH–] (c) rate = (230)(0.010)2(0.025) rate = 0.00058 M∙s–1

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 18

M∙s–1 s–1 M–1∙s–1 Summary 0th Order 1st Order 2nd Order [A] ln[A] 1 Rate Law Rate = k Rate = k[A] Rate = k[A]2 Integrated Rate Law [A] = –kt + [A]0 ln[A] = –kt + ln[A]0 Linear plot [A] ln[A] k & slope of line Slope = rate Slope = –k Slope = k M∙s–1 s–1 M–1∙s–1 Half-Life depends on [A]0 y = mx + b y = mx + b 1 [A] t t t M = ? s M = ?∙M s M = ?∙M2 s Units of k