CHAPTER 12 AP CHEMISTRY

CHEMICAL KINETICS Speed or rate of reactions - Reaction Rate Change in concentration of a reactant or product per unit of time aA + bB -----> cC + dD Rate = conc. of A at t2 - conc. of A at t1 t2 - t1 Δ [A] A = reactant or product Δt [ ] = mol/L Δ = change Page 545 - 548 Slope of tangent line = change in y Change in x Rate = -slope of tangent line

FORM OF RATE LAW Page 551-552 First order rate = k[A] k = constant, rate doubles if concentration of A doubles Rate law Any chemical reaction must be determined experimentally; it cannot be predicted R = k[reactant 1]m[reactant 2]n Page 554 exercise 12.1

INTEGRATED RATE LAW First-order rate law Integrated first-order 2N2O5(sol’n) ---> 4NO2 + O2 Rate = - Δ[N2O5] = k[N2O5] Δ t Integrated first-order In[A] = -kt + ln[A]o Reaction is first order if a plot of ln[A] versus t is a straight line Page 557-558 y = mx + b [A]o = concentration at the start of reaction Pressure can be used as a concentration half-life of a first-order t1/2 = 0.693 k Does not depend of the CONCENTRATION Page 560

SECOND-ORDER RATE LAWS Rate = - Δ[A] = k[A]2 Δ t Integrated second-order 1 = kt + 1 [A] [A]o half-life of a second-order t1/2 = 1 k[A]o Depends on the concentration Page 562 Zero-order Rate = k[A]o = k [A] = -kt + [A]o t1/2 = [A]o 2k

CONTINUE Rate laws with more than one reactant BrO3-(aq) + 5Br-(aq) + 6H+(aq) -----> 3Br2(l) + 3H2O(l) Rate = Δ[BrO3-] = k[BrO3-][Br-][H+]2 Δ t [BrO3-]o = 1.0 X 10-3 M, [Br-]o = 1.0M, [H+]o = 1.0M Page 567 table 12.6 Reaction mechanism Series of steps Has intermediates that are not seen Page 569 One step will be the rate determining one

REACTION MECHANISM Path or sequence of steps Elementary steps Must always add to give the chemical equation of the overall process Rate of steps = to rate constant multiplied by the concentration of each reactant A ---> B + C Rate = k[A] A + A -----> B + C Rate = k[A]2 Slow steps The slow step is rate-determined Overall rate cannot exceed the slowest step If the step is definitely the slowest, its rate is approximately equal to the overall reaction

CONTINUE All intermediates MUST be eliminated NO(g) + Cl2(g)  NOCl2(g) Fast NOCl2(g) + NO(g) 2NOCl(g) Slow 2NO(g) + Cl2(g) ----> 2NOCl(g) k2[NOCl2][NO] k1[NO][Cl2] = k-1[NOCl2] [NOCl2] = k1[NO][Cl2] k-1 Overall = k2[k1[NO][Cl2]][NO] [ k-1 ] = k2k1[NO]2 [Cl2]1

ACTIVATION ENERGY THERE MUST BE A CERTAIN AMOUNT OF KINETIC ENERGY FOR A REACTION TO OCCUR Ea > 0 Depends on nature of reaction Independent of temperature and concentration

ARRHENIUS EQUATION K = Ae-Ea/RT Ea= activation energy R = 8.314 J/K·mol T = absolute temperature A = frequency of collisions ln k1 = Ea(1 _ 1 ) k2 R (T2 T1)

CATALYSIS Increases the rate of reaction without being consumed Heterogenous catalysis In different phase as mixture Initial step is absorption, outside particles in a solid will have unfilled valence shells which can attach to other particles. These places are called active sites found in Enzymes Without digestive enzymes it is estimated it would take 50 years to digest a meal MnO2 Homogeneous catalysis Present in same phase