AP Chemistry B Exam Prep Session Kinetics

and for gases: Kp = (PR)r(PS)s (PA)a(PB)b aA +bB + …↔ rR +sS + . . The equilibrium expression is based upon the overall balanced equation: aA +bB + …↔ rR +sS + . . Kc = [R]r[S]s   [A]a[B]b K > 1 products favored K < 1 reactants favored   In heterogeneous equilibria only use gases & aqueous! Omitted solids; pure liquids; water (in aqueous solutions) because their [ ]’s do not change. and for gases: Kp = (PR)r(PS)s (PA)a(PB)b

Types of equilibrium problems: Kc general types Ka acid equilibrium constants, weak acids to moderate acids; strong acids Ka>>>1 Kb base equilibrium constants, weak bases to moderate; strong bases Kb>>>1 Kw ion product constant for water; basis for pH scale @ 25˚C Ksp solubility product constant, point of saturation for “insoluble” precipitation reactions Kp using pressures of gases; DO NOT use brackets, must use (P) notation for equilibrium expression.

Converting from Kp to Kc Kp using pressures of gases; DO NOT use brackets, must use (P) notation for equilibrium expression. Kp = Kc(RT)Δn Based upon Δn

Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning.

Progress of Reaction Which reagent is the reactant? How can you tell? What is the value of ∆G? Do [A] & [B] look like same ratio in terms of stoichiometry? Figure: 14-28-03UNE2 Title: Exercise 14.2 Caption: Graph of concentration versus time.

Which expression correctly states the rate of reaction? N2(g) + 3 H2(g) 2 NH3(g) Which expression correctly states the rate of reaction?

N2(g) + 3 H2(g) 2 NH3(g)

For the reaction: A 2 B, which of the following is a correct expression for the reaction rate? 1.  [A] 3.  [A] Rate Rate = =  t 2  t 2.  [B] 4.  [B] Rate = Rate = - 2  t 2  t

For the reaction: 2 A  3 B the rate of appearance of B, D[B]/Dt, is 0.30 M/s. What is the value of the rate of disappearance of A, –D[A]/Dt? 0.05 M/s 0.10 M/s 0.20 M/s 0.40 M/s 0.60 M/s

Correct Answer: 0.05 M/s 0.10 M/s 0.20 M/s 0.40 M/s 0.60 M/s  [A] [B]  [A] Rate = =  2  3 t t  [A] 2  [B] Rate = =  t 3  t  [A] 2(0.30 M /s) Rate = = = 0.20 M /s  t 3

Factors Affecting Rate: How to Speed Up a Reaction [Use Collision Theory, Kinetic Molecular Theory] increase the concentration of reactants - increase molarity of solutions - increase partial pressure of gases [collision model: more collisions] more surface area between unlike phases increase the temperature [collision model: more & harder collisions] add a catalyst

Maxwell–Boltzmann Distributions If the dotted line represents when reactions will happen (activation energy), as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases.

Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. 2006 #6

Rate cannot be determined by overall reaction. Rate law must match the rate determining step of the reaction mechanism. H2O2  2 OH- slow H2O2 + OH-  H2O + HO2 fast HO2 + OH-  H2O + O2 very fast Overall reaction? RDS? Rate law? 2009 #3

a) Which species is the catalyst? b) Which is an intermediate? A proposed mechanism for the decomposition of N2O is given below: NO + N2O  N2 + NO2 2 NO2  2 NO + O2 a) Which species is the catalyst? b) Which is an intermediate? NO NO2 N2O N2 O2

Correct Answer: NO NO2 N2O N2 O2 Note in the mechanism below that NO is regenerated. By definition a catalyst does not undergo permanent chemical change. NO + N2O  N2 + NO2 2 NO2  2 NO + O2 Intermediates are produced and consumed within the process of the reaction = NO2

Sample AP Multiple Choice Factors that affect the rate of a chemical reaction include which of the following? Frequency of collisions of reactant particles Kinetic energy of collisions of reactant particles Orientation of reactant particles during collisions II only I and II only I and III only II and III only I, II, and III

CAUTION! Requires memorization & must be determined experimentally. Rate Laws CAUTION! Requires memorization & must be determined experimentally.

Rate Laws Rate constant: same for respective reaction @ constant temp Use reactions where reverse reaction condition can be neglected (negligible) Rate = k [NO]2 [Br2] Rate constant: same for respective reaction @ constant temp Order of reaction 2 + 1 = 3rd order

Types of Rate Laws Integrated Rate Law Differential Rate Law Integrated Rate Law Shows how the rate of a reaction depends on concentrations (if I change [ ] how does rate change? Uses [initial] & initial rate data No memorizing, solve Concentrations of species in the reaction depend upon time At intervals of time, what are the concentrations Use of graphs to determine order Have to memorize!

Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these is rate = k[NO2]m[CO]n Use the following data to determine the individual and overall reaction orders. Experiment Initial Rate (mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L) 1 2 3 0.0050 0.080 0.10 0.40 0.20 PLAN: Solve for each reactant using the general rate law using the method described previously. SOLUTION: rate = k [NO2]m[CO]n First, choose two experiments in which [CO] remains constant and the [NO2] varies.

Determining Reaction Orders from Initial Rate Data Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data [NO2] 2 [NO2] 1 m = rate 2 rate 1 k [NO2]m2[CO]n2 k [NO2]m1 [CO]n1 = The reaction is 2nd order in NO2. 0.40 0.10 = m 0.080 0.0050 ; 16 = 4m and m = 2 [CO] 3 [CO] 1 n = rate 3 rate 1 = k [NO2]m3[CO]n3 k [NO2]m1 [CO]n1 The reaction is zero order in CO. 0.20 0.10 n 0.0050 = ; 1 = 2n and n = 0 rate = k [NO2]2[CO]0 = k [NO2]2

Sample AP Multiple Choice The table shows the results from a rate study of the reaction X + Y  Z. Starting with known concentration of X and Y in experiment 1, the rate of formation of Z was measured. If the reaction was first order with respect to X and second order with respect to Y, the initial rate of formation of Z in experiment 2 would be R/4 R/2 R 2R 4R Experiment [X]o [Y] o Initial rate of Formation of Z (mol L-1sec-1) 1 0.40 0.10 R 2 0.20 ?

Integrated Rate Law Order is determined by plotting data such that a straight line is obtained (1st order is most common followed by 2nd, lastly 0 order) Once order is determined, can use appropriate equation to solve for rate constant (k) or k = slope Data dictates which strategy must be used. For integrated rate law only given time vs. concentration no initial rate!

Integrated rate laws and reaction orders. 1/[A]t = kt + 1/[A]0 ln[A]t = -kt + ln[A]0 [A]t = -kt + [A]0

Figure 16.8 Graphical determination of the reaction order for the decomposition of N2O5.

Figure 16.8 Graphical determination of the reaction order for the decomposition of N2O5.

Figure 16.8 Graphical determination of the reaction order for the decomposition of N2O5.

A Sample AP Free Response Question Topic: Kinetics 2010B #6 http://apcentral.collegeboard.com/apc/public/repo sitory/ap10_frq_chemistry_formb.pdf