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Chemistry and Chemical Reactivity 6th Edition 1 John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 15 Principles of Reactivity: Chemical Kinetics Lecture written by John Kotz as modified by George Rhodes 2008 My background • • • • • 2 • • • • 2008 B.S. Chemistry ASU, 1964 Ph.D. UUtah, Chemistry, 1973 Asst. Prof., USAF Academy 1968-1970 New laser research, USAF Weapons Lab 1970-1974 Directed AEC Laser Isotope Separation Program, 1974-1976 Directed ERDA/DOE Solar Design and Construction program 1976-1978 VP Energy Programs, The BDM Corp, 19781982 Sr. VP, EMP, Energy, BDM 1982-1989 VP, Technology, Quatro, Inc, 1989-1996 Additional background • Pres/CEO, Dynamic Resonance Systems, 1996-1999 • Ex. VP & Chief Operating Officer, Avistar, Inc. (PNM Resources), 1999-2006 • Chief Scientist, PNM Resources, 2006 to present • Visiting Faculty, NAU 2006-Present • Guest Scientist, Los Alamos National Laboratory 1970-2000 • Named as Eminent Scholar, NMIT, 1989 • DOE Award for Excellence in Technology Transfer 1993 • 11 Patents (2 pending), over 50 publications • Commercial pilots license, swimming (WSI), squash, golf, PADI certified diver, fly fishing… 2008 3 4 First, let’s review algebra • • • • • • (a) a+b=b+a (c) ab=ba (e) (ab)c=a(bc) (g) abac=a(b+c) (i) a+0=a (k) a1=a (b) a+(b+c)=(a+b)+c (d) a(b+c)=ab+ac (f) (ab)c=acbc (h) (ab)c=a(bc) (j) a . 1=a 2008 5 Forward and inverse operations • • • • • • • • (a) addition a+b=c (b) multiplication ab=c (c) power ba=c (d) power ab=c (a') subtraction b=c-a (b') division b=c/a (c') root b=a√c (d') logarithm b=logac 2008 6 Logarithms • y=logax or logax=y • Log 1000 = log (103) = 3 which is: the power to which you must raise 10 to get the number • Log 2.10 = 0.3222 or 10 0.3222 = 2.10 • and ln x = 2.303 log x • The natural logarithm is the logarithm having base e, where • This function can be defined as • • For x>0 2008 7 More on Logarithms • ln x = 2.303 log x • So when logax=y to get the antilog you simply solve x=10y , and when using natural logs • When lny=x the antilog is: y=ex • Let’s take an example • The ln of 1000 is: 6.9078 • If you divide by 2.303 you get 3 • 103 = 1000, e6.9078 = 1000 • Isn’t it great when a plan comes together? 2008 8 Definitions: Chemical • • • • • • A substance that: 1) An organic chemist turns into a foul odor; 2) an analytical chemist turns into a procedure; 3) a physical chemist turns into a straight line; 4) a biochemist turns into a helix; 5) a chemical engineer turns into a profit. 2008 9 The Exam Schedule • Exams: – July 14th – Kinetics and Chemical Equilibria – July 21st – Acids and Bases and Principles of Reactivity – July 28th – Entropy and Free Energy and Electron Transfer Reactions – August 4th – Nuclear Chemistry and Carbon and the Environment – August 5th - Final 2008 10 Advice • If you are taking another time consuming course in addition to CHM 152 and 152 Lab, consider dropping the other course if you wish to do well in chemistry. • Experience has shown that you have a very high probability of doing poorly in one or both 2008 11 The Facts • There is no way we can cover the necessary material without using PowerPoint. • We will cover over 20 pages per day in the text, plus problems • This is a math course and you must be comfortable with algebraic operations • With this background, you will learn a lot and be well prepared for future science courses. • All lectures will be posted on the Chem Dept web site at http://www.nau.edu/chem/ • Go to online course materials and select the appropriate lecture 2008 12 Chemical Kinetics Chapter 15 2008 13 Chapter Goals p-741-742 • Understand rates of reaction and the conditions affecting rates • Derive the rate equation, rate constant, and reaction order from experimental data • Use integrated rate laws • Understand collision theory of Rx rates and the role of activation energy • Relate Rx mechanisms and rate laws 2008 Thermodynamics • We can use thermodynamics to tell if a reaction is product- or reactantfavored. • But this gives us no info on HOW FAST reaction goes from reactants to products. • This is the study of chemical kinetics 14 2008 15 Chemical Kinetics • KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. • Our goal is to understand the reaction mechanisms so that we can tell how fast a reaction will take place 2008 Reaction Mechanisms The sequence of events at the molecular level that control the speed and outcome of a reaction. Br & Cl from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, 1994.) 16 Step 1: Br + O3 ---> BrO + O2 Step 2: Step 3: NET: 2008 Cl + O3 ---> ClO + O2 BrO + ClO + light ---> Br + Cl + O2 2 O3 ---> 3 O2 Reaction Rates Section 15.1 17 • Reaction rate = change in concentration of a reactant or product with time. • Three ―types‖ of rates –initial rate –average rate –instantaneous rate 2008 Determining a Reaction Rate Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc with time — can be determined from the plot. Screen 15.2 18 Dye Conc Time 2008 The Reaction Rate Reaction rate is the change in the concentration of a reactant or a product with time (M/s). 19 A D[A] rate = Dt B D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative. D[B] rate = Dt 2008 Determining a Reaction Rate 20 2008 Factors Affecting Rates Section 15.2 21 • Concentrations • and physical state of reactants and products Temperature • Catalysts 2008 Concentrations & Rates Section 15.2 22 0.3 M HCl 6 M HCl Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g) 2008 23 Factors Affecting Rates • Physical state of reactants 2008 24 Factors Affecting Rates Catalysts: catalyzed decomp of H2O2 2 H2O2 --> 2 H2O + O2 2008 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea/RT ) Ea k 25 uncatalyzed catalyzed ratecatalyzed > rateuncatalyzed 2008 Ea < Ea ‘ 26 Factors Affecting Rates • Temperature Bleach at 54 ˚C Bleach at 22 ˚C 2008 Iodine Clock Reaction 1. Iodide is oxidized to iodine H2O2 + 2 I- + 2 H+ -----> 2 H2O + I2 2. I2 reduced to I- with vitamin C I2 + C6H8O6 ----> C6H6O6 + 2 H+ + 2 IWhen all vitamin C is depleted, the I2 interacts with starch to give a blue complex. Page 705 in CCR 2008 27 Iodine Clock Reaction 28 2008 29 Concentrations and Rates To postulate a reaction mechanism, we study • reaction rate and • its concentration dependence 2008 Concentrations and Rates Take reaction where Cl- in cisplatin [Pt(NH3)2Cl3] is replaced by H2O 30 Rate of change of conc of Pt compd Am't of cisplatin reacting (mol/L) = elapsed time (t) 2008 Concentrations & Rates Rate of change of conc of Pt compd Am't of cisplatin reacting (mol/L) = elapsed time (t) Rate of reaction is proportional to [Pt(NH3)2Cl2] We express this as a RATE LAW 31 Rate of reaction = k [Pt(NH3)2Cl2] where k = rate constant k is independent of conc. but increases with T 2008 To review Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A D[A] rate = Dt 32 B D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative. D[B] rate = Dt 2008 Concentrations, Rates, & Rate Laws In general, for 33 a A + b B + c C --> x X with a catalyst C Rate = k [A]m[B]n[C]p The exponents m, n, and p • are the reaction order • can be 0, 1, 2 or fractions • can only be determined by experiment! 2008 Reaction Rates and Stoichiometry 2A B 34 Two moles of A disappear for each mole of B that is formed. 1 D[A] rate = 2 Dt aA + bB D[B] rate = Dt cC + dD 1 D[A] 1 D[B] 1 D[C] 1 D[D] rate = == = a Dt b Dt c Dt d Dt 2008 Interpreting Rate Laws Rate = k [A]m[B]n[C]p 35 • If m = 1, rxn. is 1st order in A Rate = k [A]1 If [A] doubles, then rate goes up by factor of __ • If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Doubling [A] increases rate by ________ • If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate ________ 2008 Deriving Rate Laws Derive rate law and k for CH3CHO(g) --> CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO 36 Expt. [CH3CHO] (mol/L) 1 2 3 4 2008 Disappear of CH3CHO (mol/L•sec) 0.020 0.081 0.182 0.318 0.10 0.20 0.30 0.40 Deriving Rate Laws Rate of rxn = k [CH3CHO] 37 2 Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order. Now determine the value of k. Use expt. #3 data— 0.182 mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s) Using k you can calc. rate at other values of [CH3CHO] at same T. 2008 Concentration/Time Relations What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is 38 D[A] Rate = k [A] Dtime 2008 Cisplatin Concentration/Time Relations Integrating - (∆ [A] / ∆ time) = k [A], we get 39 [A] / [A]0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law. 2008 40 Write the rate expression for the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) D[CH4] D[CO2] 1 D[O2] 1 D[H2O] rate = = == Dt Dt Dt 2 Dt 2 2008 The Rate Law 41 The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A]x[B]y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall 2008 F2 (g) + 2ClO2 (g) 2FClO2 (g) 42 rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x=1 Quadruple [ClO2] with [F2] constant Rate quadruples y=1 2008 rate = k [F2][ClO2] Concentration/Time Relations Sucrose decomposes to simpler sugars 43 Rate of disappearance of sucrose = k [sucrose] If k = 0.21 hr-1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)? Glucose 2008 Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? 44 Use the first order integrated rate law 0.0010 M ln 0.010 M = - (0.21 hr-1) t ln (0.100) = - 2.3 = - (0.21 hr-1) • time time = 11 hours How many seconds? 2008 Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N2O5(g) ---> 4 NO2(g) + O2(g) Time (min) 0 1.0 2.0 5.0 [N2O5]0 (M) 1.00 0.705 0.497 0.173 ln [N2O5]0 0 -0.35 -0.70 -1.75 45 Rate = k [N2O5] 2008 Using the Integrated Rate Law 2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5] 46 Data of conc. vs. time plot do not fit straight line. 2008 Plot of ln [N2O5] vs. time is a straight line! Using the Integrated Rate Law Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b ln [N 2O5] = - kt + ln [N 2O5]o conc at time t rate const = slope conc at time = 0 47 All 1st order reactions have straight line plot for ln [A] vs. time. (2nd order gives straight line for plot of 1/[A] vs. time) 2008 48 Properties of Reactions page 719 2008 49 Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 2.2 x 10-4 1.1 x 10-4 2.2 x 10-4 1 2 3 0.08 0.08 0.16 0.034 0.017 0.017 rate = k [S2O82-]x[I-]y y=1 x=1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) 2.2 x 10-4 M/s rate k= = = 0.08/M•s 2-][I-] [S2O8 (0.08 M)(0.034 M) 2008 50 Definitions: 1st Order Reaction • First Order Reaction: The reaction that occurs first, not always the one desired. For example, the formation of brown gunk in an organic prep. 2008 51 Learn this equation & how to use [ N ]0 ln kt [ N ]t 2008 52 An Example • A reaction is 25% complete in 42 seconds. Calculate the half life. [ N ]t kt [ N ]0 [ N ]0 ln kt [ N ]t ln or Ln 1/.75=0.287, Div by 42 and k=6.84x10-3 T1/2 =0.693/k=101 s 2008 53 Half-Life Section 15.4 HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful. 2008 54 Half-Life • Reaction is 1st order decomposition of H2O2. 2008 55 Half-Life • Reaction after 1 half-life. • 1/2 of the reactant has been consumed and 1/2 remains. 2008 56 Half-Life • After 2 half-lives 1/4 of the reactant remains. 2008 57 Half-Life • A 3 half-lives 1/8 of the reactant remains. 2008 58 Half-Life • After 4 half-lives 1/16 of the reactant remains. 2008 Section 15.4 & Screen 15.8 Half-Life 59 Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction? 2008 Half-Life Section 15.4 60 Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the halflife of this reaction? Solution [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _________ Therefore, ln (1/2) = - k • t1/2 - 0.693 = - k • t1/2 t1/2 = 0.693 / k So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35 2008 min Can you perform this operation on your calculator? Half-Life Section 15.4 61 Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g 2008 Half-Life Section 15.4 62 Radioactive decay is a first order process. Tritium ---> electron + helium 3H 0 e 3He -1 t1/2 = 12.3 years If you have 1.50 mg of tritium, how much is left after 49.2 years? 2008 Half-Life Section 15.4 63 Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution ln [A] / [A]0 = -kt [A] = ? [A]0 = 1.50 mg t = 49.2 y Need k, so we calc k from: k = 0.693 / t1/2 Obtain k = 0.0564 y-1 Now ln [A] / [A]0 = -kt = - (0.0564 y-1) • (49.2 y) = - 2.77 Take antilog: [A] / [A]0 = e-2.77 = 0.0627 0.0627 = fraction remaining 2008 Half-Life Section 15.4 64 Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution [A] / [A]0 = 0.0627 0.0627 is the fraction remaining! Because [A]0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> 0.750 mg after 1 half-life ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.094 mg after 4 2008 Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 238U --> 234Th + He 4.5 x 109 y 14C --> 14N + beta 5730 y 131I --> 131Xe + beta 8.05 d Element 106 - seaborgium 263Sg 0.9 s 65 2008 A Microscopic View of Reactions Sections 15.5 MECHANISMS 66 Mechanism: how reactants are converted to products at the molecular level. RATE LAW ----> MECHANISM experiment ----> theory 2008 Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea. 67 Reaction coordinate diagram 2008 68 Activation Energy • The useful quantity of energy available in one cup of coffee 2008 MECHANISMS & Activation Energy 69 Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene] 2008 MECHANISMS Cis Transition state Trans 70 Activation energy barrier 2008 MECHANISMS Energy involved in conversion of trans to cis butene 71 energy Activated Complex -266 kJ 4 kJ/mol trans See Figure 15.14 +262 kJ cis 2008 Mechanisms • Reaction passes thru a TRANSITION STATE where there is an that has sufficient energy to become a product. 72 activated complex ACTIVATION ENERGY, Ea = energy req’d to form activated complex. Here Ea = 262 kJ/mol 2008 MECHANISMS Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis ---> trans is EXOTHERMIC This is the connection between thermodynamics and kinetics. 73 2008 Effect of Temperature • Reactions generally occur slower at lower T. Room temperature In ice at 0 oC 74 Iodine clock reaction, Screen 15.11, and book page 705. H2O2 + 2 I- + 2 H+ --> 2 H2O + I2 2008 Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. 75 In general, differences in activation energy cause reactions to vary from fast to slow. 2008 Mechanisms 1. Why is trans-butene <--> cis-butene reaction observed to be 1st order? As [trans] doubles, number of molecules with enough E also doubles. 2. Why is the trans <--> cis reaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T. 76 2008 More About Activation Energy Arrhenius equation — Rate constant Temp (K) 77 k Ae -Ea / RT Frequency factor Activation 8.31 x 10-3 kJ/K•mol energy Frequency factor related to frequency of collisions with correct geometry. Ea 1 ln k = - ( )( ) + ln A R T 2008 Plot ln k vs. 1/T ---> straight line. slope = -Ea/R More on Mechanisms A bimolecular reaction Reaction of 78 trans-butene --> cis-butene is UNIMOLECULAR - only one reactant is involved. BIMOLECULAR — two different molecules must collide --> products Exo- or endothermic? 2008 Collision Theory Reactions require (a) activation energy and (b) correct geometry. O3(g) + NO(g) ---> O2(g) + NO2(g) 79 1. Activation energy 2. Activation energy and geometry 2008 Mechanisms O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction. 80 2008 Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] NOTE 1. Rate law comes from experiment 2. Order and stoichiometric coefficients not necessarily the same! 3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction. 81 2008 Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Proposed Mechanism Step 1 — slow Step 2 — fast Step 3 — fast HOOH + I- --> HOI + OHHOI + I- --> I2 + OH2 OH- + 2 H+ --> 2 H2O 82 Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. Rate can be no faster than rds! 2008 Mechanisms 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Step 1 — slow HOOH + I- --> HOI + OHStep 2 — fast HOI + I- --> I2 + OH- 83 Step 3 — fast 2 OH- + 2 H+ --> 2 H2O Elementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should be Rate [I-] [H2O2] — as observed!! The species HOI and OH- are reaction intermediates. 2008 Rate Laws and Mechanisms NO2 + CO reaction: Rate = k[NO2]2 84 Two possible mechanisms Two steps: step 1 Single step Two steps: step 2 2008 Ozone Decomposition over Antarctica 85 2008 2 O3 (g) ---> 3 O2 (g) 86 Ozone Decomposition Mechanism 2 O3 (g) ---> 3 O2 (g) Proposed mechanism Step 1: fast, equilibrium O3 (g) → O2 (g) + O (g) Step 2: slow O3 (g) + O (g) ---> 2 O2 (g) 2008 [O 3 ]2 Rate = k [O 2 ] CATALYSIS Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier. Dr. James Cusumano, Catalytica Inc. 87 What is a catalyst? Catalysts and the environment Catalysts and society 2008 CATALYSIS In auto exhaust systems — Pt, NiO 88 2 CO + O2 ---> 2 CO2 2 NO ---> N2 + O2 2008 CATALYSIS 89 2. Polymers: H2C=CH2 ---> polyethylene 3. Acetic acid: CH3OH + CO --> CH3CO2H 4. Enzymes — biological catalysts 2008 CATALYSIS Catalysis and activation energy 90 MnO2 catalyzes decomposition of H2O2 2 H2O2 ---> 2 H2O + O2 Uncatalyzed reaction Catalyzed reaction 2008 Iodine-Catalyzed Isomerization of cis-2-Butene 91 Figure 15.16 2008 Iodine-Catalyzed Isomerization of cis-2-Butene 92 2008 93 Practice Exercises • 2 O3 (g) 3 O2 (g) give the relative rates of disappearance and formation of the reactant and product. 1 D O3 1 D O2 Reaction Rate = - =+ 2 Dt 3 Dt 2008 94 Practice Exercises 2 15.4 if -D(H2)/ Dt = 4.5x10-4 mol/L.min, what is D(NH3)/ Dt? • N2(g) + 3H2(g) 2NH3 (g) • D[NH3 ] D[H2 ] 2 mol NH3 4.5 10–4 mol/L 2 mol NH3 =– · = · = 3.0 10–4 mol/L min Dt Dt 3 mol H2 min 3 mol H2 2008 95 Practice Exercises 3 • 15.11 The reaction of 2NO(g) + O2(g) 2NO2(g) • Experiment [NO] [O2] • i 0.010 0.010 2.5 x 10-5 • ii 0.020 0.010 1.0 x 10-4 • iii 0.010 0.020 5.0 x 10-5 2008 96 Practice Exercises 3’ • What is the order of each reaction? Ans This rate change was the result of doubling the concentration of NO. The order of dependence of NO must be second order, Looking at the first and third data sets, [NO] is constant and [O2] doubles. The rate also doubles from the first data set to the third, so the reaction is first order in O2. Write the rate equation for the reaction, 2NO+O22NO2 • Rate = k[NO]2[O2] • Calculate the rate constant • 1 D[NO] 1 Rate = 2 Dt –5 –5 = (2.5 10 mol/L s) = 1.3 10 mol/L s 2 2008 97 Practice Exercises 3‖ • Thus Rate 1.3 10–5 mol/L s k= = = 13 L2 /mol2 s [NO]2 [O2 ] (0.010 mol/L)2 (0.010 mol/L) • What is the rate in mol/L.s when (NO)=0.015 mol/L and (O2)=0.0050 mol/L? • Rate = k[NO]2[O2] = (13 L2/mol2·s)(0.015 mol/L)2(0.0050 mol/L) = 1.4 10–5 mol/L·s • When NO is reacting at 1.0x10-4mol/L.s, what is the rate that O2 is reacting and NO2 is forming? 2008 98 Practice Exercises 3‖’ • Rate at which O2 is reacting = 1.0 10–4 mol/L 1 mol O2 · = 5.0 10–5 mol/L s s 2 mol NO • Rate at which NO2 is forming= 1.0 10–4 mol/L 2 mol NO2 · = 1.0 10–4 mol/L s s 2 mol NO 2008