CHM 116 Ch 14-Kinetics

CHM 116 Ch 14-Kinetics - Overview Chapter 14 Kinetics...

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Unformatted text preview: Overview Chapter 14 Kinetics (Rates of Reactions) Chapter 15 Equilibrium Chapter 19 Thermodynamics Later chapters are applications of these topics We will look at these topics at both the macroscopic level and the microscopic level. 1 Chapter 14 Chemical Kinetics Kinetics tells us how fast the reaction will go. Kinetics is the study of rates of reactions factors that affect these rates: concentration, temperature, pressure, catalysis, surface area, solvent, environment... 2 12.1 Reaction Rates Reaction rate is a measure of how fast a reaction occurs. Some reactions are inherently fast and some are slow: Effect of Concentration Changing the concentration of a reactant can change the reaction rate: NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) HCl(aq) NaCl(aq) O(l 3 4 Effect of Surface Area Effect of Temperature A light stick glows brighter at higher temps, but it will run out sooner. Milk sours more quickly when left at room temperature. Food cooks more quickly at higher temps. 5 6 Iron Nail Steel wool Rusting Fast Rusting Slow 1 1 Catalysis The catalyst called catalase in this piece of liver causes the decomposition of H2O2 to occur faster. 14.1 Reaction Rates Must be able to measure rates to determine how to control them with concentration or temperature. H2 O 2 decomposition 7 8 14.1 Reaction Rates Rate = conc/time or -conc/time so it conc/ conc/ will have a positive value For A B A Rate = - A B = time time Rates and Stoichiometry Use reaction stoichiometry to specify reference points for rates. When reacting in ratios not equal to 1:1, one substance disappears or appears faster than the other. 5Br- (aq) + BrO3- (aq) + 6H+ (aq) 3Br2(aq) + 3H2O(l) aq) aq) aq) Rate Measurement 9 Br- disappears 5 times as fast as BrO3- [BrO3-] -1 [Br-] -1 [H+] +1 [Br2 ] -------- = -- ------ = -- ------ = -- ------ t 5 t 6 t 3 t Can't measure change in H2O because its Can' concentration is so high, it stays constant. 10 0.085 5Br- (aq) + BrO3- (aq) + 6H+ (aq) 3Br2(aq) + 3H2O(l) Reaction Rates Write equations for the rate of this reaction in terms of each reactant and product. 2N2O5(g) 4NO2(g) + O2(g) [Br-] [BrO3-] 0.017 [Br2] 0.051 11 12 2 2 Calculation of Rates 2N2O5(g) 4NO2(g) + O2(g) Start with 2.330 M solution at measure the concentration periodically. -1 ([N2O5]2 - [N2O5 ]1) -1 [N2O5] Rate = -- -------- = -- ---------------- 2 t 2 (t 2 - t 1) Rate during time period 52,400 to 72,000 sec, where the concentrations are measured to be 1.350 M and 1.110 M. 13 2N2O5 4NO2 + O2(g) Time (s) [N2O5] (M) 45oC; 0 11100 19200 31600 *52400 *72000 112700 138900 188600 2.33 2.08 1.91 1.67 1.35 1.11 0.72 0.55 0.34 14 Calculation of Rates Rate = 1/2( [N2O5] / t) 1/2( t = 72,000 52,400 = 19,600 sec [N2O5] = (1.110 M 1.350 M) = 0.240 M Rate = 1/2 (240 M/19,600 s) = 6.12 106 M/s 15 Rates The rate changes as the reaction proceeds. Usually it gets smaller with time. The rates calculated are average rates for the particular time interval. 16 Rates Can get average rate and instantaneous rate from a graph of concentration vs. time Initial Rates To get around the problem of the rate changing with time, we sometimes measure initial rates. rates. 17 18 3 3 Initial Rates If the reaction is not too fast, then -C/t is a C/ good approximation to the rate. We know concentration values much better at the beginning of the reaction than we do later.r different concentrations. What is the later. relationship between initial rate and concentration? 19 14.2 The Dependence of Rate on Concentration Rates usually vary with initial concentrations of some or all reactants, possibly products. Eg. 2 g Na2CO3 + 50 mL HCl in a flask Eg. enclosed with a balloon. Which reaction is fastest? Effect of Conc Effect of SA 1M 3M 6M 20 Rate and Concentration Why does rate generally decrease as the reaction proceeds? concentration of reactants decreases fewer molecules to collide, so collision frequency decreases Rate Laws A + B C + D We describe the dependence of rate on concentration by an equation called the rate law: Rate = k[A]x[B]y... where x & y are not stoichiometric coefficients k, x, y must be determined experimentally k is the rate constant x,y are reaction orders for the individual species x+y+... = overall reaction order (n) 21 22 Rate Law Rate = k[A]x[B]y... Orders (x, y, ...) are usually positive or negative integers Each order tells us how the rate depends on that concentration; this order must be determined experimentally; only occasionally experimentally; does it match the coefficients from the balanced reaction equation 23 How do we Determine Rate Laws? Method 1: When rate is dependent upon only 1: one species. 2N2O5 4NO2 + O2(g) [N2O5] [NO2] [O2] Rate 2.21 2.21 2.21 2.21 2.00 1.00 1.00 0.00 1.00 1.00 2.00 0.00 1.14 x 10-5 1.13 x 10-5 1.12 x 10-5 1.13 x 10-5 The rate is constant when we change [NO2] and [O2], so orders are 0 (not in rate law) 24 4 4 How do we determine rate laws? Thus, Rate = k[N2O5]x Determine Reaction Order Rate = k[N2O5]x What is the order? (What is the value of x?) [N2O5] Rate (M/s) Vary [N2O5] while measuring rate. Since k should have a constant value, calculate: k = Rate k = Rate/[N2O5] k = Rate/[N2O5]2 And see which is constant 2.00 1.79 1.51 1.23 0.92 25 1.05x10 -6 9.67x10 -6 7.83x10 6.31x10-6 -6 4.81x10 26 -5 How do we determine rate laws? Method 2: If the rate depends on the concentration of more than one substance, design the experiment so only one concentration changes appreciably. Vary one concentration at a time and measure the rate. Compare one variable at a time. Group Work Derive the rate law and value of k for the reaction, F2 + 2ClO2 2FClO2 [F2]o [ClO2]o Rate (M/s) Exp 1 Exp 2 Exp 3 0.100 0.100 0.200 0.0100 0.0400 0.0100 0.0012 0.0048 0.0024 R2/R1 = 4 and C2/C1 = 4, so 1st order in ClO2 R3/R1 = 2 and C3/C1 = 2, so 1st order in F2 Rate = k[F2][ClO2] and k = 1.2 M-1s-1 27 28 Units on k The units on k depend on the overall order for the reaction. Determine the units using dimensional analysis. What are the units on k for: Zero Order 1st Order 2nd Order 29 Group Work Determine the rate law and the value of the rate constant for the following reaction: 2I-(aq) + 2VO2+(aq) + 4H+(aq) I2(aq) + 2VO2+(aq) + 2H2O(l) [I-] 0.00200 0.00400 0.00200 0.00200 [VO2+] 0.0100 0.0100 0.0200 0.0100 [H+] 0.100 0.100 0.100 0.0500 Rate 26.0 52.1 51.9 6.50 30 Exp 1 Exp 2 Exp 3 Exp 4 5 5 Group Work Determine the rate law and the value of the rate constant for the following reaction: 2I-(aq) + 2VO2+(aq) + 4H+(aq) I2(aq) + 2VO2+(aq) + 2H2O(l) [I-] [VO2+] [H+] Rate 0.00200 0.0100 0.100 26.0 Exp 1 0.00400 0.0100 0.100 52.1 Exp 2 0.00200 0.0200 0.100 51.9 Exp 3 0.00200 0.0100 0.0500 6.50 Exp 4 0.00200 0.0100 0.150 ? Exp 5 31 Rate Laws Rate Laws in the form Rate = k[A]x[B]y tell us about the rate of a reaction at a specific instant, and about the dependence of initial concentration on intitial rate. This form of the rate law does not tell us information about time, such as "how long it will take for a specific concentration to react completely, or half way, ..." Integrated rate laws will give us this information. 32 14.4 The Change of Concentration with Time How do we determine rate laws? The dependence of concentration on reaction rate for each reactant is either zero order, 1st order, or 2nd order (usually). Zero Order: Rate = -[A]/t = k[A]0 = k [A]/ 1st Order: Rate = -[A]/t = k[A]1 [A]/ 2nd order: Rate = -[A]/t = k[A]2 [A]/ Converting these equations to linear equations as a function of time will allow us to graph the three possibilities (Method 3) 33 14.3 The Change of Concentration with Time How do we determine rate laws? Zero Order in A: (n=0) Rate = -[A]/t = k[A]0 = k [A]/ -[A]/t = k [A]/ Rearranging and taking the integral of both sides gives the integrated rate law: law: [A] = [A]o - kt Example: Mg + 2HCl MgCl2 + H2 34 How do we determine rate laws? If zero order ([H2]/t = k), a graph of [H2] ( ]/ vs. time should be linear. [H2] = [H2]o + kt Why +kt in +kt this case? How do we Determine Rate Laws? 1st Order in A (n=1) Rate = -[A]/t = k[A] [A]/ Rearranging and taking the integral of both sides gives the integrated rate law: law: ln [A] = ln [A]o - kt 35 36 6 6 How do we Determine Rate Laws? 1st Order in A (n=1) ln [A] = ln [A]o - kt 2N2O5 4NO2 + O2(g) If n = 1, a graph of ln [N2O5] vs. time should be linear. How do we determine rate laws? 2nd Order in A (n=2) Rate = -[A]/t = k[A]2 [A]/ Rearranging and taking the integral of both sides gives the integrated rate law: law: 1/[A] = 1/[A]o + kt 37 38 How do we determine rate laws? 2nd Order in A (n=2) 1/[A] = 1/[A]o + kt 2UO2+ + 4H+ U4+ + UO22+ + 2H2O If n = 2, a graph of 1/[UO2+] vs. time should be linear. How do we determine rate laws? Summary: n = 0 -[A]/t = k [A]/ n = 1 -[A]/t = k[A] [A]/ n = 2 -[A]/t = k[A]2 [A]/ [A] = [A]o - kt ln [A] = ln [A]o - kt 1/[A] = 1/[A]o + kt Method 3 for determining rate law: Graph [A] vs. t, ln [A] vs. t, and 1/[A] vs. t and see which graph is linear. 39 40 What is the order? What is the order? 41 42 7 7 How do we determine rate laws? Method 4: Measure half-life (t1/2) - the time 4: halfto consume half of the current reactant - reduce its concentration to 1/2 How do we determine rate laws? Half Lives Zero Order Half Life n = 0, [A] = [A]o - kt [A] - [A]o = -kt [A]o - [A] = kt At t1/2, [A] = 1/2[A]o [A]o - 1/2[A]o = k t1/2 1/2[A]o = k t1/2 t1/2 = [A]o/2k t1/2 decreases as the reaction proceeds 43 44 How do we determine rate laws? 1st Order Half Life n = 1, ln[A] = ln[A]o - kt ln[A] ln[A] - ln[A]o = -kt ln[A] ln[A]o - ln[A] = kt ln[A] ln ([A]o / [A]) = kt At t1/2, [A] = 1/2[A]0 ln ([A]o / 1/2[A]o) = k t1/2 ln2 = k t1/2 t1/2 = ln 2/k = 0.693/k t1/2 is constant throughout the reaction 45 How do we determine rate laws? 2nd Order Half Life n = 2, 1/[A] = 1/[A]o + kt Rearrange and substitute [A] = 1/2[A]0 at t1/2 1/[A]o = kt1/2 n = 2, t1/2 = 1/k[A]o t1/2 increases as the reaction proceeds 46 Summary of Half-Life Equations n = 0, t1/2 = [A]o/2k t1/2 decreases as the reaction proceeds Group work: 2N2O5 4NO2 + O2 1st order Approximate the k Value n = 1, t1/2 = ln 2/k = 0.693/k t1/2 is constant throughout the reaction n = 2, t1/2 = 1/k[A]o t1/2 increases as the reaction proceeds Use these equations to calculate k or t1/2 from one another. Often used for radioactivity - Chapter 21 47 48 8 8 First-Order Problem Group Work The rate constant (k) for a 1st order reaction was determined to be 0.0462 s-1. Predict how long it should take for the reaction to be a) 50% complete (1/2 unreacted) unreacted) b) 75% complete (1/4 unreacted) unreacted) c) 87.5% complete (1/8 unreacted) unreacted) Types of Problems Use integrated rate laws to calculate: half life from k k from half life Conc., after time t, given initial conc. time it takes for a certain amount to react 49 50 Group Work Consider the decomposition of 3.00 M H2O2, which follows first order kinetics with a rate constant of 0.042 min-1 at 70C. 70 1. What is the half life? 2. What conc. of H2O2 is left after 40.0 min? 3. How long will it take for the conc. of H2O2 to decrease to 0.100 M? 14.4 Temperature and Rate Examples N2 and O2 coexist in air, but combine at high temperatures (in auto engines) to form NOx Rates generally increase with increasing T 2Cr3+ + 6Mn3+ + 7H2O CrO42- + 6Mn2+ + 14H+ 51 cold hot cold hot 52 Rates and Temperature k increases with temperature, causing the increase in rate 2NO2 2NO + O2 Rate = k[NO2]2 Often k increases ~ x 2 for 10oC T increase 53 Rates and Temperature What must the reactants do in order for a reaction to occur? Consider the following reaction as an example: 2O3(g) 3O2(g) 54 9 9 Collision Theory explains why k increases with T Collision Theory: For reaction to occur, 1. molecules must collide; more collisions higher collide; rate 2. colliding molecules must be oriented properly to react 3. properly oriented colliding molecules must have sufficient energy (which explains the T dependence) (The average kinetic energy of molecules increases with temperature.) Gas simulation Collision Theory Not all collisions lead to products. 55 56 Group Work NO + O3 O2 + NO2 Average Kinetic Energy The number of molecules with sufficient kinetic energy to react increases as the temperature increases. Which collisions are likely to be effective? Not all molecules have the same kinetic energy. 57 58 Activation Energy Why don't all properly-oriented collisions lead don' properlyto products? There must be sufficient energy to break reactant bonds. The activation energy, Ea, is the minimum energy required to initiate a chemical reaction. Ea is often referred to as the energy barrier for a reaction. 59 Energy Barrier 60 10 10 Energy-Reaction Coordinate Activation Energy CH3NC CH3CN Ea Explains Methane Combustion 61 62 Activation Energy The energy barrier is called the activation energy, Ea. This is the difference between the transition state energy and the average reactant energy. Related to the energy change of the reaction: H = Ea,f - Ea,r O3 + NO O2 + NO2 H = -199.8 kJ/mol, Ea,f = 10.7 kJ/mol What is Ea,r ? Ea,r = 10.7 - (-199.8) = 210.5 kJ/mol 63 O3 + NO O2 + NO2 250 Ea,f=10.7 kJ 200 150 E 100 Ea,r=210.5 kJ 50 0 Reaction Coordinate 64 O3 + NO 250 O2 + NO2 250 O3 + NO O2 + NO2 200 200 150 150 E 100 E 100 50 50 0 0 Reaction Coordinate 65 Reaction Coordinate 66 11 11 O3 + NO O2 + NO2 250 250 O3 + NO O2 + NO2 200 200 150 150 E 100 E 100 50 50 0 0 Reaction Coordinate 67 Reaction Coordinate 68 O3 + NO O2 + NO2 250 250 O3 + NO O2 + NO2 200 200 150 150 E 100 E 100 50 50 0 0 Reaction Coordinate 69 Reaction Coordinate 70 Activation Energy Exothermic reaction Activation Energy Endothermic reaction 71 72 12 12 Arrhenius Equation Relates k to T and Ea and is used to calculate Ea k = A e-Ea/RT A is frequency factor, T in Kelvins, R = 8.314 J/mol K 14.5 Reaction Mechanisms Mechanism is a detailed pathway for a reaction Proposed mechanism must be consistent with the rate law and other experimental evidence Consider two possible mechanisms for the reaction: SN1 or SN2 73 Measured experimentally by measuring k at different temperatures ln k = lnA - Ea/RT Graph ln k vs. 1/T CH3Cl + OH- CH3OH + Cl- 74 Mechanisms Usually have several mechanisms consistent with a given rate law Mechanism identifies the elementary reactions (one-step reactions) that combine to (onemake up the overall reaction. Elementary steps must add up to give the balanced chemical equation. Molecularity: number of reactant molecules Molecularity: in an elementary reaction unimolecular, bimolecular, termolecular (more is rare) 75 Mechanisms Overall reaction may be elementary: CH3NC CH3CN Rate = k[CH3NC] For an elementary reaction orders = coefficients O3 + NO O2 + NO2 Rate = k[O3][NO] This reaction may be elementary. 76 Multistep Mechanisms Many rate laws are more complex, so the overall reaction cannot be an elementary reaction. These have multistep mechanisms. One step in multistep mechanism is slower than the others called the rate-determining ratestep. Multistep Mechanisms Tl3+ + 2Fe2+ Tl+ + 2Fe3+ Rate = k[Tl3+][Fe2+]2/[Fe3+] Tl3+ + Fe2+ Tl2+ + Fe3+ fast Tl2+ + Fe2+ Tl+ + Fe3+ slow (rate determining) Intermediate: substance produced in one step and consumed in a later step (Tl2+) Ozone mechanism Hydrazine mechanism 77 78 13 13 Multistep Mechanisms Group Work S2O82- + I- 2SO42- + I+ I+ + I- I2 Which step is rate determining? What is the intermediate? 1st step; I+ 79 Multistep Mechanisms Group Work slow fast S2O82- + I- 2SO42- + I+ I+ + I- I2 slow fast S2O82- + 2I- 2SO42- + I2 (overall or net equation) 80 Group Work O3(g) + Cl(g) ClO(g) + O2(g) Cl(g ClO(g ClO(g) + O3(g) Cl(g) + 2O2(g) Cl(g ClO(g Identify any intermediates. Write equation for the overall (net) reaction. 14.6 Catalysis Catalyst: substance that changes the rate of reaction but is not permanently changed itself Metals catalyze many reactions of gases; called heterogeneous catalysis Mechanism: absorption, reaction, desorption H2 O 2 decomposition 81 82 Heterogeneous Catalysis Catalysis Catalyst changes the reaction mechanism and lowers the activation energy C2 H4 + H 2 C2 H6 Catalytic Converter: NOx to N2 and O2 CO to CO2 and O2 CO + O2 83 84 14 14 Homogeneous Catalysis In homogeneous catalysis, everything is in solution (or in gas phase) Identify the intermediate(s) and catalyst(s) in the following mechanism NO + O2 NO3 NO3 + NO 2NO2 NO2 + SO2 NO + SO3 NO2 + SO2 NO + SO3 -----------------------------------2SO2 + O2 2SO3 Homogeneous Catalysis In homogeneous catalysis, everything is in solution (or in gas phase) Identify the intermediate(s) and catalyst(s) in the following mechanism NO + O2 NO3 NO3 + NO 2NO2 NO2 + SO2 NO + SO3 NO2 + SO2 NO + SO3 -----------------------------------2SO2 + O2 2SO3 85 86 Enzymes as Catalysts Enzymes have cavities that just fit the reacting molecules and act to bring the reactants together in the proper orientation. anim 87 15 15 ...
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This note was uploaded on 04/16/2008 for the course CHM 116 taught by Professor Unknown during the Spring '08 term at ASU.

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