1/13/2015 14.1 Reaction Rates Ch. 14: Chemical Kinetics Chemical Kinetics is the study of how fast reactions take place. Reactions can be fast (reaction of sodium metal with water) or very slow (rusting of iron). Why do you need to study kinetics? Drug delivery: time-release capsules Pollution control Food processing Chemistry: Atoms First Julia Burdge & Jason Overby 1 Reaction Rates 2 14.2 Collision Theory Reaction rate (or speed) = change in amount of Most reactions occur as a result of collisions substance per unit of time. between reacting molecules. Collision theory: rate of reaction is directly proportional to the number of molecule collisions per time. What factors can we change in a lab to alter See Figure 14.4 (pp. 554-555) the rate of a reaction? Concentration Temperature Catalysts Surface area of solids 3 Collision Theory Effective Collisions: molecules must collide with sufficient kinetic energy and in the correct orientation in order to make products. Cl + NOCl → Cl2 + NO Correct orientation to facilitate reaction An effective collision results in reaction. 5 4 Collision Theory Ineffective collisions: incorrect orientation of molecules will NOT result in the formation of products; molecules will simply bounce off of each other. Incorrect orientation does not favor reaction An ineffective collision results in no reaction.6 1 1/13/2015 Activation Energy Activation Energy Activation Energy (Ea): energy barrier Only a small fraction of molecules have enough molecules have to overcome in order to react. The high energy state is the transition state. kinetic energy to overcome the activation energy. Ea is different for each reaction. Reactions with low Ea are faster because more molecules can overcome the Ea. At higher temperatures, a larger fraction of molecules have enough kinetic energy to overcome Ea. 7 Reaction Energy Profile 8 Reaction Energy Profile Is the reaction shown on the energy profile Enthalpy (H) is heat of reaction Endothermic: Energy absorbed H > 0 Exothermic: Energy released H < 0 below endo- or exothermic? What is the value of Ea? 9 14.3 Measuring Reaction Progress and Expressing Reaction Rate 10 Reaction Rates Reaction: A B For the reaction: A B, we can measure rate of disappearance of reactants: [A] decreases ( - sign) and appearance of products: 11 12 2 1/13/2015 Reaction Rates Reaction Rates Br2(aq) + HCOOH(aq) → 2Br –(aq) + 2H+(aq) + CO2(g) Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g) First 50 seconds: First 100 seconds: 13 14 Reaction Rates N2O5(g) NO2(g) + Reaction Rates O2(g) Rates are related by coefficients. O2 will appear ¼ as fast as NO2. Can use coefficients as mol-mol ratio. 2 N2O5(g) 4 NO2(g) + O2(g) This is called the “rate expression”. We can use this expression (or equation) to determine the rate of each substance and the rate of reaction over a given time period. If N2O5 disappears at - 0.20 M/s, what is the rate of appearance of NO2? Of O2? Two methods: Stoichiometry and rate expression What is the rate of reaction? 15 16 Reaction Rates Reaction Rates Write rate expressions for the following: Write the rate expression for the following reaction: 2HI(g) H2(g) + I2(g) CO2(g) + 2H2O(g) → CH4(g) + 2O2(g) 5Br - (aq) + BrO3- (aq) + 6H+ (aq) 3Br2(aq) + Solution: Use the equation below to write the rate expression. 3H2O(l) C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) SiO3 + 4C SiC + 3CO Worked examples: 14.1, 14.2 17 18 3 1/13/2015 Reaction Rates – Practice! Measuring Reaction Rates 5Br - (aq) + BrO3- (aq) + 6H+ (aq) 3Br2(aq) + 3H2O(l) Average Rate: rate of substance over a given time interval Rate = (M2 – M1) / (t2 – t1) If the rate of appearance of Br2 is 1.2 x 10-3 M/s, what is the rate of disappearance of Br -? 1.2 x 10-3 M/s * (5 mol Br -/3 mol Br2) = - 2.0 x 10-3 M/s Instantaneous rate: rate at a specific time Determined by measuring slope of tangent at a point C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g) If rate of appearance of ethanol is 4.2 x 10-2 M/s, what is rate Initial rate: rate at time = 0 Determined by measuring slope of tangent from t = 0 of disappearance of glucose? - 2.1 x 10-2 M/s SiO3 + 4C SiC + 3CO These rates will all have different values. Rates are not If the rate of appearance of CO is +0.78 M/s, what is the rate constant; they change over time. If they were constant, graphs would be straight lines. with respect to C? +1.0 M/s 19 20 14.4 Dependence of Reaction Rate on Reactant Concentration – Rate Laws Measuring Reaction Rates Rate law relationship between concentration of each Initial reactant and the reaction rate. Instantaneous (pink) Average (orange and blue) aA + bB cC + dD rate = k[A]x[B]y Rate is the rate of reaction k is the rate constant [A] and [B] are reactant concentrations x and y are orders of reactants (NOT necessarily the same as coefficients). 21 22 Rate Laws Rate Laws 5Br Rate Laws have to be determined from experimental data (concentrations and reaction rates), using initial rates (at time = 0) Br2(aq) + HCOOH(aq) 2Br - (aq) + 2H+ (aq) + CO2(g) rate = OR rate= k[Br2] k[Br2]1[HCOOH]0 - (aq) + BrO3- (aq) + 6H+ (aq) 3Br2(aq) + 3H2O(l) Experiment [Br-](M) [BrO3-](M) [H+](M) Initial Rate (M/s) 1 0.10 0.010 0.15 1.2 x 10–3 2 0.10 0.020 0.15 4.8 x 10–3 3 0.30 0.010 0.15 3.6 x 10–3 4 0.10 0.020 0.30 4.8 x 10–3 Need two trials where only one reactant concentration changes (all others are constant). Compare change in rates to change in concentration to 23 determine the order of the reactant (the exponent in the rate law). 24 4 1/13/2015 Rate Laws 5Br - (aq) + BrO3- (aq) + 6H+ (aq) Rate Laws 3Br2(aq) + 3H2O(l) Experiment [Br-](M) [BrO3-](M) [H+](M) Initial Rate (M/s) 1 0.10 0.010 0.15 1.2 x 10–3 2 triples 0.10 0.020 0.15 3 0.30 0.010 [H+] 0.15 [BrO3-] constant constant 0.020 0.10 0.30 [Br-] 4 [Br-] triples, rate triples: Br- is 1st Rate triples Shortcut for determining orders: If the concentration doubles and the rate doesn’t change, then the concentration has no effect on rate and the reactant is 0 order If the concentration doubles and the rate doubles, then the reactant is 1st order. What would happen to rate if the concentration of a 1st order reactant tripled? If the concentration doubles and rate quadruples, then the reactant is 2nd order. What would happen to rate if the concentration of a 2nd order reactant tripled? 4.8 x 10–3 3.6 x 10–3 4.8 x 10–3 order (x = 1) Which two trials should be used for BrO3-? Which two for H+? 25 26 Rate Laws Rate Laws Experiment [Br-](M) [BrO3-](M) [H+](M) Initial Rate (M/s) 1 0.10 0.010 0.15 1.2 x 10–3 2 0.10 0.020 0.15 4.8 x 10–3 3 0.30 0.010 0.15 3.6 x 10–3 4 0.10 0.020 0.30 4.8 x 10–3 1) A rate law is an equation that is always defined in terms of reactant concentrations, never products. 2) The exponents (orders of reactants) in a rate law must be determined from a table of experimental data. rate = k[Br-][BrO3-]2 3) Comparing changes in individual reactant concentrations to changes in rate allows you to find the order of the reactant. Now we can solve for the rate constant. How? What information do we need to solve for k? Rate = 120 M-2s-1 [Br-][BrO3-]2 Three important things to remember about rate laws: 27 Exp’l Determination of Rate Laws Determine the rate law and calculate the rate constant (with units): 2NO + Cl2 2NOCl Experiment [NO] [Cl2] Rate (M/s) 1 2 3 4 5 0.10 M 0.10 M 0.10 M 0.20 M 0.30 M 0.10 M 0.20 M 0.30 M 0.30 M 0.30 M 1.7x10-4 6.7x10-4 1.5x10-3 3.0x10-3 4.5x10-3 Rate = 0.17 M-2s-1 [NO] [Cl2]2 28 Rate Laws – Group Practice S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq) Determine the rate law and calculate the rate constant, including its units. Experiment [S2O82–](M) [I–](M) Initial Rate (M/s) 1 0.080 0.034 2.2 x 10–4 2 0.080 0.017 1.1 x 10–4 3 0.16 0.017 2.2 x 10–4 rate = 0.081 M-1s-1 [S2O82–] [I–] 30 5 1/13/2015 14.5 Dependence of Reactant Conc. on Time – Integrated Rate Laws Rate Laws – Practice Determine the rate law and calculate the rate constant for: 2O3(g) 3O2(g) [O3] (M) 0.00600 0.00300 0.00150 In a 1st order ( A B ), Rate (M/s) 5.03 x 10-7 1.28 x 10-7 3.08 x 10-8 Rate also can be written, rate = k[A] Setting them equal: What is the rate if [O3] = 0.00500 M? Hint: What other information is needed to solve Integrate to get 1st order integrated rate law: this? Rate = 0.0140 M-1s-1 [O3]2; rate = 3.49 x 10-7 M/s Worked Example 14.3 31 Integrated Rate Laws 32 First Order IRL First order integrated rate law: • ln is the natural logarithm In linear format: ln[A]t = –kt + ln[A]0 y = mx + b Math note: ln (5/7) = ln 5 - ln 7 ln (x) = 0.23; What is x? x = 1.3 33 IRL – Half Life 34 Second Order IRL Two options for rates of second order reactions: Rate = k[A]2 OR Rate = k[A][B] Half-life: the time required for the amount of a reactant to decrease by one half (for 1st order). 2nd option is too complex, we’ll only work with the first example. Rate = k[A]2 = -[A]/t Integrate: 1/[A]t = kt + 1/[A]o (linear eqn) 35 36 6 1/13/2015 Second Order IRL IRL – Summary Table 37 Integrated Rate Laws – Graphing 38 IRL – Practice The half life for the reaction 2A → B is 8.06 Make three plots: [A] vs. time (zero order) ln [A] vs time (first order) 1/[A] vs. time (second order) minutes at 110oC. The reaction is first order in A. How long will it take for [A] to decrease from 1.25 M to 0.71 M? The one that is most linear determines the k = 8.6x10-2 min-1; t = 6.58 minutes order of the reactant (the other graphs should be curves). 39 IRL – Practice 40 IRL – Practice If the half life is 18.2 min for a first order reactant A second order reaction has a half life of 699 with initial concentration of 2.4 M, what is the concentration of A after 7.5 minutes? k = 0.0381 min-1, [A]t = 1.8 M seconds. If the initial reaction concentration is 0.0355 M, what will the reactant concentration be after 855 seconds? k = 0.040239 M-1s-1, [A]t = 0.0160 M If a first order reaction has a rate constant of 4.73 x 102 s-1, how long would it take for 75% to react? t = 2.93 x 10-3 s Worked Example: 14.7 Worked Examples 14.4, 14.5, 14.6 41 42 7 1/13/2015 14.6 Dependence of Reaction Rate on Temperature Arrhenius Equation The Arrhenius Equation can be used to relate the The initial equation can be modified to give a temperature of a reaction to the rate constant. linear variation. Ea is the activation energy (in kJ/mol). R is the gas constant (8.314 J/mol K). T is the absolute temperature (in K). A represents the collision frequency and is called the frequency factor (a constant for each reaction). This variation works well for sets of 4 or more data points. 43 Worked Example 14.8 Arrhenius Equation • Worked Example 14.8: Rate constants for the reaction CO(g) + NO2(g) → CO2(g) + NO(g) were measured at four different temperatures. The data are shown in the table. Determine the activation energy for the reaction. T (K) 288 298 308 318 k (M-1·s-1) 0.0521 0.101 0.184 0.332 1/T (1/K) 0.00347 0.00336 0.00325 0.00314 44 Solution 0 -0.5 -1 -1.5 -2 ln k -2.95 -2.29 -1.69 -1.10 -2.5 y = -5590.9x + 16.47 -3 -3.5 0.0031 m = -Ea / R 0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035 Ea = m * -R = -5590.0 K * -8.314 J / K·mol Ea = 46,482 J/mol = 46.5 kJ/mol Worked Example 14.9, 14.10 45 46 Worked Example 14.10 Arrhenius Equation A certain first-order reaction has an activation energy of 83 kJ/mol. If the rate constant for this reaction is 2.1×10-2 s-1 at 150oC, what is the rate constant at 300oC? The Arrhenius Equation can also be used to determine the rate constant after a change in temperature (i.e., 2 data points) or to find Ea given two k’s and T’s: Strategy Rearrange and solve for k2 using the following k1 k2 = E 1 1 a R e T2 – T2 Ea = 8.3×104 J/mol, T1 = 423 K, T2 = 573 K, R = 8.314 J/K·mol, and k1 = 2.1×10-2 s-1. Solution 2.1×10-2 s-1 k2 = e 47 8.3×104 J/mol 8.314 J/K·mol 1 1 – 573 K 423 K The rate constant of 300oC is 1.0×101 s-1. = 1.0×101 s-1 48 8 1/13/2015 14.7 Reaction Mechanisms Reaction Mechanisms A balanced chemical equation does not indicate how a reaction actually takes place. Given elementary steps in a mechanism, we can: A chemical reaction takes place through a series of simpler steps called elementary steps (typically 2 or more). These elementary steps determine the step-wise reaction mechanism for the overall reaction. Determine the molecularity of each step Write the overall (or net) equation Use relative rates (slow or fast) to identify the ratedetermining step Identify the intermediate (if it exists) Is produced and then consumed in a later step Identify the catalyst (if it exists) Is consumed and then produced in a later step (recycled) 49 Write the rate law of each step Reaction Mechanisms Multistep Mechanisms Each step in a mechanism is defined by its molecularity: the number of reactant molecules. One step in a multistep mechanism will be slower unimolecular (one reactant molecule) Step 1 (slow): Br2 + NO Br2NO A → products rate = k[A] 50 than the others. The slow step is the ratedetermining step (RDS). 1st order Step 2 (fast): Br2NO + NO 2BrNO bimolecular (two reactant molecules) A + B → products rate = k[A][B] 2nd order 1) Write the overall equation. A + A → products rate = k[A]2 2nd order 2) What is the molecularity of each step? termolecular (three reactant molecules) – rare! 3) Which step is the rate-determining step? 4) What is the intermediate? The catalyst? 51 52 Rate Laws: Elementary Steps Rate Laws: Mechanisms For elementary steps, the rate law can be written Group Work: S2O82- + I- 2SO42- + I+ from the coefficients in the equation. The molecularity of the step = the overall order of the step. I+ + I- I2 Step 1 (slow): 2NO2 (g) NO3(g) + NO(g) Step 2 (fast): NO3(g) + CO(g) NO2(g) + CO2(g) Determine: overall equation, molecularity of each step, RDS, intermediate, and catalyst. slow fast Which step is rate determining? What is the intermediate? What is the rate law for each step? What is the overall rate law? Worked Example 14.11, 14.12 Write the rate law for each step. 53 54 9 1/13/2015 Rate Laws: Mechanisms Rate Laws: Mechanisms If the 1st step is slow, the rate law of the overall reaction Step 1: is the same as the rate law of the first step. If the 2nd step is slow, rate laws are more complicated to determine. (slow) Step 2: H2O2 + I– H2O2 + Overall reaction: Step 1: Tl3+ + Fe2+ Tl2+ + Fe3+ 2H2O2 Step 2: Tl2+ + Fe2+ Tl+ + Fe3+ 2H2O + O2 55 56 14.8 Catalysis Catalysis A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. H2O2 + I– H2O2 + IO– Overall reaction: H 2O + O 2 + I – Rate = k[H2O2][I–] Which step is rate-determining? Second Step 2: H2O + IO– Which step is slow? Overall rate law: Step 1: (slow) IO– 2H2O2 Energy profile without a catalyst (left) and with a catalyst (right). Figure 14.17. H2O + IO– H2O + O2 + I– 2H2O + O2 A catalyst speeds up a reaction by providing a set of elementary steps with more favorable kinetics than those that exist in its absence; it usually does this by lowering the activation energy. 57 58 Catalysis Catalysis Catalysts can be classified as Catalysts can also be classified as Heterogeneous: the catalyst and reactants are in different phases (e.g., catalytic converter) Homogeneous : the catalyst and reactants are in the same phase. The advantages of homogeneous catalysts: Reactions can be carried out under atmospheric conditions Can be designed to function selectively Are generally cheaper 59 60 10

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