Chapter 14 – Chemical Kinetics

Chapter 14 – Chemical Kinetics Chemical Kinetics – Rates of Reactions Chemical Kinetics is the study of the rate of reaction. “How fast” does it take place? Very Slow Reactions Very Fast Reactions Acid/Base Combustion Rusting Decomposition of H2O2 Radioactive Decay Explosive Decomposition 1 Collision Theory Reaction rates are not constant. They are actually dependant on the conditions under which the reaction takes place. How can we make a reaction go faster? How do we increase the rate of a reaction? Rule #1: For a reaction to take place, reactants must collide!! # of particles But even if they do collide, it does not guarantee a reaction!! Energy These two reactants cannot react because they are not in contact with one another. They may eventually run into each other, but they need time to diffuse Number of particles that actually react Collision Theory Rule #2: When molecules collide, they must be in an optimal orientation. RXN! (maybe) Unfruitful Collisions (no reactions) Correct orientation for a reaction Rule #3: When molecules collide, they must have a certain minimum amount of energy in order for the reaction to occur. Although this collision seems to be in the correct orientation, the reactants do not have enough energy Not enough energy (no reactions) RXN! Enough energy for a reaction 2 Collision Theory and Activation Energy (Ea) Energy *Activated Complex/Transition State Reactants The activation energy (Ea) for a reaction is the energy barrier that must be overcome for the reaction to take place. Similar to a roller coaster… Ea Products ΔH Multi-step reaction *1 *2 Reaction Progress Ea1 A multi-step reaction has two Ea and two transition states Ea2 ΔH Exothermic vs. Endothermic Reactions Energy Energy Ea ΔH Ea ΔH Reaction Progress Products have lower energy than reactants Energy released during reaction Reaction Progress Products have higher energy than reactants Energy absorbed during reaction -ΔH +ΔH EXOTHERMIC ENDOTHERMIC 3 Concentration and Rate Lower concentration of reactants Higher concentration of reactants Higher concentration of reactants allow more opportunities for collisions. Surface Area and Rate Can only collide with particles on the surface of the solid If the particles are made finer, there are many more opportunities for collisions Higher surface area of reactants increase the rate of a reaction 4 Temperature and Rate Why does temperature affect rate? •More collisions: Since particles move faster at higher temperatures there is more opportunity to interact/collide. •More energy: More particles will have enough energy to overcome Ea Distribution of energies of particles at different temperatures # of particles T1 Area = number of particles that react at lower temp Ea Area = number of particles that react at higher temp T2 Notice that increasing the temperature does not change the Ea Lower Energy Higher Energy More reactant particles reach Ea = faster reaction Catalysts and rate *Activated Complex/Transition State Since the reaction takes a different route, a different transition state and different mechanism will be involved Reactants Ea Ea ΔH Reaction Progress Products Notice that the ΔH of the catalyzed and unanalyzed reactions are the same •Catalysts allow a different path for the reaction to tale place. A path with a lower Ea. •Catalysts are regenerated in a reaction. That is, they are not consumed 5 Catalysts Catalysts allow more particles to react per unit of time Catalyzed Ea (new rxn route) Ea # of particles Distribution of particles that reach Ea in a catalyzed vs. non-catalyzed reaction Catalysis lowers Ea allowing more particles to react (less energy is needed) Area = number of particles that react at lower temp Energy More reactant particles reach Ea = faster reaction Area = number of particles that react at higher temp Biological Catalysts / Catalysis Many reactions in the human body are catalyzed by proteins called enzymes. Enzymes catalyze reactions by binding to reactant particles called substrates. Allows unfavorable reactions to take place under biological conditions. Kinetics of enzymes are much more complicated!! 6 Chemical Kinetics What does a graphical representation of different reaction rate look like? Amount of reactant Let’s say we have two different decomposition reactions; one faster than the other. Reaction A Slower Reaction Reaction B Faster Reaction time While “fast” and “slow” are good qualitative words when comparing rates, how do we quantitatively determine this?   [reactant(s)]  [product(s)] Rateof reaction  t t Most often, especially for aqueous solutions, we define “amount” as concentration (M or mol/L) denoted by brackets, [ ]. Time can be in any units. Very often, we use seconds (s). How do we determine rate of reaction? We can measure the concentrations of reactants or products various ways. In some solutions, we can use visible spectroscopy to determine concentration of reactants or products, usually expressed in M or mol/L. Gases can be expressed in partial pressures (atm). Rateof reaction   [reactant(s)]  [product(s)]  t t Δ = final - initial A → B 1.00M t = 0s Rate  0.50M t = 25s  (0.50 M  1.00 M) 0.50 M M   0.020 (25s - 0s ) 25s s Rate  0.25M t = 50s  (0.25M  0.50 M) 0.25 M M   0.010 (50s - 25s ) 25s s 7 Chemical Kinetics – Rates of Reactions Rateof reaction A → B Time (min) [Reactant] (M) 0 5.00 .50 2.50 1.0 1.25 1.5 0.63 2.0 0.31 2.5 0.16 3.0 0.08   [reactant(s)]  [product(s)]  t t Avg Rate Δ = final - initial  (2.50 M  5.00 M) 2.50 M M   5.0 min (0.50 min- 0.00 min) 0.50 min Avg Rate  (0.16 M  0.63 M) 0.47 M M   0.47 min (2.5min- 1.5 min) 1.0 min Avg Rate   (0.08 M  0.16 M) 0.08 M M   0.2 min (3.0 min - 2.5 min) 0.5 min We can calculate an “average” rate of the reaction between any two data points during the reaction Overall Avg Rate  (5.00 M  0.08 M) 4.92 M M   1.6 min (0 min - 3.0 min) 3.0 min Chemical Kinetics – Rates of Reactions 5.0 M/min Δ Amount of reactant 6.00 Δ Time Concentration (M) 5.00 4.00 3.00 1.6 M/min 2.00 1.00 0.00 0 0.5 1 1.5 2 2.5 3 Time (min) 0.73 M/min What about an instantaneous rate? Let’s say at 1.25 minutes? 8 Chemical Kinetics – Rates of Reactions 0.6 [Phenyl Acetate] Time (s) (M) 0.0 0.550 15.0 0.420 30.0 0.310 45.0 0.230 60.0 0.170 75.0 0.120 90.0 0.085 [Phenyl Acetate] (M) 0.5 0.4 0.3 0.2 0.1 0 0 20 40 60 80 100 Time (s) • What is the overall average reaction rate 5.17x10-3 M/s • What is the average rate from t=0.0s to t=15.0s 8.67x10-3 M/s • Should the average from t=15.0s to 30.0s be higher or lower than t=0.0s to t=15.0s??? The rate from t=15.0s to 30.0s should be slower Rate of Disappearance of Reactants and Appearance of Products aA + bB → c C “Generic” rate of reaction 1  Δ[A]  1  Δ[B]  1  Δ[C]  Rateof reaction        a  Δt  b  Δt  c  Δt  1  Δ[A]  1  Δ[B]  1  Δ[C]  Rateof reaction        a  Δt  b  Δt  c  Δt  Rate of change in A Notice that the reactants have a “-” and the products do not. Rate of change in B Rate of change in C If the coefficients are different, the rates will be different 9 Rate of disappearance of reactants and appearance of products based on coefficients 1A + 2B → 1C 1  Δ[A]  1  Δ[B]  1  Δ[C]  Rateof reaction        1  Δt  2  Δt  1  Δt  If C is appearing at 5.0x10-5 M/s at what rate is A disappearing? 1  Δ[A]  1  Δ[C]       1  Δt  1  Δt  1  Δ[A]  1 5     5.0x10 M/s  1  Δt  1  Δ[A]  5     5.0x10 M/s  Δt  If A is disappearing at -1.0 M/min at what rate is B disappearing? 1  Δ[A]  1  Δ[B]       1  Δt  2  Δt   1  1.0 M/s    1  Δ[B]  1 2  Δt   Δ[B]   2 1.0 M/s       Δt   Δ[B]      2.0 M/s  Δt  If B is disappearing at -0.44 M/min at what is the (generic) reaction rate? 1  Δ[B]  Rateof reaction    2  Δt  Rateof reaction  1 - 0.44 M/min 2 Rateof reaction 0.22 M/min Rate of Disappearance of Reactants and Appearance of Products In the reaction 2N2O5(g) → 4NO2(g) + O2(g) the appearance of NO2 is 3.12x10-1 M/s. What is the rate of disappearance of N2O5? 1.56 x10-1 M/s In the synthesis of Ammonia, N2(g) + 3H2(g)→ 2NH3(g), if the rate of disappearance of H2 is 4.5x10-4 M/s, what is the rate of appearance of ammonia, NH3? 3.0x10-4 M/s 10 Chemical Kinetics and Concentration – Rate Law Each reaction can be expressed by what is called a “Rate Law” The rate law describes how each reactant’s concentration has an effect on the reaction rate •[A] = concentration (M) of reactant A •[B] = concentration (M) of reactant B •k = rate constant (depends on other units) •Rate = usually in M/s or mol/L·s •x and y = “order” of each reactant General Rate Law Rate = k[A]x[B]y… Only reactants that affect the rate of the reaction are included in the rate law. In other words, not every reactant that appears in the balanced equation will be in the rate law! Determining Rate Law – Determining the Order of each reactant While it seems like it should work, we should never determine the Rate Law based on the balanced equation!! This is because reactions can have multiple intermediate steps not accounted for in the balanced equation. We can determine the Rate Law using experimentation data of initial rates. A+B→Z Trial Initial [A] Initial [B] Initial Rate 1 0.100M 0.300M 1.00 M/s 2 0.100M 0.600M 2.00 M/s 3 0.300M 0.300M 9.00 M/s •First, find two trials where one reactant has the same concentration. This why we can cancel the effect of that reactant has on the rate (For example, trials 1 &2). •Plug everything you know into the rate equation. Rate = k[A]x[B]y x Trial 2 Trial 1  2.00 M/s   0.100M   0.600M   k     1.00 M/s   0.100M   0.300M  y 2  k 1x 2y •Ignore the reactant with the same concentration and ignore k, for now. 2  2y y =1 •Solve for the exponent 11 Determining Rate Law – Determining the Order of Each Reactant Now, we go through the same process for A: So Far… Rate = k [A]x [B] We normally do not show the “1” as an exponent x Trial 3 Trial 1  9.00 M/s   0.300M   0.300M   k     1.00 M/s   0.100M   0.300M  y Therefore, the rate law can be written as: Trial Initial [A] Initial [B] Initial Rate 1 0.100M 0.300M 1.00 M/s 2 0.100M 0.600M 2.00 M/s 3 0.300M 0.300M 9.00 M/s 9  k 3x 1y 9  3x x=2 The reaction is second-order with respect to “A” and first-order with respect to “B” Rate = k [A]2 [B] 2 + 1 = 3…This reaction is 3rd order overall Determining Rate Law – Short Cut Trial Initial [A] Initial [B] Initial Rate Trial Initial [A] Initial [B] Initial Rate 1 0.100M 0.300M 1.00 M/s 1 0.100M 0.300M 1.00 M/s 2 0.100M 0.600M 2.00 M/s 2 0.100M 0.600M 2.00 M/s 3 0.300M 0.300M 9.00 M/s 3 0.300M 0.300M 9.00 M/s 0.600M 0.300M 2.00 Ms 1.00 Ms 0.300M 0.100M 9.00 Ms 1.00 Ms concentration doubles rate doubles concentration triples rate nonuples (9x) 2y  2 1st order Therefore, the rate law can be written as: 3x  9 2nd order The reaction is second-order with respect to “A” and first-order with respect to “B” Rate = k [A]2 [B] 2 + 1 = 3…This reaction is 3rd order overall 12 Determining Rate Law – Determining the Rate Law Constant (k) Since k does not change during the reaction, we can use any trial to calculate the value of k Rate = k [A]2 [B] Trial 1 Rate = k [A]2 [B] Initial [A] Initial [B] Initial Rate 1 0.100M 0.300M 1.00 M/s 2 0.100M 0.600M 2.00 M/s 3 0.300M 0.300M 9.00 M/s 1.00 M/s = k [0.100]2 [0.300] k  333 Trial 2 Trial 1 or M-2s -1 M2s 2.00 M/s = k [0.100]2 [0.600] k  333 Cancel units (1.00 M/s) k (0.100M)2 (0.300M) k (2.00 M/s) (0.100M)2 (0.600M) 1 or M-2s -1 M2s Rate Law Rate 333 M12s [A] [B] 2 • • • • A = 2nd-order B = 1st-order Overall reaction = 3rd order k = 333 1/M2s Determining Rate Law Practice Use the following data to determine the rate law and the value of k for the reaction between NO and O2. Trial 1 2 3 4 5 Trial Initial [NO] Initial [O2] Initial Rate (M) (M) (M/s) 0.020 0.010 0.028 0.020 0.020 0.057 0.020 0.040 0.114 0.040 0.020 0.227 0.010 0.020 0.014 Initial [A] Initial [B] Initial [C] (M) (M) (M) Rate (M/s) 1 0.10 0.10 0.10 1.8x10-6 2 0.30 0.10 0.10 5.4x10-6 3 0.30 3.00 0.10 4.86x10-3 4 0.10 0.10 3.00 1.8x10-6 13 Determining Rate Law – Tricks Up My Sleeve Trial 1 2 3 Initial [A] Initial [B] (M) (M) Rate (M/s) 1.0 1.0 1.67 1.0 1.2 2.00 1.3 1.0 3.67 1.2M 1.0M concentration changes 1.2x Trial 1 2 3 2.00 Ms 1.67 Ms rate changes 1.2x Initial [A] Initial [B] (M) (M) Rate (M/s) 1.0 1.0 1.67 1.0 1.2 2.00 1.3 1.0 3.67 1.3M 1.0M 3.67 Ms 1.67 Ms concentration changes 1.3x rate changes 2.2x 1.2y  1.2 1.3x  2.2 1st order WTF?! Determining Rate Law – Tricks Up My Sleeve #1 ln(ax) = x ln(a) 1.3x  2.2 Take natural log of both sides ln 1.3  ln 2.2 x x ln 1.3  ln 2.2 x 0.262  0.778 solve for “x” Rearrange using ln(ax) = x ln(a) x = 2.96 ≈ 3 x ln 1.3  ln 2.2 14 Determining Rate Law Group Practice Trial 1 2 3 Initial PA Initial PB Rate (atm) (atm) (atm/min) 5.97 0.25 1.33 0.50 0.25 7.81x10-4 0.50 7.62 2.38x10-2 Trial 1 2 3 Initial [A] Initial [B] Rate (M) (M) (M/s) 0.10 0.20 2.5x10-4 0.10 0.10 1.2x10-4 0.30 0.30 1.1x10-3 *This one may initially look impossible, but it’s not. Hint: Think about what you already know…. Integrated Rate Laws - Rate and Time With the help of calculus, we can use another form of the rate law that allows us to calculate how concentration of a reactant changes with time. It is called the integrated rate law. Remember… 1st-order rate law Rate = k[A] 2nd-order rate law Rate = k[A]2 [A] l n t   kt [A]0 -or- l n[A]t   kt  l n[A]0 1 1  kt  [A]t [A]0 What about a reaction where the concentration of the reactants that do not affect the rate. Does it exist? Yup. 0th-order rate law Rate = k [A]t  [A]0  kt 15 Integrated Rate Laws – 1st [A] l n t   kt [A]0 1 1  kt  [A]t [A]0 order •A certain decomposition reaction is 1st order overall. What is the rate constant if the concentration of A goes from an initial concentration of 5.55M to 1.11M in 3.0 minutes?  1.11M  ln     k (3.0min)  5.55M  [A]t  [A]0  kt -1.609   k(3.0min) k = 0.54min-1 •How long would it take for the concentration to reach 50% of its original concentration?  50%  1 ln     (0.536 min ) t  100%  - 0.693   (0.536 min 1 )t t = 1.3min Using % directly in the equation will only work with 1st order since k doesn’t include a concentration unit. For other orders (2nd and 0th), use actual concentrations in the equations. Integrated Rate Laws – 1st order More natural log tricks [A] l n t   kt [A]0 1 1  kt  [A]t [A]0 [A]t  [A]0  kt •What would be the concentration after 13 minutes?  x  -1 l n    ( 0.536 mi n )(13mi n)  5.55M  e ln(x) = x e  x  ln    5.55M  e-(0.536min  x  -1 ln     (0.536 min )(13min)  5.55M  x l n   l nx - l ny y How do we get rid of natural log (ln)? -1 )(13min) x  9.42x10 -4 5.55M x = 0.0052M CAUTION! x l nx l n  DOES NOT  l ny y l nx - l n5.55M   (0.536 mi n-1 )(13mi n) l nx   5.254 x = 0.0052M 16 2nd Integrated Rate Laws – • order [A]t  [A]0  kt A second-order decomposition has a rate constant of 0.050 M-1min-1. Assuming you start with 3.2M, how long will it take for only 5.0% of the reactant to remain.  5.0%  3.2M   0.16M  100.0%  Unlike 1st order, you cannot simply use %’s. You must use specific concentrations or amounts for your reactant. So.. • [A] l n t   kt [A]0 1 1  kt  [A]t [A]0 1 1  (0.050 M -1min 1 ) t  0.16M 3.2M 5.94  (0.050 M-1min 1 ) t t  120min If the same reaction took 625s to reach 0.222M, what was the original concentration? Watch for time units. MUST be the same in k and for t 1 1  (0.050 M -1min 1 )(10.4 min)  0.222M [A]0 3.98  625s 1min  10.4 min 60s 1 [A]0 [A]0  0.25M Half-lives The time it takes for a reactant to reach half of its initial amount 16M 8M 4M 50 % 25 % 2M 1M Original concentration 100 % 12.5 % t1/2 t1/2 t1/2 1 half-life 2 half-lives 3 half-lives 6.25 % t1/2 4 half-lives General Equation for Half-life 1 [A]t1 /2  [A]0   2 x Where “x” = number of half-lives 17 Half lives 7 7 6 6 1st Order 4 1st half-life 3 2nd half-life 2 4 1st half-life 3 2 3rd half-life 1 2nd Order 5 [A] (mol/L) [A] (mol/L) 5 4thhalf-life 2nd half-life 3rd half-life 1 0 4thhalf-life 0 0 0.5 1 1.5 2 2.5 3 0 0.5 1 Time (s) Half lives are equal t 1/2 1.5 2 2.5 3 Time (s) Half-lives increase over time t 1/2 l n(2) k 1 k[A]0 Integrated Rate Laws – Half lives 7 6 0th Order [A] (mol/L) 5 4 1st half-life 3 2nd half-life 2 3rd half-life 1 4thhalf-life 0 0 0.5 1 1.5 2 2.5 3 Time (s) Half lives decrease t 1/2 [A]0 2k 18 Half-life equations General Equation for Half-life 1 [A]t  [A]0   2 x Where “x” = number of half-lives Half-lives are related to the rate constant, k, by the following equations 1st Order  [A]0    2  ln    kt [A]0 2nd Order 1 1  kt  [A]0  [A]0     2  0th Order  [A]0     [A]0  kt  2  t 1/2 l n(2) k t 1/2 1 k[A]0 t 1/2 [A]0 2k Half-Life Examples A certain reaction has a rate constant of 1.25 hr -1. What is the half life of this reaction in hours? Minutes? t 1/2 l n(2) k t 1/2 1 k[A]0 t 1/2 [A]0 2k 0.555hr How many half lives will it take for the same reaction above to go from 6.0M to 0.020M? 8.2 half-lives If a 2nd order reaction has a half life of 3.1 days, how many hours will it take for 10.0% of the reactant to disappear if the concentration starts at 1.0M? 19 Graphing Integrated Rate Laws slope Equation of a straight line ln[A]0 y-intercept ln[A]t y = m x+b l n[A]t   k t  l n[A]0 1st-order rate law ln[A]t vs. t time 2nd-order rate law 1  [A]t k t  1 [A]0 1/[A]t 1/[A]0 1/[A] t vs. t time [A]t [A]0 0th-order rate law [A]t   k t  [A]0 [A] t vs. t time Determining order by graphing To determine the “order” of the reactant, you must graph the data three different ways: Time vs. Conc.; Time vs. ln(Conc.); and Time vs. 1/Conc. Time (s) 0 1 2 3 4 5 6 7 8 [A] (M) 1 0.5 0.333333 0.25 0.2 0.166667 0.142857 0.125 0.111111 Time vs. Conc. Time (s) 0 1 2 3 4 5 6 7 8 ln[A] 0 -0.69315 -1.09861 -1.38629 -1.60944 -1.79176 -1.94591 -2.07944 -2.19722 Time vs. ln(Conc.) Time (s) 1/[A] 1/M 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 Time vs. 1/(Conc.) 20 Multistep Reactions – Reaction Mechanisms NO2 + CO → NO + CO2 NO2 + CO ? NO + CO2 This reaction happens in multiple steps. But what are they? And how can we figure it out? First we need to learn about these steps. Multistep Reactions – Reaction Mechanisms Reactions that have multiple steps are made of individual steps called elementary steps. Elementary steps are single (concerted) steps. Step 1: 2NO2 → NO + NO3 Step 2: NO3 + CO → NO2 + CO2 Intermediates are chemical species that are created during the reaction, but not a part of the overall reaction equation. Intermediates are created in an elementary step and consumed in another NO2 + CO → NO + CO2 Elementary steps must add up to the overall reaction equation. Step 1: 2NO2 → N2O4 Step 2: N2O4 + CO2 → 2NO2 + CO2 These are not plausible steps because the steps add up to a different overall reaction CO2 → CO2 21 Molecularity and Rate Laws of Elementary steps Rate laws for elementary steps can be determined directly from the balanced equation The molecularity of an elementary step is based on the rate law for that step Molecularity Equation Rate Law Unimolecular A → Product(s) Rate = k[A] Bimolecular A + A → Product(s) A + B → Product(s) Rate = k[A]2 Rate = k[A][B] Termolecular A + A +A → Product(s) A + B + B → Product(s) A + B + C → Product(s) Rate = k[A]3 Rate = k[A][B]2 Rate = k[A][B][C] Since elementary steps are single steps, rate laws for each step can be determined from the balanced chemical equation Step 1: 2NO2 → NO + NO3 Step 2: NO3 + CO → NO2 + CO2 Elementary Step Rate = k1 [NO2]2 Elementary Step Rate = k2[NO3][CO] Both of these elementary steps are bimolecular Overall reaction rates of multi-step reactions Two-step reaction Elementary Steps 6,000 kisses per hour Slow Step 500,000 kisses per hour Fast Step How many kisses can be processed per hour overall??? The overall reaction can go no faster than the slow step. This step is called the rate-determining or rate-limiting step. Just remember that reaction rates are not static; reaction rates are variable based on concentration. 22 Multiple-step Reactions – Elementary Steps – Rate-limiting Step 1st Slow-step 1st Overall Chemical Formula A + A→ X 2A→ X B+X→C B+X→C 2A + B → C Rate 1 < Rate 2 The elementary-step formulas must add up to the overall chemical formula Rate laws for elementary steps can be determined directly from the balanced equation 2A→ X + C B+X→C Step 1 Rate Law Step 2 Rate Law Rate = k1 [A]2 Rate = k2 [B][X] If the slow, rate-determining, step is first, the overall rate law is the same as the slow step Step 1 Rate Law Rate = k1 [A]2 = Overall Rate Law Rate = k [A]2 Notice that the concentration of “B” does not affect the rate of reaction Overall reaction rates of multi-step reactions Elementary Steps Two-step reaction 500,000 kisses per hour Fast RXN 10,000 kisses per hour Slow RXN How many kisses can be processed per hour overall??? 500,000 kisses per hour Although it seems like the overall rate should = the slow step, in reality, reaction rates are variable based on concentrations of reactants. 23 Multiple-step Reactions – Elementary Steps – Rate-limiting Step 2nd Slow-step 2nd…3rd… Overall Chemical Formula A + A→ X 2A→ X B+X→C B+X→C 2A + B → C Rate 1 > Rate 2 The elementary-step formulas will add up to the overall chemical formula 2A→ X + C B+X→C Step 1 Rate Law Step 2 Rate Law [A]2 Rate = k2 [B][X] Rate = k1 However, unlike the rate-limiting step being first, If the slow, ratedetermining, step is second…it’s much more complicated…… Multiple-step Reactions – Elementary Steps – Rate-limiting Step 2nd If the slow step is second, the bottleneck creates an equilibrium in the first step. 2A ⇌ X Slow-step 2nd…3rd… Step 1 Rate Law 2A ⇌ X Rate = k1 [A]2 = k-1[X] B+X→C Equilibrium Step 2 Rate Law Rate 1 > Rate 2 Rate = k2 [B][X] Some products in Step 1 revert back into reactants before continuing to step 2 Step 1 Rate Law Rate = k1 [A]2 = k-1[X] Solve for the intermediate [X]  Intermediates cannot be in the overall rate law. The overall rate law cannot be based solely on the Step 2 rate law. We must combine both rate laws. k1[A]2 k1 Substitute  k [A]2   Rate k2 [B] 1  k1  Overall Rate Law Rate k[A]2 [B] Combine k’s 24 Multiple-step Reactions – Elementary Steps – Rate-limiting Step Notice that the same reaction with the same elementary steps can give us different overall rate laws because of the location of the slow step. 2A + B → C Slow step 2nd…3rd… Slow-step 1st 2A → X Rate 1 2A ⇌ X B+X→C < Rate 2 B+X→C Rate 1 Overall Rate Law > Rate 2 Overall Rate Law Rate = k [A]2 Rate k[A]2 [B] How do we know which one is correct for this reaction? Experimentally Determined Rate Law Multiple-step Reactions – Elementary Steps NO2 + CO → NO + CO2 has an experimentally determined rate law of NO2 + CO ? Rate = k[NO2]2 NO + CO2 Which of the following is most likely to represent the mechanism for the reaction? NO2 + CO → NO + CO2 2NO2 → NO + NO3 (slow) NO3 + CO → NO2 + CO2 (fast) 2NO2 → NO + NO3 (fast) NO3 + CO → NO2 + CO2 (slow) 25 Catalysts #2 – Mechanisms Catalyzed Ea (new rxn route) Ea 2H2O2 → 2H2O + O2 # of particles H2O2 + I- → IO- +H2O H2O2 + IO- → I- + H2O + O2 Notice that the catalyst, I-, is regenerated in the end. There is no net change in catalyst in the reaction. Energy Area = number of particles that react at lower temp More reactant particles reach Ea = faster reaction The intermediate, IO-, is completely used in the reaction (as it should be). Area = number of particles that react at higher temp Arrhenius Equation Svante Arrhenius determined that activation energy (Ea) of a reaction is related to the reaction constant. - Ea k  Ae R T That is…at any temperature the lower the activation energy, the faster a reaction will proceed. lnk  lnA - lnk  y = Ea RT Ea 1  lnA R T m x + b 26 Arrhenius Equation We can also do calculations without graphing, but combining two equations into one. If we know both k1 and k2, along with T1 and T2 we calculate the energy of activation lnk1  lnA - Ea RT1 ln E lnk2  lnA - a RT2 k1 E a  1 1      k 2 R  T2 T1  R  8.314 molJK In a certain equation, the rate constant at 701K was measured as 2.57M-1s-1 and that at 895 K was measured as 556M-1s-1. Find the activation energy of the reaction. ln 1 556 Ms Ea 1   1     1 2.57 Ms 8.314 molJ K  701K 895K  5.38   Ea 3.09x10 -4 8.314 molJ K 1 K  1.45x105 J/mol 27