CHEM 217 Chemical Kinetics PDF

CHEM 217 MODULE 3 Chemical Kinetics  Defining Rate rate is how much a quantity changes in a given period of time the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) so the rate of your car has units of mi/hr...

CHEM 217 MODULE 3 Chemical Kinetics  Defining Rate rate is how much a quantity changes in a given period of time the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) so the rate of your car has units of mi/hr  distance Speed =  time 2  Defining Reaction Rate the rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time or product concentration increases for reactants, a negative sign is placed in front of the definition Δ concentration Rate = Δ time Δ [product] Δ [reactant] Rate = =− Δ time Δ time 3  Reaction Rate Changes Over Time as time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases. at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium. 4  at t = 0 [A] = 8 [B] = 8 at t = 0 [C] = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 at t = 16 [B] = 4 [X] = 7 [C] = 4 [Y] = 7 [Z] = 1 Rate = − A  =− (A2 − A1 ) X  (X2 − X1 ) t (t 2 − t 1 ) Rate = − =− t (t 2 − t 1 ) Rate = − (4 − 8) = 0.25 (7 − 8) = 0.0625 (16 − 0) Rate = − (16 − 0) C (C2 − C1 ) Rate = = Z (Z2 − Z1 ) t (t 2 − t1 ) Rate = = t (t 2 − t 1 ) (4 − 0) = 0.25 Rate = (16 − 0) Rate = (1 − 0) = 0.0625 5 (16 − 0) at t = 16 at t = 16 [A] = 4 [X] = 7 [B] = 4 [Y] = 7 [C] = 4 [Z] = 1 at t = 32 at t = 32 [A] = 2 [X] = 6 [B] = 2 [Y] = 6 [C] = 6 [Z] = 2 A  (A2 − A1 ) Rate = − X  =− (X2 − X1 ) Rate = − =− t (t 2 − t 1 ) t (t 2 − t 1 ) Rate = − (2 − 4) = 0.125 (6 − 7 ) = 0.0625 (16 − 0) Rate = − (16 − 0) C (C2 − C1 ) Rate = t = (t 2 − t1 ) Z (Z2 − Z1 ) Rate = = (6 − 4) = 0.125 t (t 2 − t 1 ) Rate = (16 − 0) Rate = (2 − 1) = 0.0625 6 (16 − 0)  at t = 32 at t = 32 [A] = 2 [X] = 6 [B] = 2 [Y] = 6 [C] = 6 [Z] = 2 at t = 48 at t = 48 [A] = 0 [X] = 5 [B] = 0 [Y] = 5 [C] = 8 [Z] = 3 A  (A2 − A1 ) Rate = − X  =− (X2 − X1 ) Rate = − =− t (t 2 − t 1 ) t (t 2 − t 1 ) Rate = − (0 − 2) = 0.125 (5 − 6) = 0.0625 (16 − 0) Rate = − (16 − 0) C (C2 − C1 ) Rate = t = (t 2 − t1 ) Z (Z2 − Z1 ) Rate = = (8 − 6) = 0.125 t (t 2 − t 1 ) Rate = (16 − 0) Rate = (3 − 2) = 0.0625 7 (16 − 0) Hypothetical Reaction Red → Blue Time Number Number (sec) Red Blue in this reaction, 0 100 0 one molecule of Red turns 5 84 16 into one molecule of Blue 10 71 29 the number of molecules 15 59 41 will always total 100 20 50 50 25 42 58 the rate of the reaction can 30 35 65 be measured as the speed of 35 30 70 loss of Red molecules 40 25 75 over time, or the speed of 45 21 79 gain of Blue molecules 50 18 82 8 over time HypotheticalConcentration Reaction Red → Blue vs Time for Red -> Blue 100 100 90 84 80 71 82 79 Number of Molecules 70 75 59 70 60 65 50 Number Red 50 58 42 Number Blue 50 40 35 41 30 30 25 21 29 18 20 10 16 0 0 5 10 15 20 25 30 35 40 45 50 Time (sec) 9 Hypothetical Reaction Red → Blue Rate of Reaction Red -> Blue 4.5 4 3.5 5, 3.2 Rate,  [Blue]/ t 3 10, 2.6 2.5 15, 2.4 2 20, 1.8 25, 1.6 1.5 30, 1.4 1 35, 1 40, 1 45, 0.8 50, 0.6 0.5 0 0 10 20 30 40 50 time, (sec) 10  Reaction Rate and Stoichiometry in most reactions, the coefficients of the balanced equation are not all the same H2 (g) + I2 (g) → 2 HI(g) for these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient [H 2 ] [I 2 ]  1  [HI] Rate = − =− = +  t t  2  t 11 Average Rate the average rate is the change in measured concentrations in any particular time period linear approximation of a curve the larger the time interval, the more the average rate deviates from the instantaneous rate 12 Hypothetical Reaction Red → Blue Avg. Rate Avg. Rate Avg. Rate Time Number Number (5 sec (10 sec (25 sec (sec) Red Blue intervals) intervals) intervals) 0 100 0 5 84 16 3.2 10 71 29 2.6 2.9 15 59 41 2.4 20 50 50 1.8 2.1 25 42 58 1.6 2.3 30 35 65 1.4 1.5 35 30 70 1 40 25 75 1 1 45 21 79 0.8 50 18 82 0.6 0.7 1 13  H2 I2 Time HI [H2], M[HI], M -[H2]/t The average rate is the change in the (s) concentration in a given time period. 0.000 1.000 0.000 10.000 0.819 0.362 0.0181 Stoichiometry tells us that for every 1 20.000 0.670 0.660 0.0149 mole/L of H2 used, 2 moles/L of HI are made. 30.000 0.549 0.902 0.0121 40.000 0.449 1.102 0.0100 Assuming a 1 L container, at 10 s, we used 50.000 0.368 1.264 0.0081 0.181 moles of H2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles 60.000 0.301 1.398 0.0067 70.000 0.247 1.506 0.0054 80.000 0.202 1.596 0.0045 At 60 s, we used 0.699 moles of H2. 90.000 0.165 1.670 0.0037 Therefore the amount of HI made is 100.000 0.135 1.730 0.0030 2(0.699 moles) = 1.398 moles In the first 10 s, the [H2] is -0.181 M, so the rate − 0.181 M − is 10.000 s M = 0.0181 14 s Instantaneous Rate the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function for you calculus fans 15 H2 (g) + I2 (g) → 2 HI (g) Using [H2], the instantaneous rate at 50 s is: − 0.28 M Rate = − 40 s M Rate = 0.0070 s Using [HI], the instantaneous rate at 50 s is:  1  0.56 M Rate =    2  40 s M Rate = 0.0070 s 16 Ex 13.1 - For the reaction given, the [I−] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and Predict the rate of change in concentration of H+ the([H+]/ t) H2O2 (aq) + 3 I−(aq) + 2 H+(aq) → I3−(aq) + 2 H2O(l) Solve the equation  1  (0.868 M − 1.000 M ) −  1  [I ] for the Rate (in Rate = −   = −  terms of the change  3  t  3 10 s M in concentration of Rate = 4.40 10-3 s the Given quantity) + Solve the equation  1  [H ] Rate = −   of the Rate (in terms  2  t of the change in the [H + ] concentration for the = −2(Rate ) t quantity to Find) for the unknown value [H + ] t ( = −2 4.40  10-3 M s )= − 8.80 10-3 M s  Measuring Reaction Rate in order to measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time there are two ways of approaching this problem (1) for reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration, or (2) for reactions that happen over a very long time, sampling of the mixture at various times can be used when sampling is used, often the reaction in the sample is stopped by a quenching technique 18 Continuous Monitoring polarimetry – measuring the change in the degree of rotation of plane-polarized light caused by one of the components over time spectrophotometry – measuring the amount of light of a particular wavelength absorbed by one component over time the component absorbs its complimentary color total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction 19  Sampling gas chromatography can measure the concentrations of various components in a mixture for samples that have volatile components separates mixture by adherence to a surface drawing off periodic aliquots from the mixture and doing quantitative analysis titration for one of the components gravimetric analysis 20 Methods of Measuring Reaction Rates A. Physical measurements 1. Continuous measurements 2 Initial rate measurements (Clock reactions) B. Chemical measurements (Titration) 1. Continuous measurements Experiment is done in ONE take. The reaction rates are determined by measuring continuously a convenient property which is directly proportional to the concentration of any one reactant or product of the reaction mixture. Properties to be measured : – Gas volume / Gas pressure / Mass / Color intensity / Electrical conductivity Advantages of physical measurements 1. Suitable for fast reactions. 2. Small sample size 3. More accurate than chemical method (titration) 4. No interruption → continuous measurements 5. Can be automated. Disadvantages of physical measurements 1. More sophisticated 2. More expensive 3. More specific – only suit a limited number of reactions.  Quenching methods: Temperature  Cooling the reaction mixture rapidly in ice. Diluting the reaction mixture with a sufficient amount of cold water or an appropriate solvent. Concentration  Removing one of the reactants or the catalyst (if any) by adding another reagent. H+ as catalyst CH3COCH3 + I2 CH3COCH2I + HI The reaction is quenched by adding to it NaHCO3(aq) that removes the catalyst. HCO3−(aq) + H+(aq) → H2O(l) + CO2(g) B. Chemical Measurements (Titration Methods) 1. Start a reaction with all reaction conditions but one fixed. 2. Withdraw and quench fixed amounts of the reaction mixture at different times. 3. Titrate the quenched samples to determine the concentration of one of the reactants or products.  H+ as catalyst CH3COCH3 + I2 CH3COCH2I + HI Titrated with standard solution of Na2S2O3(aq) using starch as indicator (added when the end point is near) 2− 2− − 2S2O3 (aq) + I2(aq) → S4O6 (aq) + 2I (aq) Colour change at the end point : deep blue to colourless The excess S2O32−(aq) would react with H+ to give a cloudy mixture with a pungent smell. S2O32−(aq) + 2H+(aq) → S(s) + SO2(g) + H2O(l) Advantages of titrimetric method 1. Only simple apparatus are required. 2. Can be applied to a great variety of slow reactions. Disadvantages of physical measurements 1. Not suitable for fast reactions. It takes time to withdraw samples and perform titration. 2. Reactions are disturbed – NOT continuous 3. Time consuming – NOT automated Factors Affecting Reaction Rates  Kinetics kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. experimentally it is shown that there are 4 factors that influence the speed of a reaction: nature of the reactants, temperature, catalysts, concentration 31  Factors Affecting Reaction Rate Nature of the Reactants nature of the reactants means what kind of reactant molecules and what physical condition they are in. small molecules tend to react faster than large molecules; gases tend to react faster than liquids which react faster than solids; powdered solids are more reactive than “blocks” more surface area for contact with other reactants certain types of chemicals are more reactive than others e.g., the activity series of metals ions react faster than molecules no bonds need to be broken 32 CaCO3(aq) + 2H+(excess) → CaCl2(aq) + H2O(l) + CO2(g) Rate involving powdered solid reactant is higher Reason: higher chance of contact between reactant particles 0.5 g powder 0.5 g granule  Factors Affecting Reaction Rate Temperature increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles for many reactions there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later 35 Effect of temperature Applicable to ALL reactions T → K.E. of particles  → Collision frequency  (minor effect) and No. of particles with K.E. > Ea  (major effect) → No. of effective collisions  → Rate of reaction  Rate Rate of reaction  exponentially with temperature −Ea Rate  e RT In general, a 10oC  in T doubles the rate. T / C  Factors Affecting Reaction Rate Catalysts catalysts are substances which affect the speed of a reaction without being consumed. most catalysts are used to speed up a reaction, these are called positive catalysts catalysts used to slow a reaction are called negative catalysts. homogeneous = present in same phase heterogeneous = present in different phase 39 Effect of Catalyst A catalyst is a substance that alters the rate of a chemical reaction by providing an alternative reaction pathway with a different activation energy. A positive catalyst speeds up a reaction by providing an alternative reaction pathway with a lower Ea. A negative catalyst slows down a reaction by providing an alternative reaction pathway with a higher Ea.  Factors Affecting Reaction Rate Reactant Concentration generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the gas higher pressure = higher concentration concentration of solutions depends on the solute to solution ratio (molarity) 41 Effect of concentration 2.0 M HCl (b) 1.0 M HCl (c) 0.5 M HCl Reaction rate: (a) > (b) > (c) Effect of concentration Time for reaction to complete: t1 < t2 < t3 Higher [HCl(aq)] → Faster reaction [X]  → Reactant particles are more crowded → Collision frequency  → Number of effective collisions  → Reaction rate  Effect of pressure Only applicable to reactions involving gaseous reactants. Pressure  → Reactant particles are more crowded → Collision frequency  → No. of effective collisions  → Rate of reaction  The Rate Law the Rate Law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants the rate of a reaction is directly proportional to the concentration of each reactant raised to a power for the reaction aA + bB → products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant Rate = k[A] [B] n m 47  Reaction Order the exponent on each reactant in the rate law is called the order with respect to that reactant the sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction: 2 NO(g) + O2(g) → 2 NO2(g) is Rate = k[NO]2[O2] The reaction is second order with respect to [NO], first order with respect to [O2], and third order overall 48 Sample Rate Laws Reaction Rate Law CH3CN → CH3NC Rate = k[CH3CN] CH3CHO → CH4 + CO Rate = k[CH3CHO]3/2 2 N2O5 → 4 NO2 + O2 Rate = k[N2O5] H2 + I2 → 2 HI Rate = k[H2][I2] Tl + Hg2 → Tl + 2 Hg +3 +2 +1 +2 Rate = k[Tl+3][Hg2+2][Hg+2]-1 The reaction is autocatalytic, because a product affects the rate. Hg2+ is a negative catalyst, increasing its concentration slows the reaction. 49 Reactant Concentration vs. Time A → Products 50 Zero Order Reactions Rate = k[A]0 = k constant rate reactions [A] = -kt + [A]0 graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0 t ½ = [A0]/2k when Rate = M/sec, k = M/sec [A]0 [A] time 51  First Order Reactions Rate = k[A] ln[A]0 ln[A] = -kt + ln[A]0 graph ln[A] vs. time gives straight line with ln[A] slope = -k and y-intercept = ln[A]0 used to determine the rate constant t½ = 0.693/k the half-life of a first order reaction is constant the when Rate = M/sec, k = sec-1 time 52  Half-Life the half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value the half-life of the reaction depends on the order of the reaction n= number of half lives [A]t =(1/2)n[A]O 53 Rate Data for C4H9Cl + H2O → C4H9OH + HCl Time (sec) [C4H9Cl], M 0.0 0.1000 50.0 0.0905 100.0 0.0820 150.0 0.0741 200.0 0.0671 300.0 0.0549 400.0 0.0448 500.0 0.0368 800.0 0.0200 10000.0 0.0000 54 C4H9Cl + H2O → C4H9OH + 2 HCl Concentration vs. Time for the Hydrolysis of C 4H9Cl 0.12 0.1 0.08 concentration, (M) 0.06 0.04 0.02 0 0 200 400 600 800 1000 time, (s) 55 C4H9Cl + H2O → C4H9OH + 2 HCl Rate vs. Time for Hydrolysis of C 4H9Cl 2.5E-04 2.0E-04 1.5E-04 Rate, (M/s) 1.0E-04 5.0E-05 0.0E+00 0 100 200 300 400 500 600 700 800 time, (s) 56  C4H9Cl + H2O → C4H9OH + 2 HCl LN([C4H9Cl]) vs. Time for Hydrolysis of C 4H9Cl 0 -0.5 slope = -2.01 x 10-3 -1 k= -1.5 2.01 x 10-3 s-1 LN(concentration) -2 -2.5 0.693 -3 t1 = y = -2.01E-03x - 2.30E+00 2 k -3.5 0.693 = -4 2.01 10 −3 s -1 -4.5 = 345 s 0 100 200 300 400 500 600 700 800 time, (s) 57  Second Order Reactions Rate = k[A]2 1/[A] = kt + 1/[A]0 graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A]0 used to determine the rate constant t½ = 1/(k[A0]) when Rate = M/sec, k = M-1∙sec-1 58 Rate Data For 2 NO2 → 2 NO + O2 Partial Pressure Time (hrs.) NO2, mmHg ln(PNO2) 1/(PNO2) 0 100.0 4.605 0.01000 30 62.5 4.135 0.01600 60 45.5 3.817 0.02200 90 35.7 3.576 0.02800 120 29.4 3.381 0.03400 150 25.0 3.219 0.04000 180 21.7 3.079 0.04600 210 19.2 2.957 0.05200 240 17.2 592.847 0.05800  Rate Data Graphs For 2 NO2 → 2 NO + O2 Partial Pressure NO2, mmHg vs. Time 100.0 90.0 80.0 70.0 Pressure, (mmHg) 60.0 50.0 40.0 30.0 20.0 10.0 0.0 0 50 100 150 200 250 Time, (hr) 60  Rate Data Graphs For 2 NO2 → 2 NO + O2 ln(PNO2) vs. Time 4.8 4.6 4.4 4.2 4 ln(pressure) 3.8 3.6 3.4 3.2 3 2.8 2.6 2.4 0 50 100 150 200 250 Time (hr) 61  Rate Data Graphs For 2 NO2 → 2 NO + O2 1/(PNO2) vs Time 0.07000 0.06000 1/PNO2 = 0.0002(time) + 0.01 Inverse Pressure, (mmHg-1 ) 0.05000 0.04000 0.03000 0.02000 0.01000 0.00000 0 50 100 150 200 250 62 Time, (hr)  Determining the Rate Law can only be determined experimentally graphically rate = slope of curve [A] vs. time if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope initial rates by comparing effect on the rate of changing the initial concentration of reactants one at a time 63 64 Practice - Complete the Table and Determine the Rate Equation for the Reaction A → 2 Prod [A], (M) [Prod], (M) Time (sec) ln([A]) 1/[A] 0.100 0 0 0.067 50 0.050 100 0.040 150 0.033 200 0.029 250 65 Practice - Complete the Table and Determine the Rate Equation for the Reaction A → 2 Prod [A], (M) [Prod], (M) Time (sec) ln([A]) 1/[A] 0.100 0 0 -2.3 10 0.067 0.066 50 -2.7 15 0.050 0.100 100 -3.0 20 0.040 0.120 150 -3.2 25 0.033 0.134 200 -3.4 30 0.029 0.142 250 -3.5 35 66  [A] vs. Time 0.12 0.1 concentration, M 0.08 0.06 0.04 0.02 0 0 50 100 150 200 250 time, (s) 67  LN([A]) vs. Time -2 -2.2 -2.4 Ln(concentration) -2.6 -2.8 -3 -3.2 -3.4 -3.6 -3.8 0 50 100 150 200 250 time, (s) 68  1/([A]) vs. Time y = 0.1x + 10 40 35 -1 30 inverse concentration, M 25 20 15 10 5 0 0 50 100 150 200 250 time, (s) 69 Practice - Complete the Table and Determine the Rate Equation for the Reaction A → 2 Prod -[A] the reaction is second order, Rate = = 0.1 [A]2 t 70 Ex. 13.4 – The reaction SO2Cl2(g) → SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10-4 s-1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0 = 0.0225 M Given: [SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1 Find: [SO2Cl2] Concept Plan: [SO2Cl2]0, t, k [SO2Cl2] Relationships: for a 1st order process : ln[A] = kt + ln[A] 0 ln[SO 2Cl 2 ] = kt + ln[SO 2Cl 2 ]0 ( ) Solution: ln[SO 2Cl 2 ] = 2.90 10- 4 s -1 (865 s ) + ln (0.0225) ln[SO 2Cl 2 ] = −0.251 − 3.79 = −4.04 [SO 2Cl 2 ] = e(-4.04) = 0.0175 M Check: the new concentration is less than the original, as expected  Initial Rate Method another method for determining the order of a reactant is to see the effect on the initial rate of the reaction when the initial concentration of that reactant is changed for multiple reactants, keep initial concentration of all reactants constant except one zero order = changing the concentration has no effect on the rate first order = the rate changes by the same factor as the concentration doubling the initial concentration will double the rate second order = the rate changes by the square of the factor the concentration changes doubling the initial concentration will quadruple the rate 72  Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) given the data below. Write a general rate Expt. Initial Initial Initial Rate law including all Number [NO Number [NO22], ], (M) (M) [CO], (M) [CO], (M) (M/s) (M/s) reactants 1. 0.10 0.10 0.0021 Examine the data and 2. 0.20 0.10 0.0082 find two experiments 3. 0.20 0.20 0.0083 in which the concentration of one 4. 0.40 0.10 0.033 reactant changes, but the other Rate = k[NO2 ] [CO] n m concentrations are the Comparing Expt #1 and Expt #2, the [NO2] same changes but the [CO] does not 73 Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) given the data below. Determine by Expt. Initial Initial Initial Rate what factor the Number [NO2], (M) [CO], (M) (M/s) concentrations 1. 0.10 0.10 0.0021 and rates change 2. 0.20 0.10 0.0082 in these two experiments. 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 [NO2 ]expt 2 0.20 M Rate expt 2 0.0082 M s = =2 = 4 [NO2 ]expt 1 0.10 M Rate expt 1 M 0.0021 s 74  Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) given the data below. Determine to Expt. Initial Initial Initial Rate what power the Number [NO2], (M) [CO], (M) (M/s) concentration 1. 0.10 0.10 0.0021 factor must be 2. 0.20 0.10 0.0082 raised to equal the rate factor. 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 [NO2 ]expt 2 0.20 M n =  [NO2 ]expt 2  = 2 Rate Rate expt 2 expt 2 = 0.0082 M s 4 [NO2 ]expt 1 0.10 M   = M Rate expt 1 0.0021 s  [NO2 ]expt 1  Rate expt 1   2n = 4 n=2 75  Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) given the data below. Repeat for the other Initial Initial Expt. Initial reactants [NO2], Rate Number (M) [CO], (M) (M/s) [CO]expt 3 0.20 M 1. 0.10 0.10 0.0021 = =2 2. 0.20 0.10 0.0082 [CO]expt 2 0.10 M 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 m  [CO]expt 3  Rate expt 3 0.0083 M s   = Rate expt 3 = 1  [CO]expt 2  Rate expt 2 Rate expt 2 0.0082 M s   2m = 1 m=0 76 Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) given the data below. Substitute the Expt. Initial Initial Initial Rate exponents into Number [NO2], (M) [CO], (M) (M/s) the general rate 1. 0.10 0.10 0.0021 law to get the 2. 0.20 0.10 0.0082 rate law for the reaction 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 n = 2, m = 0 Rate = k[NO2 ] [CO] n m Rate = k[NO 2 ] [CO] 2 0 Rate = k[NO 2 ] 2 77 Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) → NO(g) + CO2(g) given the data below. Substitute the Expt. Initial Initial Initial Rate concentrations Number [NO2], (M) [CO], (M) (M/s) and rate for any 1. 0.10 0.10 0.0021 experiment into 2. 0.20 0.10 0.0082 the rate law and solve for k 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 Rate = k[NO 2 ]2 for expt 1 0.0021 M s = k (0.10 M )2 0.0021 M s k = 78 = 0. 21 M -1 s -1 0.01 M 2 Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1 →  +   given the data below. Expt. Initial Initial Initial Rate, No. [NH4+], M [NO2-], M (x 10-7), M/s 1 0.0200 0.200 10.8 2 0.0600 0.200 32.3 3 0.200 0.0202 10.8 4 0.200 0.0404 21.6 79 Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1 →  +   given the data below. Expt. Initial Initial Initial Rate, Rate = k[NH4+]n[NO2−]m [NH4+], M [NO2-], M No. (x 10-7), M/s Rate = k[NH 4 + ][NO 2 − ] 1 0.0200 0.200 10.8 for expt 1 10.8 10-7 M s = k (0.0200 M )(0.200 M ) 2 0.0600 0.200 32.3 10.8  10-7 M s 3 0.200 0.0202 10.8 k= = 2.70  10 − 4 M -1 s -1 4 0.200 0.0404 21.6 4.00  10-3 M 2 + Expt 2 0.0600 − Expt 4 0.0404 For [NH 4 ], = =3 For [NO 2 ], = =2 Expt 1 0.0200 Expt 3 0.0202 Expt 2 32.3 10 −7 Expt 4 21.6 10− 7 Rate, = =3 Rate, = =2 Expt 1 10.8 10 − 7 Expt 3 10.8 10 − 7 Rate Factor = [NH +4 ]n 3 = 3n Rate Factor = [NO−2 ]m 2 = 2m  n = 1, first order  m = 1, first order 80 The Effect of Temperature on Rate changing the temperature changes the rate constant of the rate law Svante Arrhenius investigated this relationship and showed that:  −RTEa  k = A e     where T is the temperature in kelvins R is the gas constant in energy units, 8.314 J/(mol∙K) A is a factor called the frequency factor Ea is the activation energy, the extra energy needed to start the molecules reacting 81 82 Energy Profile for the Isomerization of Methyl Isonitrile the collision frequency is the number of molecules that approach the peak in a given period of time the activation energy is the difference in energy between the reactants and the activated complex 83 Activation Energy and the Activated Complex energy barrier to the reaction amount of energy needed to convert reactants into the activated complex aka transition state the activated complex is a chemical species with partially broken and partially formed bonds always very high in energy because partial bonds 84 Isomerization of Methyl Isonitrile methyl isonitrile rearranges to acetonitrile in order for the reaction to occur, the H3C- N bond must break; and a new H3C-C bond form 85  Energy Profile for the Isomerization of Methyl Isonitrile As the reaction begins, the C-N bond weakens enough for the CN group to start to rotate the activated complex is a chemical species with partial bonds the collision frequency is the number of molecules that approach the peak in a given period of time 86  The Arrhenius Equation: The Exponential Factor the exponential factor in the Arrhenius equation is a number between 0 and 1 it represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier the higher the energy barrier (larger activation energy), the fewer molecules that have sufficient energy to overcome it that extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy of the molecules therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier therefore, increasing the temperature will increase the reaction rate 87 88 Arrhenius Plots the Arrhenius Equation can be algebraically solved to give the following form: − Ea 1 ln( k ) =   + ln ( A) R T this equation is in the form y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line (-8.314 J/mol∙K)(slope of the line) = Ea, (in Joules) ey-intercept = A, (unit is the same as k) 89 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) → O2(g) + O(g) given the following data: Temp, K k, M-1∙s-1 Temp, K k, M-1∙s-1 600 3.37 x 103 1300 7.83 x 107 700 4.83 x 104 1400 1.45 x 108 800 3.58 x 105 1500 2.46 x 108 900 1.70 x 106 1600 3.93 x 108 1000 5.90 x 106 1700 5.93 x 108 1100 1.63 x 107 1800 8.55 x 108 1200 3.81 x 107 1900 1.19 x 109 90 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) → O2(g) + O(g) given the following data: use a spreadsheet to graph ln(k) vs. (1/T) 91 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) → O2(g) + O(g) given the following data: Ea = m∙(-R) (  ) Ea = 1.12 10 4 K  8.314 J   = 9.3110 mol K  4 J mol solve for Ea kJ Ea = 93.1 mol A = ey-intercept A = e 26.8 = 4.36 1011 solve for A A = 4.36 1011 M -1 s −1 92 Arrhenius Equation: Two-Point Form if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:  k 2  Ea  1 1  ln   =  −   k1  R  T1 T2  Rx ln k1 R x ln k1 (R T1 T2 ) x ln k1 k2 k2 k2 Ea = = = 1 − 1 T1 − T2 (T1 − T2 ) T2 T1 T1 T2 93 Ex. 13.8 – The reaction NO2(g) + CO(g) → CO2(g) + NO(g) has a rate constant of 2.57 M-1∙s-1 at 701 K and 567 M-1∙s-1 at 895 K. Find the activation energy in kJ/mol Given: T1 = 701 K, k1 = 2.57 M-1∙s-1, T2 = 895 K, k2 = 567 M-1∙s-1 Find: Ea, kJ/mol Concept Plan: T1, k1, T2, k2 Ea k  E  1 1  ln  2  = a  −  Relationships:  k1  R  T1 T2   567 M s   1 1  -1 -1 Ea = − ln  Solution:  2.57 M -1 s -1  8.314 J   701 K 895 K   mol K 5.3965 = Ea J 8.314 mol K −1 (3.09 2  10 −4 K -1 ) 1.45 105 J mol = 145 mol kJ = Ea Check: most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable Activation Energy Exothermic reaction related to the rate of reaction Activation energy, Ea = energy required to start the reaction  Activation Energy Endothermic reaction related to the rate of reaction Most reactions have positive Ea since energy is absorbed to break bonds in reactant particles. Arrhenius Equation -Ea k = Ae RT Since rate = k[A]a[B]b... At fixed concentrations, rate depends on k which in turn depends on temperature (T) and the nature of the reaction (A and Ea) A depends on the nature of the reaction and varies with T Determination of Activation Energy Using Two Rate Constants − Ea RT1 k1 = A1 e − Ea RT2 k2 = A2 e Ea  1 1  Ea  1 1  k1 A1  −  R  T2 T1   − R  T2 T1  = e e  k2 A2 Ea  1 1  k1  − R  T2 T1  Ea  1 1  ln = lne  =  −  k2 R  T2 T1   Collision Theory of Kinetics for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other. once molecules collide they may react together or they may not, depending on two factors - 1. whether the collision has enough energy to "break the bonds holding reactant molecules together"; 2. whether the reacting molecules collide in the proper orientation for new bonds to form. 99  Effective Collisions collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions the higher the frequency of effective collisions, the faster the reaction rate when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state 100 Effective Collisions Kinetic Energy Factor for a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide it can form the activated complex 101 Effective Collisions Orientation Effect 102 Collision Theory and the Arrhenius Equation A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z)  −RTEa  − Ea k = A e  = pze RT    103 Orientation Factor the proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form the more complex the reactant molecules, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations leading to reaction have p slightly less than 1 for most reactions, the orientation factor is less than 1 for many, p 1 in which an electron is transferred without direct collision 104  Energy Profile Transition State Theory Transition State Theory - focuses on what happens after the collisions have started. Energy profile - shows the variation of the potential energy of the reaction mixture as the reaction proceeds. P.E. reaction coordinate Advantages of Transition State Theory 1. Ea and A can be calculated A  Zp  the steric factor p can be predicted 2. It explains why the reaction pathway is specific. 3. It is applicable to gaseous and aqueous reactions. Reaction Mechanisms we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism 109  An Example of a Reaction Mechanism Overall reaction: H2(g) + 2 ICl(g) → 2 HCl(g) + I2(g) 110 Mechanism: 1) H2(g) + ICl(g) → HCl(g) + HI(g) 2) HI(g) + ICl(g) → HCl(g) + I2(g) the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps Elements of a Mechanism Intermediates H2(g) + 2 ICl(g) → 2 HCl(g) + I2(g) 1) H2(g) + ICl(g) → HCl(g) + HI(g) 2) HI(g) + ICl(g) → HCl(g) + I2(g) notice that the HI is a product in Step 1, but then a reactant in Step 2 since HI is made but then consumed, HI does not show up in the overall reaction materials that are products in an early step, but then a reactant in a later step are called intermediates 111  Molecularity the number of reactant particles in an elementary step is called its molecularity a unimolecular step involves 1 reactant particle a bimolecular step involves 2 reactant particles though they may be the same kind of particle a termolecular step involves 3 reactant particles though these are exceedingly rare in elementary steps 112 Rate Laws for Elementary Steps each step in the mechanism is like its own little reaction – with its own activation energy and own rate law the rate law for an overall reaction must be determined experimentally but the rate law of an elementary step can be deduced from the equation of the step H2(g) + 2 ICl(g) → 2 HCl(g) + I2(g) 1) H2(g) + ICl(g) → HCl(g) + HI(g) Rate = k1[H2][ICl] 2) HI(g) + ICl(g) → HCl(g) + I2(g) Rate = k2[HI][ICl] 113 Rate Laws of Elementary Steps 114  Rate Determining Step in most mechanisms, one step occurs slower than the other steps the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy the rate law of the rate determining step 115 determines the rate law of the overall reaction Another Reaction Mechanism NO2(g) + CO(g) → NO(g) + CO2(g) Rateobs = k[NO2]2 1) NO2(g) + NO2(g) → NO3(g) + NO(g) Rate = k1[NO2]2 slow 2) NO3(g) + CO(g) → NO2(g) + CO2(g) Rate = k2[NO3][CO] fast The first step is slower than the second step because its activation energy is larger. The first step in this mechanism is the rate determining step. The rate law of the first step is the same as the rate law of the overall reaction. 116  Validating a Mechanism in order to validate (not prove) a mechanism, two conditions must be met: 1. the elementary steps must sum to the overall reaction 2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law 117  Mechanisms with a Fast Initial Step when a mechanism contains a fast initial step, the rate limiting step may contain intermediates when a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related and the product is an intermediate substituting into the rate law of the RDS will produce a rate law in terms of just reactants 118 Show that the proposed mechanism for the reaction 2 O3(g) → 3 O2(g) matches the observed rate law Rate = k[O3]2[O2]-1 k1 O3(g)  O2(g) + O(g) Fast k-1 O3(g) + O(g) → 2 O2(g) Slow Rate = k2[O3][O] for Step 1 Rateforward = Ratereverse Rate = k2 [O3 ][O] k1[O3 ] = k−1[O2 ][O] k1 Rate = k2 [O3 ] [O3 ][O 2 ]-1 k1 k−1 [O] = [O3 ][O 2 ]−1 k2 k1 k−1 Rate = [O3 ]2 [O 2 ]-1 k−1 119  Reaction Mechanism and Rate Law The number of reactant particles that takes part in each elementary step is called the molecularity of that step. Unimolecular – one particle collides with the wall of the vessel or the excess solvent Bimolecular – two particles collide together Termolecular – three particles collide together simultaneously (very rare) Effect of Catalysts on Rates of Reactions Catalysts catalysts are substances that affect the rate of a reaction without being consumed catalysts work by providing an alternative mechanism for the reaction with a lower activation energy catalysts are consumed in an early mechanism step, then made in a later step mechanism without catalyst mechanism with catalyst O3(g) + O(g) → 2 O2(g) V. Slow Cl(g) + O3(g)  O2(g) + ClO(g) Fast ClO(g) + O(g) → O2(g) + Cl(g) Slow 122 Ozone Depletion over the Antarctic 123  Energy Profile of Catalyzed Reaction polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals 124 Catalysts homogeneous catalysts are in the same phase as the reactant particles Cl(g) in the destruction of O3(g) heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system 125 Types of Catalysts 126 Catalytic Hydrogenation H2C=CH2 + H2 → CH3CH3 127 Working Principle of Catalysts and their Effects on Reaction Rates Catalysis  Catalytic action Catalysts alter the rates of reaction, 1. but remain chemically unchanged at the end of the reaction 2. by providing new, alternative reaction pathways with different activation energies.  Working Principle of Catalysts and their Effects on Reaction Rates Positive catalyst: Provides an alternative reaction pathway with a lower activation energy Lower Ea ➔ Greater fraction of molecules with K.E. greater than or equal to Ea ➔ Reaction proceeds faster Ea’ Working Principle of Catalysts and their Effects on Reaction Rates Negative catalyst: Provides an alternative reaction pathway with a higher activation energy Higher Ea ➔ Smaller fraction of molecules with K.E. greater than or equal to Ea ➔ Reaction proceeds slower Ea” Working Principle of Catalysts and their Effects on Reaction Rates With catalysts, the contour diagrams and thus the energy profiles are totally different from those without  Catalyst Homogeneous Heterogeneous Catalyst Catalyst Reactants & catalyst Reactants & catalyst are are in the same phase NOT in the same phase Characteristics of Catalysts 1. For a given reversible reaction, k1 Reactants Products K-1 catalysts affect the rates of forward reaction and backward reaction to the same extent.  Characteristics of Catalysts 2. Catalysts are chemically unchanged at the end of reactions but may undergo physical changes. E.g. Lumps of MnO2 used in the decomposition of H2O2 become powdered at the end of the reaction. 3. Only small quantity is sufficient to catalyze a reaction because catalysts can be regenerated. However, if the catalysts are involved in the rate equation, higher concentrations may affect the rate more. 4. The effect of heterogeneous catalysts depends on the surface area available for the catalytic action. Surface area of solid catalyst   number of reaction sites   catalytic activity E.g. Finely divided Fe powder is used as the catalyst in Haber process.  Characteristics of Catalysts 5. Catalytic actions are specific especially in biological systems. E.g. Enzymatic actions are highly specific. 6. The efficiency of a catalyst is often enhanced by adding promoters. Promoters have no catalytic actions on their own. E.g. Fe2O3, KOH, Al2O3 in Haber process 7. The efficiency of a catalyst can be lowered by adding poisons or inhibitors. Catalyst poisons are specific in action. E.g. Arsenic impurities may poison Pt but not V2O5 in Contact process 8. Transition metals or compounds/ions containing transition metals show marked catalytic activities. E.g. Pt, Ni, Fe, V2O5, MnO2, Mn2+ Fe3+ The catalytic actions are due to the presence of low-lying partially filled d-orbitals. Heterogeneous Catalysis – Adsorption Occur on the surface of the catalyst. Reactants are adsorbed on the surface, forming new bonds with the catalyst while weakening bonds in reactants 2. Products, once formed, are desorbed from the surface, Autocatalysis Catalysis in which the product acts as the catalyst of the reaction 2MnO4−(aq) + 16H+(aq) + 5C2O42−(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + H+(aq) + I−(aq) Examples of heterogeneous catalysis MnO2(s) 2H2O2(aq) ⎯⎯⎯⎯→ 2H2O(l) + O2(g) Fe(s) 3H2(g) + N2(g) ⎯⎯→ 2NH3(g) Al2O3/SiO2(s) C8H18(g) ⎯⎯⎯⎯⎯⎯→ C4H10(g) + C4H8(g) Ni(s) CH2=CH2(g) + H2(g) ⎯⎯→ CH3–CH3(g) Enzymes because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate protein molecules that catalyze biological reactions are called enzymes enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction 139 Enzyme-Substrate Binding Lock and Key Mechanism 140 Enzymatic Hydrolysis of Sucrose 141 Thank you