Chapter

Chapter

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 13 Chemical Kinetics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall 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 Tro, Chemistry: A Molecular Approach 2 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 Tro, Chemistry: A Molecular Approach 3 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 concentration Rate time [product] [reactant] Rate time time Tro, Chemistry: A Molecular Approach 4

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. Tro, Chemistry: A Molecular Approach 5 at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 ]

[ ) ][2C A ]1) [C[]A ] ([C]2([A 1 Rate= == (t2 (tt21)t1) tt ) [ []Z X ]1) [Z[]X ] ([Z]2([X 2 ] 1 Rate= == (tt21)t1) tt (t2 (4( 40 )8) Rate= .25 == 0.0 25 (1(61 6 0)0)

(1( 70 )8) Rate= .0625 == 0.0 0625 (1(61 6 0)0) Tro, Chemistry: A Molecular Approach 6 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6

[Y] = 6 [Z] = 2 ] [ ) ][2C A ]1) [C[]A ] ([C]2([A 1 Rate= == (t2 (tt21)t1) tt ) [ []Z X ]1) [Z[]X ] ([Z]2([X 2 ] 1 Rate= == (tt21)t1) tt (t2 (6( 2 4)4) Rate= .125 == 0.0 125

(1(61 6 0)0) (2(6 1)7) .0625 Rate= == 0.0 0625 (1(61 6 0)0) Tro, Chemistry: A Molecular Approach 7 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8

at t = 48 [X] = 5 [Y] = 5 [Z] = 3 ] [ ) ][2C A ]1) [C[]A ] ([C]2([A 1 Rate= == (t2 (tt21)t1) tt ) [ []Z X ]1) [Z[]X ] ([Z]2([X 2 ] 1 Rate= == (tt21)t1) tt (t2 (8( 06 )2) .125 Rate=

== 0.0 125 (1(61 6 0)0) (3( 5 2)6) .0625 Rate= == 0.0 0625 (1(61 6 0)0) Tro, Chemistry: A Molecular Approach 8 Hypothetical Reaction Red Blue Time (sec) Number Number Red Blue 0

100 0 5 84 16 10 71 29 15 59 41 20 50 50 25 42 58

30 35 65 35 30 70 40 25 75 45 21 79 50 18 82 in this reaction, one molecule of Red turns

into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules 9 over time Hypothetical Reaction Red Blue Concentration vs Time for Red -> Blue 100 100 84 90 Number of Molecules 80 71 70 59 60

65 50 70 75 79 82 Number Red Number Blue 58 42 50 50 40 35 41 30 30 25

21 29 20 10 18 16 0 0 5 10 15 20 25 30 35 40 45 50

Time (sec) Tro, Chemistry: A Molecular Approach 10 Hypothetical Reaction Red Blue Rate of Reaction Red -> Blue 4.5 4 3.5 Rate, [Blue]/ t 5, 3.2 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 Tro, Chemistry: A Molecular Approach 20 30 time, (sec) 40 50 11 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 Tro, Chemistry: A Molecular Approach 12 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

Tro, Chemistry: A Molecular Approach 13 Hypothetical Reaction Red Blue Avg. Rate Avg. Rate Avg. Rate (5 sec Number Number intervals) Red Blue Time (sec) 0 100 0 5 84 16 3.2 10 71 29

2.6 15 59 41 2.4 20 50 50 1.8 25 42 58 1.6 30 35 65

1.4 35 30 70 1 40 25 75 1 45 21 79 0.8 50 18 82 0.6

(10 sec intervals) (25 sec intervals) 2.9 2.1 2.3 1.5 1 0.7 1 14 H2 I2 Time (s) [H2], M 0.000 1.000 10.000 0.819 20.000 0.670 30.000 0.549 40.000 0.449 50.000 0.368 60.000 0.301 70.000 0.247 80.000 0.202 90.000 0.165

100.000 0.135 HI Avg. Rate, M/s Avg. Rate, M/s -[H2]/t [HI], M 1/2 [HI]/t 0.000 StoichiometryThe tells us thatrate for is average every 1 mole/L H0.0181 0.362 0.0181 theof change 2 used, in the 2 moles/L HI are 0.0149 made. in a 0.660 0.0149of concentration given time period. Assuming 0.902

0.0121a 1 L container, 0.0121 at 10 s, we used 0.181 moles of In the first 10 s, the 1.102 0.0100 H2. 0.0100 Therefore the amount of [H2] is0.0081 -0.181 M, 1.264 0.0081 HI made is 2(0.181 moles) = so the rate is 1.398 0.0067 0.0067 0.362 moles 0.181M 0.699 moles At 60 s, we used 1.506 0.0054 0.0054 10.000 s of H0.0045 the0.0045 amount

M 2. Therefore 0.0181 1.596 s of HI made is 2(0.699 moles) 1.670 0.0037 0.0037 = 1.398 moles 15 1.730 0.0030 0.0030 2.000 1.800 1.600 1.400 1.200 Concentration vs. Time for I --> H 2HI 2+ 2 average rate in a given

time period = slope of the line connecting the [H2] points; and +slope of the line for [HI] the average rate for the first 80 10 s is 0.0108 40 0.0181 M/s 0.0150 1.000 [H2], M [HI], M 0.800 concentration, (M) 0.600 0.400 0.200 0.000 0.000 10.000 20.000 30.000

Tro, Chemistry: A Molecular Approach 40.000 50.000 60.000 time, (s) 70.000 80.000 90.000 100.000 16 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 Tro, Chemistry: A Molecular Approach 17

H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is: Rate 0.28 M 40 s Rate 0.0070 M s Using [HI], the instantaneous rate at 50 s is: 1 0.56 M Rate 2 40 s Rate 0.0070 M s 18 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 the [H+]. H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3(aq) + 2 H2O(l) Solve the equation

for the Rate (in terms of the change in concentration of the Given quantity) Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value 1 [I ] 1 ( 0.868 M 1.000 M ) Rate 10 s 3 t 3 Rate 4.40 10-3 M s 1 [H ] Rate 2 t [H ] 2( Rate ) t

[H ] M M 2 4.40 10-3 8.80 10-3 s s t ( ) 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 Tro, Chemistry: A Molecular Approach 20 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 Tro, Chemistry: A Molecular Approach 21 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 Tro, Chemistry: A Molecular Approach 22

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 Tro, Chemistry: A Molecular Approach 23 Factors Affecting Reaction Rate Temperature increasing temperature increases reaction rate chemists rule of thumb - for each 10C 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 Tro, Chemistry: A Molecular Approach

24 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 how catalysts work will be examined later Tro, Chemistry: A Molecular Approach 25 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) Tro, Chemistry: A Molecular Approach 26

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 and homogeneous catalysts as well 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 n m Rate k[A] [B] Tro, Chemistry: A Molecular Approach 27 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 Tro, Chemistry: A Molecular Approach 28 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] +3 +2

+1 Tl + Hg2 Tl + 2 Hg +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. Tro, Chemistry: A Molecular Approach 29 Reactant Concentration vs. Time A Products Tro, Chemistry: A Molecular Approach 30 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 its initial value the half-life of the reaction depends on the order of the

reaction Tro, Chemistry: A Molecular Approach 31 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 = [A0]/2k when Rate = M/sec, k = M/sec [A]0 [A] Tro, Chemistry: A Molecular Approach slo p time e= -k

32 First Order Reactions Rate = k[A] ln[A] = -kt + ln[A]0 graph ln[A] vs. time gives straight line with slope = -k and y-intercept = ln[A]0 used to determine the rate constant t = 0.693/k the half-life of a first order reaction is constant the when Rate = M/sec, k = sec-1 Tro, Chemistry: A Molecular Approach 33 ln[A]0 slo p ln[A] e= k time Tro, Chemistry: A Molecular Approach 34

Half-Life of a First-Order Reaction Is Constant Tro, Chemistry: A Molecular Approach 35 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 Tro, Chemistry: A Molecular Approach 36 C4H9Cl + H2O C4H9OH + 2 HCl Concentration vs. Time for the Hydrolysis of C 4H9Cl 0.12 concentration, (M) 0.1

0.08 0.06 0.04 0.02 0 0 200 Tro, Chemistry: A Molecular Approach 400 600 time, (s) 800 1000 37 C4H9Cl + H2O C4H9OH + 2 HCl Rate vs. Time for Hydrolysis of C4H9Cl 2.5E-04 Rate, (M/s) 2.0E-04

1.5E-04 1.0E-04 5.0E-05 0.0E+00 0 100 Tro, Chemistry: A Molecular Approach 200 300 400 time, (s) 500 600 700 800 38 C4H9Cl + H2O C4H9OH + 2 HCl LN([C4H9Cl]) vs. Time for Hydrolysis of C 4H9Cl

0 slope = -2.01 x 10-3 k= 2.01 x 10-3 s-1 -0.5 LN(concentration) -1 -1.5 -2 -2.5 -3 t1 2 y = -2.01E-03x - 2.30E+00 -3.5 -4 -4.5 0 100 200

300 400 500 600 700 800 0.693 k 0.693 2.0110 3 s -1 345 s time, (s) Tro, Chemistry: A Molecular Approach 39 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

t = 1/(k[A0]) when Rate = M/sec, k = M-1sec-1 Tro, Chemistry: A Molecular Approach 40 k = e p o sl 1/[A] l/[A]0 time Tro, Chemistry: A Molecular Approach 41 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

2.847 0.05800 42 Rate Data Graphs For 2 NO2 2 NO + O2 Partial Pressure NO2, mmHg vs. Time 100.0 90.0 80.0 Pressure, (mmHg) 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0.0 0 50 Tro, Chemistry: A Molecular Approach 100 150 Time, (hr)

200 250 43 Rate Data Graphs For 2 NO2 2 NO + O2 ln(P NO2) vs. Time 4.8 4.6 4.4 4.2 ln(pressure) 4 3.8 3.6 3.4 3.2 3 2.8 2.6 2.4 0 50 Tro, Chemistry: A Molecular Approach 100

150 Time (hr) 200 250 44 Rate Data Graphs For 2 NO2 2 NO + O2 1/(PNO2) vs Time 0.07000 0.06000 Inverse Pressure, (mmHg -1 ) 1/PNO2 = 0.0002(time) + 0.01 0.05000 0.04000 0.03000 0.02000 0.01000 0.00000 0 50 Tro, Chemistry: A Molecular Approach

100 150 Time, (hr) 200 250 45 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 Tro, Chemistry: A Molecular Approach 46 Tro, Chemistry: A Molecular Approach 47 Practice - Complete the Table and

Determine the Rate Equation for the Reaction A 2 Prod [A], (M)[Prod], (M) Time (sec)ln([A]) 0.100 0 0 0.067 50 0.050 100 0.040 150 0.033 200 0.029 250 Tro, Chemistry: A Molecular Approach 1/[A]

48 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 Tro, Chemistry: A Molecular Approach 49 [A] vs. Time 0.12 0.1 0.08 0.06 concentration, M 0.04 0.02 0 0 50 100 150

200 250 time, (s) Tro, Chemistry: A Molecular Approach 50 LN([A]) vs. Time -2 -2.2 -2.4 -2.6 -2.8 -3 -3.2 Ln(concentration) -3.4 -3.6 -3.8 0

50 100 150 200 250 time, (s) Tro, Chemistry: A Molecular Approach 51 1/([A]) vs. Time y = 0.1x + 10 40 35 -1 30 25 20 15

10 inverse concentration, M 5 0 0 50 100 150 200 250 time, (s) Tro, Chemistry: A Molecular Approach 52 Practice - Complete the Table and Determine the Rate Equation for the Reaction A 2 Prod the reaction is second order, Tro, Chemistry: A Molecular Approach -[A]

Rate = = 0.1 [A]2 t 53 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: Find: Concept Plan: Relationships: Solution: [SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1 [SO2Cl2] [SO2Cl2]0, t, k [SO2Cl2] for a 1st order process : ln[A] kt ln[A]0 ln[SO 2Cl 2 ] kt ln[SO 2Cl 2 ]0 ( ) 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 Tro, Chemistry: A Molecular Approach 55 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 law including all reactants

Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same Expt. Initial Initial Initial Expt. Initial Initial Rate Rate Initial Number (M/s) [NO22], ], (M) (M) [CO], Number [NO [CO], (M) (M) (M/s) 1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082 3. 4. 0.20 0.40 0.20 0.10 n 0.0083 0.033 m Rate k[NO 2 ] [CO] Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not 56 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 what factor the concentrations and rates change

in these two experiments. [NO 2 ]expt 2 [NO 2 ]expt 1 Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20

0.0083 4. 0.40 0.10 0.033 0.20 M 2 0.10 M Tro, Chemistry: A Molecular Approach Rateexpt 2 Rateexpt 1 0.0082 M s 4 M 0.0021 s 57 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 what power the

concentration factor must be raised to equal the rate factor. [NO2 ]expt 2 [NO 2 ]expt 1 Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20

0.20 0.0083 4. 0.40 0.10 0.033 0.20 M n M Rate 0 . 0082 expt 2 2 s [NO 2 ]expt 2 Rate expt 2 4 0.10 M M Rateexpt 1 0.0021 s

[NO 2 ]expt 1 Rate expt 1 2 n 4 n 2 Tro, Chemistry: A Molecular Approach 58 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 reactants Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2.

0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 m 0[CO] M Rate Rate 0 . 0083 expt

3 expt 3 . 20 M expt 3 s [CO] 2 Rate 1 expt 2 Rateexpt 0.0082 M s [CO]expt 2 0.10 M expt2 2 2 m 1 m 0 Tro, Chemistry: A Molecular Approach 59 [CO]expt 3 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 exponents into the general rate law to get the rate law for the reaction

Expt. Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4.

0.40 0.10 0.033 n0 m 2 n = 2, mRate =0 Rate 2k][NO k[NO [CO] 2 ] [CO] 2 Rate k[NO 2 ] Tro, Chemistry: A Molecular Approach 60 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 concentrations and rate for any experiment into the rate law and solve for k Expt.

Initial Initial Rate Initial Number [NO2], (M) [CO], (M) (M/s) Tro, Chemistry: A Molecular Approach 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 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 -1 -1 k 0 . 21 M s 0.01 M 2 61 Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1 given the data below. Expt. Initial Initial Initial Rate, [NH4+], M [NO2-], M (x 10-7), M/s No.

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 Tro, Chemistry: A Molecular Approach

62 Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1 given the data below. Expt. No. Initial [NH4+], M Initial [NO2-], M Initial Rate, (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 Expt 2 0.0600 For [NH 4 ], 3 Expt 1 0.0200 Expt 2 32.3 10 7 Rate, 3 7 Expt 1 10.8 10 Rate Factor [NH 4 ]n 3 3n n 1, first order

Rate = k[NH4+]n[NO2]m Rate k[NH 4 ][NO 2 ] for expt 1 10.8 10-7 M s k ( 0.0200 M )( 0.200 M ) 10.8 10-7 M s k 2.70 10 4 M -1 s -1 4.00 10-3 M 2 Expt 4 0.0404 For [NO 2 ], 2 Expt 3 0.0202 Expt 4 21.6 10 7 Rate, 2 7 Expt 3 10.8 10 Rate Factor [NO 2 ]m 2 2 m m 1, first order 63 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: Ea k A e RT where T is the temperature in kelvins R is the gas constant in energy units, 8.314 J/(molK) A is a factor called the frequency factor Ea is the activation energy, the extra energy needed to start the molecules reacting Tro, Chemistry: A Molecular Approach 64 Tro, Chemistry: A Molecular Approach 65 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 Tro, Chemistry: A Molecular Approach 66

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 Tro, Chemistry: A Molecular Approach 67 Energy Profile for the Isomerization of Methyl Isonitrile As the reaction the activated collision activation frequency complex energy begins, the C-N is athe chemical number difference species of in bond weakens molecules energy

with partial between that bonds the enough for the approach reactants the peak the toin a CNand group givenstart activated period complex of time to rotate 68 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 Tro, Chemistry: A Molecular Approach 69 Tro, Chemistry: A Molecular Approach 70 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/molK)(slope of the line) = Ea, (in Joules) ey-intercept = A, (unit is the same as k)

Tro, Chemistry: A Molecular Approach 71 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-1s-1 Temp, K k, M-1s-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 Tro, Chemistry: A Molecular Approach 72 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) Tro, Chemistry: A Molecular Approach 73 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) solve for Ea A = ey-intercept solve for A ( )

J 4 J Ea 1.12 10 4 K 8.314 9.3110 mol K mol kJ Ea 93.1 mol A e 26.8 4.36 1011 A 4.36 1011 M -1 s 1 Tro, Chemistry: A Molecular Approach 74 Arrhenius Equation: Two-Point Form if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used: Ea k2 ln k1 R

Rx Ea 1 T2 1 1 T1 T2 k1 k1 k1 ( R T1 T2 ) x ln ln R x ln k2 k2 k2 1 T1 T2 ( ) T T 1 2

T1 T1 T2 Tro, Chemistry: A Molecular Approach 75 Ex. 13.8 The reaction NO2(g) + CO(g) CO2(g) + NO(g) has a rate constant of 2.57 M-1s-1 at 701 K and 567 M-1s-1 at 895 K. Find the activation energy in kJ/mol Given: Find: Concept Plan: T1 = 701 K, k1 = 2.57 M-1s-1, T2 = 895 K, k2 = 567 M-1s-1 Ea, kJ/mol T1, k1, T2, k2 Ea k E ln 2 a k1 R Relationships: -1 -1 Ea 567 M s ln

Solution: 2.57M -1 s -1 8.314 molJ K 5.3965 1.45 105 1 1 T1 T2 1 1 701 K 895 K Ea 4 -1 ( ) 3 . 0 9

2 10 K J 1 8.314 mol K J mol kJ 145 mol Ea Check: most activation energies are tens to hundreds of kJ/ mol so the answer is reasonable 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. Tro, Chemistry: A Molecular Approach 77 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 Tro, Chemistry: A Molecular Approach 78 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 Tro, Chemistry: A Molecular Approach 79 Effective Collisions Orientation Effect Tro, Chemistry: A Molecular Approach 80

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 k A e Tro, Chemistry: A Molecular Approach Ea pze RT 81 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 there are some reactions that have p > 1 in which an electron is transferred without direct collision Tro, Chemistry: A Molecular Approach 82 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 Tro, Chemistry: A Molecular Approach 83

An Example of a Reaction Mechanism Overall reaction: H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) Mechanism: H2(g) + ICl(g) HCl(g) + HI(g) 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 Tro, Chemistry: A Molecular Approach 84 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

Tro, Chemistry: A Molecular Approach 85 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 Tro, Chemistry: A Molecular Approach 86 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] Tro, Chemistry: A Molecular Approach 87 Rate Laws of Elementary Steps Tro, Chemistry: A Molecular Approach 88 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 determines the rate law of the overall reaction Tro, Chemistry: A Molecular Approach 89 Another Reaction Mechanism

NO2(g) + CO(g) NO(g) + CO2(g) 1) NO2(g) + NO2(g) NO3(g) + NO(g) 2) NO3(g) + CO(g) NO2(g) + CO2(g) Rateobs = k[NO2]2 Rate = k1[NO2]2 slow 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. Tro, Chemistry: A Molecular Approach 90 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 Tro, Chemistry: A Molecular Approach 91 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 Tro, Chemistry: A Molecular Approach 92 An Example k1 2 NO(g) N2O2(g) k-1 H2(g) + N2O2(g) H2O(g) + N2O(g) Fast Slow Rate = k2[H2][N2O2] H2(g) + N2O(g) H2O(g) + N2(g) Fast

2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2 for Step 1 Rateforward = Ratereverse k1[NO]2 k 1[N 2O 2 ] k1 [N 2O 2 ] [NO]2 k 1 Tro, Chemistry: A Molecular Approach Rate k2 [H 2 ][N 2O 2 ] k1 Rate k2 [H 2 ] [NO2 ]2 k 1 k2k1 Rate [H 2 ][NO2 ]2 k 1 93 Ex 13.9 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) k-1 O3(g) + O(g) 2 O2(g) Slow for Step 1 Rateforward = Ratereverse k1[O3 ] k 1[O2 ][O] k1

[O] [O3 ][O2 ] 1 k 1 Tro, Chemistry: A Molecular Approach Fast Rate = k2[O3][O] Rate k2[O3 ][O] k1 Rate k2[O3 ] [O3 ][O2 ]-1 k 1 k2 k1 Rate [O3 ]2 [O 2 ]-1 k 1 94 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) Cl(g) + O3(g) O2(g) + ClO(g) Fast ClO(g) + O(g) O2(g) + Cl(g) Slow V. Slow Tro, Chemistry: A Molecular Approach 95 Ozone Depletion over the Antarctic Tro, Chemistry: A Molecular Approach 96 Energy Profile of Catalyzed Reaction polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals Tro, Chemistry: A Molecular Approach

97 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 cars exhaust system Tro, Chemistry: A Molecular Approach 98 Types of Catalysts Tro, Chemistry: A Molecular Approach 99 Catalytic Hydrogenation H2C=CH2 + H2 CH3CH3 Tro, Chemistry: A Molecular Approach 100 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 Tro, Chemistry: A Molecular Approach 101 Enzyme-Substrate Binding Lock and Key Mechanism Tro, Chemistry: A Molecular Approach 102 Enzymatic Hydrolysis of Sucrose Tro, Chemistry: A Molecular Approach 103

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