CHEM 2880 - Kinetics Determining Rate Order As mentioned earlier, the rate order of a reaction can only be determined experimentally. A number of methods can be used to determine both the reaction order and the rate constant. 1. Integration method - the concentration (of reagent or product) is measured at various time intervals, and the integrated rate law equations are plotted. The order is determined by the plot with the best linear fit (most consistent k value). 2. Isolation method - if more than one reactant is involved, the reaction is run several times, each time changing the initial concentration of one of the reagents. Any corresponding change in the rate is due to the reagent that changed concentration and the corresponding order can be determined. Once the order wrt to each reagent has been determined the overall rate can be calculated. 3. Reagents in excess - this method can be used to reduce the order of a reaction (as mentioned earlier), and make complicated rate laws easier to analyze. 40 CHEM 2880 - Kinetics 4. Differential method - for an nth-order reaction, the rate law is: taking logarithms of both sides yields: Thus a plot of log v vs log[A] will be linear with a slope of n and an intercept of log k. This method often involves measuring the initial rate of reaction at several different initial concentrations. This has the advantages of: avoiding any complications caused by build up of products; and measuring rate at a time when the concentration is known most accurately. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 459. 41 CHEM 2880 - Kinetics Examples: Integration method: The following pharmacokinetic data were determined for the elimination of beta adrenergic blocking agents (beta blockers, used to treat hypertension) from the blood. The drug was administered to the patients and their blood plasma was monitored for the drug over a period of time. Determine the order of reaction and rate constant for this process. t (min) 30 60 120 150 240 360 480 c (ng/ml) 699 622 413 292 152 60 24 From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 252. 42 CHEM 2880 - Kinetics Using the integration method requires making plots for each of zero-, 1st-, and 2nd-order. The respective plots are c vs t, ln c vs t and 1/c vs t. The resulting table and plot look like: t c ln c 1/c*100 30 699 6.55 0.143 60 622 6.43 0.161 120 413 6.02 0.242 150 292 5.68 0.342 240 152 5.02 0.658 360 60 4.09 1.667 480 24 3.18 4.167 Note: The 1/c values have been multiplied by 100 so that the ln c and 1/c plots are on the same scale. The ln c vs t is the best linear fit, indicating a 1st-order process. From the slope of the line, the rate constant is determined to be 7.6 x 10-3 s-1. The elimination of drugs from the body is usually a first order process. 43 CHEM 2880 - Kinetics Isolation method: For the acid catalyzed bromination of acetone: The reaction rate was determined for different initial concentrations of the reagents. Determine the rate law and the rate constant. + [CH33COCH ] [Br2] [H ] Rate of dis- appearance of Br2 (M) (M) (M) (M s-1) 1 0.30 0.050 0.050 5.7 x 10-5 2 0.30 0.10 0.050 5.7 x 10-5 3 0.30 0.050 0.10 1.2 x 10-4 4 0.40 0.050 0.20 3.1 x 10-4 5 0.40 0.050 0.050 7.6 x 10-5 From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 507. 44 CHEM 2880 - Kinetics For this method, we want to compare the reaction rates between two different runs where the concentration of only one reagent has changed from one run to the next. Comparing runs 1 and 2, the concentrations of [H+] and [CH33COCH ] are constant, the [Br2] increases by a factor of 2, and the rate does not change. Therefore the rate is not dependant on the [Br2] and the reaction is zero-order in [Br2]. Comparing runs 1 and 3, the concentrations of [Br2] and + [CH33COCH ] are constant, the [H ] increases by a factor of 2, and the rate increases by a factor of 2.2. Therefore the rate is directly proportional to the [H+] and the reaction is 1st-order in [H+]. Comparing runs 1 and 5, the concentrations of [H+] and [Br23] are constant, the [CH COCH3] increases by a factor of 1.3, and the rate increases by a factor of 1.3. Therefore the rate is directly proportional to the st [CH33COCH ] and the reaction is 1 -order in [CH33COCH ]. The rate law for this reaction is: + v = k [CH33COCH ][H ] 45 CHEM 2880 - Kinetics Using the rate law, the rate constant for each run can be calculated and the average rate constant determined. + [CH33COCH ] [H ] Rate k (M) (M) (M s-1) (M-1 s)-1 1 0.30 0.050 5.7 x 10-5 3.8 x 10-3 2 0.30 0.050 5.7 x 10-5 3.8 x 10-3 3 0.30 0.10 1.2 x 10-4 4.0 x 10-3 4 0.40 0.20 3.1 x 10-4 3.9 x 10-3 5 0.40 0.050 7.6 x 10-5 3.8 x 10-3 -3 Avg 3.86 x 10 46 CHEM 2880 - Kinetics Differential method: For the binding of glucose to the enzyme hexokinase, the -1 -1 following initial rate (v0) data (in mol L s )were obtained: [glucose] (mmol L-1) 1.00 1.54 3.12 4.02 1.34 5.0 7.6 15.5 20.0 3.00 7.0 11.0 23.0 31.0 10.0 21.0 34.0 70.0 96.0 [Enzyme] (mmol L-1) Determine the rate law and the rate constant. From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 248. 47 CHEM 2880 - Kinetics The data provided are initial rate constants for different initial concentrations for the two reagents, glucose and hexokinase. The rate law willbe of the form: ab v00 = k[glucose] [hexokinase]0 For a constant [hexokinase]0, the order of reaction wrt to glucose can be determined using: logv00 = log k’ + alog[glucose] b where k’ = k[hexokinase]0 Plotting log v00 vs log[glucose] for each [hexokinase]0 yields straight lines with slopes of a and intercepts of log k’. log k’ = log k + blog[hexokinase]0 Plotting log k’ vs log[hexokinase]0 yields a straight line with a slope of b and an intercept of log k. 48 CHEM 2880 - Kinetics The table with the data converted to logs looks like: log [glucose] (mmol L-1) log[hexokinase] k’ -3.00 -2.81 -2.51 -2.40 (mmol L-1) 1.34 0.699 0.881 1.19 1.30 -2.87 3.69 3.00 0.845 1.04 1.36 1.49 -2.52 4.04 10.0 1.32 1.53 1.85 1.98 -2.00 4.56 [hexokinase] (mmol L-1) The plot of this data looks like: The three lines all have slopes of 1, so the reaction is 1st order wrt to glucose. The intercepts are tabulated above. 49 CHEM 2880 - Kinetics The plot of log k’ vs log[hexokinase]0 looks like: The slope and intercept are 1 and 6.55 respectively. Therefore b is 1and k is 106.55 = 3.6 x 106. Thus the rate law for this reaction is: v = 3.6 x 106[glucose][hexokinase] 50 CHEM 2880 - Kinetics More Examples: Sucrose (table sugar) undergoes hydrolysis (reaction with water) to produce fructose and glucose: 6 C122HO211 + H2O C61H 26O (fructose) + C61H 26O (glucose) “This reaction has particular significance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so candy made with this combination is chewier and bot brittle as crystalline sucrose is. Sucrose is dextrorotatory (+) whereas the mixture of glucose and fructose resulting from inversion is levorotatory (-). Thus, a decrease in the concentration of sucrose will be accompanied by a proportional decrease in the optical rotation.” Given the following kinetic data, show that the reaction is first order and find k. t (min) 0 7.20 18.0 27.0 4 optical rotation (") +24.08° +21.40° +17.73° +15.01° -10.73° From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 508. 51 CHEM 2880 - Kinetics The total change in optical rotation from t=0 to t=4, is " " ( 0 - 4). This is proportional to the total change in " [sucrose]. The change in optical rotation at time=t, ( 0 - " t) is proportional to the change in [sucrose] at t. thus the [sucrose] remaining at t is: "" " " " " (0 - 4) - ( 0t - ) = t - 4 Since it is a 1st-order reaction: should be linear, and a plot of vs t will have a slope of -k. 52 CHEM 2880 - Kinetics " " " " " t ln(( t- 4)-( 0- 4)) (min) 0.00 +24.08 0.00000 7.20 +21.40 -0.08011 18.0 +17.73 -0.20141 27.0 +15.01 -0.30186 4 -10.73 The plot is linear, so the reaction is 1st-order and k = 1.11 x 10-2 s-1.