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
th 4. Differential method - for an n -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
Examples: CHEM 2880 - Kinetics 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
Using the integration method requires making plots for st nd each of zero-, 1 -, and 2 -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. st The ln c vs t is the best linear fit, indicating a 1 -order process. From the slope of the line, the rate constant is -3-1 determined to be 7.6 x 10 s. The elimination of drugs from the body is usually a first order process. 43
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. + [CH3COCH 3] [Br 2] [H ] Rate of disappearance of Br 2-1 (M) (M) (M) (M s ) 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
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 [CH3COCH 3] are constant, the [Br 2] increases by a factor of 2, and the rate does not change. Therefore the rate is not dependant on the [Br 2] and the reaction is zero-order in [Br ]. 2 Comparing runs 1 and 3, the concentrations of [Br 2] and + [CH3COCH 3] 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 st + reaction is 1 -order in [H ]. + Comparing runs 1 and 5, the concentrations of [H ] and [Br 2] are constant, the [CH3COCH 3] 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 [CH3COCH 3] and the reaction is 1 -order in [CH COCH ]. 3 3 The rate law for this reaction is: + v = k [CH3COCH 3][H ] 45
Using the rate law, the rate constant for each run can be calculated and the average rate constant determined. + [CH3COCH 3] [H ] Rate k -1-1 -1 (M) (M) (M s ) (M s ) -5-3 1 0.30 0.050 5.7 x 10 3.8 x 10-5 -3 2 0.30 0.050 5.7 x 10 3.8 x 10-4 -3 3 0.30 0.10 1.2 x 10 4.0 x 10-4 -3 4 0.40 0.20 3.1 x 10 3.9 x 10-5 -3 5 0.40 0.050 7.6 x 10 3.8 x 10 Avg 3.86 x 10-3 46
Differential method: For the binding of glucose to the enzyme hexokinase, the -1-1 following initial rate (v 0) data (in mol L s )were obtained: -1 [glucose] (mmol L ) 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 96.0-1 [Enzyme] (mmol L ) 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
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: a b v = k[glucose] [hexokinase] 0 0 0 For a constant [hexokinase] 0, the order of reaction wrt to glucose can be determined using: logv = log k + alog[glucose] where k = k[hexokinase] b 0 0 Plotting log v 0 vs log[glucose] 0 for each [hexokinase] 0 yields straight lines with slopes of a and intercepts of log k. 0 log k = log k + blog[hexokinase] 0 0 Plotting log k vs log[hexokinase] yields a straight line with a slope of b and an intercept of log k. 48
The table with the data converted to logs looks like: -1 log [glucose] (mmol L ) log[hexokinase] k -1 (mmol L ) -3.00-2.81-2.51-2.40 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 4.56-1 [hexokinase] (mmol L ) The plot of this data looks like: The three lines all have slopes of 1, so the reaction is 1 st order wrt to glucose. The intercepts are tabulated above. 49
The plot of log k vs log[hexokinase] 0 looks like: The slope and intercept are 1 and 6.55 respectively. 6.55 6 Therefore b is 1and k is 10 = 3.6 x 10. Thus the rate law for this reaction is: 6 v = 3.6 x 10 [glucose][hexokinase] 50
More Examples: Sucrose (table sugar) undergoes hydrolysis (reaction with water) to produce fructose and glucose: C12H22O 11 + H2O C6H12O 6 (fructose) + C6H12O 6 (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 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
The total change in optical rotation from t=0 to t=, is ( 0 - ). 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 0 t t st Since it is a 1 -order reaction: should be linear, and a plot of will have a slope of -k. vs t 52
t ln(( t- )-( 0- )) (min) 0.00 +24.08 0.00000 7.20 +21.40-0.08011 18.0 +17.73-0.20141 27.0 +15.01-0.30186-10.73 st The plot is linear, so the reaction is 1 -order and -2-1 k = 1.11 x 10 s. 53
A certain 1 -order reaction is 34.5% complete in 49 min st at 298 K. What is its rate constant? From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 500. 54
When the concentration of A in the reaction A B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to 4.0 min at 25 C. Calculate the order of the reaction and the rate constant. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 500. When [A] 0 decreased by a factor of 2, the t ½ increased by a factor of 2. Therefore t ½ is inversely proportional to [A] and n=2. 0 55
The progress of a reaction in the aqueous phase was monitored by the absorbance of a reactant at various times. Determine the order of reaction and the rate constant. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 500. Time (s) Abs ln Abs 1/Abs 0 1.67 0.5128 0.5988 54 1.51 0.4121 0.6623 171 1.24 0.2151 0.8065 390 0.847-0.1661 1.181 720 0.478-0.7381 2.092 1010 0.301-1.201 3.322 1190 0.216-1.532 4.630 The ln Abs vs t plot has the best linear fit. Therefore the st -3-1 reaction is 1 -order and the rate constant is 1.7 x 10 s. 56
Iodoacetamide and N-acetylcysteine react with 1:1 stoichiometry. The following data were collected at 298-1 K for the reaction of 1.00 mmol L N-acetylcysteine with -1 1.00 mmol L iodoacetamide. From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 262. t (s) [ ] (mm) ln [ ] 1/[ ] 10 0.770-0.261 1.30 20 0.580-0.545 1.72 40 0.410-0.892 2.44 60 0.315-1.16 3.17 100 0.210-1.56 4.76 150 0.155-1.86 6.45 The 1/[ ] vs t plot has the best linear fit. Therefore the nd reaction is 2 -order and the rate constant is -2-1 -1-1 -1 3.7 x 10 mm s or 37 M s. 57
Effect of Temperature CHEM 2880 - Kinetics Intuitively most people would think that an increase in temperature would increase the rate of a reaction. x y Looking at a typical rate law v = k[a] [B], changing the temperature would not effect the concentrations of A and B, so any effect of temperature must reflect an effect on k. Thus k must increase with increasing temperature. However, temperature can effect reaction rate constants in a number of ways as illustrated below: 58
Plot a is the most common situation where rate constants increase with increasing temperatures. Plot b follows a similar trend, but eventually reaches a temperature where the rate constant reaches a maximum and then starts to decrease with further increases in temperature. A typical example here might be an enzyme-catalyzed reaction. As long as the enzyme remains in its native state, the rate constant increases with increasing temperature, but eventually a temperature is reached where the enzyme will denature, and the rate constant starts to decrease. Plot c is quite uncommon, but has been observed for reactions where one of the steps in the reaction is strongly exothermic and this effect outweighs any increase in the rate constant. Plot d illustrates the trend in rate constants for chain reactions. The rate gradually increases with temperature until a threshold is reached where the propagation reactions become significant. Any further increase in temperature causes a more dramatic increase in the rate constant. 59
The Arrhenius Equation CHEM 2880 - Kinetics The Arrhenius equation was derived empirically and describes the temperature dependence of many rate constants. Where A is the pre-exponential factor, E a is the activation energy, R is the gas constant and T is the temperature. Note: the units of E and R must match. a Taking the logarithm of both sides yields: A plot of ln k vs 1/T, or Arrhenius plot, has an intercept of lna and a slope of -E /R. Alternatively, if k 1, T 1 and k 2, T 2 are known: a Thus the activation energy for a reaction can be determined by measuring the rate constant at different temperatures. 60
A rough rule for reactions in solution near room temperature is that k increases 2-3x for every 10 C increase in temperature. Pre-exponential factor (A) has the same units as k represents the frequency of collisions between reagent molecules - more collisions means more chances for reaction to occur is somewhat temperature dependent, but less so than Ea and can be considered temperature independent over short (50 K) temperature ranges 61
Activation Energy (E a) represents the minimum energy required for initiation of the reaction the exponential term exp(e a/rt) is the fraction of collisions that have energy E a the magnitude of the activation energy reflects the sensitivity of the reaction to temperature Note: The reaction coordinate represents the change of the reactant into products as the reaction takes place. The Arrhenius equation is an empirical equation. We will look at two theories: Collision Theory and Transition State Theory which attempt to explain the effect of temperature on chemical kinetics. 62
Collision Theory of Chemical Kinetics Is based on the kinetic theory of gases and in its simplest form applies only to bimolecular reactions in the gas phase. Consider the reaction bimolecular, elementary reaction: M+N P for the reaction to occur M and N must collide the collision must involve some minimum kinetic energy the rate is proportional to the fraction of collisions having KE = E a this is reflected in the exponential term of the Arrhenius equation the rate is proportional to the frequency of collisions successful collisions require the correct orientation of molecules (i.e. steric factor) this is reflected in the pre-exponential term of the Arrhenius equation 63
The exponential term: CHEM 2880 - Kinetics The energy distribution of the molecules is represented by the Maxwell-Boltzmann distribution. As the temperature increases the most probable speed increases (the maximum in the curve), but the spread of the speed increases more markedly. The fraction of molecules with the highest speeds increases much faster with temperature than the average speeds. 64
The pre-exponential term: CHEM 2880 - Kinetics In collision theory: A = Pz where P = steric factor and z = collision frequency. z can be calculated from the kinetic theory of gases. The steric factor, P, takes into account that the molecules must have the correct orientation for the reaction to occur. For example: P values are generally very difficult to estimate accurately. Calculated rate constants of bimolecular gas-phase 11-1 -1 reactions: ~ 10 mol L s for small molecules. 2 13-1 -1 A = 10 to 10 mol L s for simple reactions. Few reactions proceed much faster than the rates calculated from simple collision theory, but many are orders of magnitude slower. 65
The overall result for Collision Theory is the modified Arrhenius equation: Drawbacks of Collision Theory based on kinetic theory of gases - assumes reagents are hard spheres ignores structure of molecules cannot be used to calculate activation energies Transition-State Theory overcomes some of these drawbacks. 66
Transition-State Theory CHEM 2880 - Kinetics developed by Eyring et al. in the 1930's focuses on the relationship between the reagents and the transition state allows for accurate calculation of rate constants applicable to gas and solution phase reactions assumes reagents are in constant equilibrium with transition state The transition state MN has a lifetime of only a few molecular vibrations, but can sometimes be observed -15-12 using femtosecond (10 s) or picosecond (10 s) spectroscopy. 67
K is the equilibrium constant between the reagents and the transition state and k is a universal rate constant: These can then be related back to the observable rate constant: thus: k = k K Statistical thermodynamics tells us: where k B is the Boltzmann constant and h is Planck s constant. Thus K can be determined experimentally from the rate constant. 68
For any equilibrium G = RTlnK. Thus for the equilibrium between the reagents and the transition state: G = RTlnK Since K can be calculated from experimental rate constants, this gives us a way to determine the free energy of activation. G reverse - G forward = G rxn Given G forward from the rate constant and G rxn, G can also be calculated. reverse 69
Also: G = H - T S So: Compare this to the Arrhenius equation: a Given the E and A, the enthalpy and entropy changes for activation can be determined. 70
Examples The rate of bacterial hydrolysis of fish muscle is twice as great at 2.2 C as at -1.1 C. Estimate the E a for this reaction. Is there any relation to the problem of storing fish for food? From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 503. Using the Arrhenius equation: This activation energy is relatively large, and therefore the reaction is quite temperature sensitive. Thus refrigeration is an essential and effective method for preserving fish. 71
st -4-1 The rate constant of a 1 -order reaction is 4.60 x10 s at -1 350 C. If the activation energy is 104 kj mol, calculate -4-1 the temperature at which its rate constant is 8.80 x 10 s. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 504. Again, using the Arrhenius equation: 72
Determine A and E a from the following data: From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 257. -1-1 -1 T (K) k (L mol s ) 1/T (K ) lnk 300 7.90E+06 0.00333 15.9 350 3.00E+07 0.00286 17.2 400 7.90E+07 0.00250 18.2 450 1.70E+08 0.00222 19.0 500 3.20E+08 0.00200 19.6 Plotting ln k vs 1/T: Slope = -E a/r and intercept = lna -1-1 -2767.2 = -E a/8.314 J K mol 25.11 = lna -1 10 E = 23 kj mol A = 8.0 x 10 a 73
The rate constant of a reaction increases by a factor of 1.23 when the temperature is increased form 20 C to 27 C. What is the activation energy of the reaction? From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 263. Using the Arrhenius equation: 74
Make an appropriate Arrhenius plot for the following data for the binding of an inhibitor to the enzyme carbonic anhydrase and calculate the activation energy for the reaction. From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 263. 6-1 -1-1 T (K) k (10 Lmol s ) 1/T (K ) lnk 289.0 1.04 0.00346 13.9 293.5 1.34 0.00341 14.1 298.1 1.53 0.00335 14.2 303.2 1.89 0.00330 14.5 308.0 2.29 0.00325 14.6 313.5 2.84 0.00319 14.9 Plotting lnk vs 1/T: Slope = -E a/r -3620.8= -E a/8.314 J K mol E = 30.1 kj mol -1 a -1-1 75
Food rots about 40 times more rapidly at 25 C than when it is stored at 4 C. Estimate the overall activation energy for the processes responsible for decomposition. From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 263. Using the Arrhenius equation: 76
The pre-exponential factor and activation energy for the uni-molecular reaction CH2HC (g) CH3CN (g) 13-1 -1 are 4.0 x 10 s and 272 kj mol, respectively. Calculate the values of H, S and G at 300 K. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 479. -1 H = E a = 272 kj mol G = H - T S -1-3 -1-1 = 272 kj mol - (300K)(15.4 x 10 kj K mol ) -1 = 267 kj mol 77
Estimate the activation Gibbs energy for the decomposition of urea in the reaction + 2- (NH 2) 2CO (aq) + 2H2O (l) 2NH 4 (aq) + CO 3 (aq) for which the pseudo-first-order rate constant is -7-1 -7-1 1.2 x 10 s at 60 C and 4.6 x 10 s at 70 C. From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 293. Using : k = k K, k =kbt/h and G = -RTln K 60 C 70 C. -7-1 -7-1 k 1.2 x 10 s 4.6 x 10 s T 333 K 343 K 12 12 k =kbt/h 7.34 x 10 7.56 x 10-20 -20 K = k/k 1.63 x 10 6.07 x 10-1 -1 G = -RTln K 126 kj mol 126 kj mol 78
Calculate the Gibbs energy, enthalpy and entropy of activation (at 300 K) for the binding of an inhibitor to the enzyme carbonic anhydrase by using the following data: From: Physical chemistry for the life sciences, P.W. Atkins, W.H. Freeman and Company, 2005, p. 293. 6-1 -1 T (K) k (10 Lmol s ) 1/T lnk 289.0 1.04 0.00346 13.9 293.5 1.34 0.00341 14.1 298.1 1.53 0.00335 14.2 303.2 1.89 0.00330 14.5 308.0 2.29 0.00325 14.6 313.5 2.84 0.00319 14.9 Slope = -E a/r = -36020.8 K H = E a -1-1 = -(-36020.8 K)(8.314 J K mol ) = 30.1 kj mol -1 79
intercept = lna = 26.403 A = 2.93 x 10 11 CHEM 2880 - Kinetics G = H - T S -1-3 -1-1 = 30.1 kj mol - (300K)(-25.9 x 10 kj K mol ) -1 = 37.9 kj mol 80
The thermal isomerization of cyclopropane to propene in -4-1 the gas phase has a rate constant of 5.95 x 10 s at 500 C. Calculate the value of G for the reaction. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 504. Using : k = k K, k =kbt/h and G = -RTln K -4-1 k 5.95 x 10 s T 500 C = 773 K 13 k =kbt/h 1.70 x 10-17 K = k/k 3.49 x 10-1 G = -RTln K 243 kj mol 81
The rate of the elctron-exchange reaction between naphthalene and its anion radical are diffusion controlled: - - C10H 8 + C10H 8 C10H 8 + C10H8 The reaction is bimolecular and second order. Using the rate constants below, calculate the values of Ea, H, S and G at 307 K for the reaction. From: Physical Chemistry for the Chemical and Biological Sciences, R. Chang, University Science Books, 2000, p. 504. 9-1 -1 T (K) k (10 Lmol s ) 1/T lnk 307 2.71 0.00326 21.7 299 2.40 0.00334 21.6 289 1.96 0.00346 21.4 273 1.43 0.00366 21.1 Slope = -E a/r = -1592.1 K H = E a -1-1 = -(-1592.1 K)(8.314 J K mol ) = 13.2 kj mol -1 82
intercept = lna = 26.912 A = 4.87 x 10 11 CHEM 2880 - Kinetics G = H - T S -1-3 -1-1 = 13.2 kj mol - (307K)(-21.9 x 10 kj K mol ) -1 = 20.0 kj mol 83
Catalysis CHEM 2880 - Kinetics catalysts are species that increase the rate of a reaction but are not consumed by the reaction forms an intermediate with the reagent(s) in the first step of the mechanism is released in the product-forming step does not appear in the overall reaction catalysts lower the G of the reaction by providing a different mechanism for the reaction which increases the rate of reaction lowering E a improving collision efficiency by improving the orientation of the molecules helping to break bonds can also have a concentration effect the increase in rate applies to both the forward and reverse reactions used throughout history in food-preparation and wine-making 84
catalysts effect the G not the overall G rxn. Equilibrium may be achieved more quickly, but the thermodynamic equilibrium constant is not effected 85
Three types of catalysts: CHEM 2880 - Kinetics Heterogenous reactants and catalyst are in different phases usually gas/solid or liquid/solid examples: Haber synthesis of ammonia - solid iron Ostwald synthesis of nitric acid - solid Pt/Rh hydrogenation of liquid fatty acids to saturated fatty acids - solid Pd, Pt or Ni Homogenous reactants and catalyst are in the same phase Enzyme subset of homogenous very complex type of catalyst 86
Example The activation energy for the decomposition of hydrogen -1 peroxide in solution is 76 kj mol. When iodide ions are -1 added, the activation energy falls to 57 kj mol. Assuming that the pre-exponential factor does not change upon addition of a catalyst, calculate the change in the rate constant. The enzyme catalase reduces the activation energy for the -1 same reaction to 8 kj mol which corresponds to a rate 15 increase of 10. 87