Effect of Concentration Integrated rate laws and reaction order ln[a] t = -kt + ln[a] 0
Effect of Concentration 1/[A] t = kt + 1/[A] 0 12-2
Effect of Concentration 12-3 [A] t = -kt + [A] 0
Effect of Concentration Graphical determination of the reaction order for the decomposition of N 2 O 5.
Effect of Concentration For the chemical reaction A C, a plot of 1/[A] t versus time was found to give a straight line with a positive slope. What is the order of reaction? Answer: second order 12-5
Effect of Concentration For the reaction X + Y Z, the reaction rate is found to depend only upon the concentration of X. A plot of 1/X verses time gives a straight line. What is the rate for this reaction? Answer: rate = k [X] 2 12-6
Effect of Concentration The following data were obtained for the gas phase decomposition of nitrogen dioxide at 300 C NO2 NO + ½ O2 Time [NO2], M 0 sec 0.0100 50 0.00787 100 0.00649 200 0.00481 300 0.00380 Is the reaction first order or second order? Answer: second order (plot of 1/[NO2] vs. time is linear 12-7
Effect of Concentration The thermal decomposition of acetaldehyde, CH 3 CHO CH 4 + CO, is a second-order reaction. The following data were obtained at 518 C. time, s Pressure CH 3 CHO, mmhg 0 364 42 330 105 290 720 132 Calculate the rate constant for the decomposition of acetaldehyde from the above data. Answer: k = 6.7 10 6/mmHg s 12-8
Worksheet # 10-5 1. The gas-phase conversion of 1,3-butadiene to 1,5-cyclooctadiene, 2C 4 H 6 C 8 H 12 was studied, providing data for the plot shown at the right. a. Explain how this plot confirms that the reaction is second order. b. Calculate the second-order rate constant, k. c. Determine the initial concentration of 1,3- butadiene in this experiment. 2. In the gas phase at 500. C, cyclopropane reacts to form propene. a. Explain how this plot confirms that the reaction is first order. b. Calculate the first-order rate constant, k. c. Determine the initial concentration of cyclopropane in this experiment [y-intercept = - 2.30] 12-9
Worksheet # 10 5: Answers 1. The gas-phase conversion of 1,3-butadiene to 1,5- cyclooctadiene, 2C 4 H 6 C 8 H 12 was studied, providing data for the plot shown at the right. a. Explain how this plot confirms that the reaction is second order. Plot of 1/[A] vs time b. Calculate the second-order rate constant, k. Answer = 0.04 /Ms c. Determine the initial concentration of 1,3-butadiene in this experiment. Answer = 0.025 M 2. In the gas phase at 500. C, cyclopropane reacts to form propene. a. Explain how this plot confirms that the reaction is first order. Plot of ln [A] vs time b. Calculate the first-order rate constant, k. Answer = 0.00175 /s c. Determine the initial concentration of cyclopropane in this experiment. Answer = 0.10 M 12-10 12-10
Effect of Concentration An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Zero Order First Order Second Order Rate law rate = k rate = k [A] rate = k [A] 2 Units for k mol/l*s 1/s L/mol*s Integrated rate law in straight-line form Plot for straight line Slope, y-intercept [A] t = ln[a] t = 1/[A] t = k t + [A] 0 -k t + ln[a] 0 k t + 1/[A] 0 [A] t vs. t ln[a] t vs. t 1/[A] t = t k, [A] 0 -k, ln[a] 0 k, 1/[A] 0 Half-life [A] 0 /2k ln 2/k 1/k [A] 0 12-11
Effect of Temperature Increasing the temperature of the reaction increases the number of effective collisions by both increasing the total number of collisions (because molecules move faster) and increasing the fraction of collisions that are effective (because the average kinetic energy is higher). Figure 12.9 12-12
Effect of Temperature Figure 12.9 12-13
Effect of Temperature The dependence of possible collisions on the product of reactant concentrations. A A B B 4 collisions Add another molecule of A A A A B B 6 collisions A B Add another molecule of B A B A B
Effect of Temperature The Effect of E a and T on the Fraction (f) of Collisions with Sufficient Energy to Allow Reaction E a (kj/mol) f (at T = 298 K) 50 1.70x10-9 75 7.03x10-14 100 2.90x10-18 T f (at E a = 50 kj/mol) 25 0 C(298K) 1.70x10-9 35 0 C(308K) 3.29x10-9 45 0 C(318K) 6.12x10-9
Effect of Temperature The effect of temperature on the distribution of collision energies
Collision Energy Effect of Temperature Collision Energy Energy-level diagram for a reaction ACTIVATED STATE E a (forward) E a (reverse) REACTANTS PRODUCTS The forward reaction is exothermic because the reactants have more energy than the products.
Effect of Temperature An energy-level diagram of the fraction of collisions exceeding E a.
Sample Problem Suppose the collision rate between molecules A and B at 25 C is 10,000 collisions per second. The number of effective collisions at this same temperature is 100 collisions per second. How will each of the following affect the total number of collisions and the fraction of effective collisions between molecules A and B? The temperature decreases The concentration of reactant B decreases 12-19
Sample Problem When the temperature decreases Average kinetic energy decreases Average velocity of the molecules decreases Should decrease the number of collisions and the fraction of effective collisions! Therefore, we expect the total number of collisions to be less than 10,000 and the fraction of effective collisions to be less than 100. 12-20
Sample Problem When the concentration of reactant B decreases The number of collisions between A and B decreases Less B molecules can make contact with A molecules. The fraction of effective collisions should remain the same The average kinetic energy and temperature of the molecules emains the same. Therefore, we expect the total number of collisions to be less than 10,000 and the number of effective collisions to be less than 100, but still one-tenth of the total number of collisions (same fraction as before). 12-21
Figure 16.10 Effect of Temperature Dependence of the rate constant on temperature
Effect of Temperature The Arrhenius Equation k Ae Ea RT ln k = ln A - E a /RT where k is the kinetic rate constant at T E a is the activation energy R is the energy gas constant T is the Kelvin temperature A is the collision frequency factor ln k 2 k 1 = - E a R 1 1 - T 2 T 1 R = 8.314 J/mol*K
Effect of Temperature Graphical determination of the activation energy ln k = -E a /R (1/T) + ln A
Sample Problem The decomposition of hydrogen iodide, 2HI(g) H 2 (g) + I 2 (g) has rate constants of 9.51x10-9 L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find E a. SOLUTION: ln k 2 k 1 = - E a R 1 1 - E a = - R ln k 2 1 1 - T k 2 1 T 2 T 1 T 1-1 E a = - (8.314J/mol*K) ln 1.10x10-5 L/mol*s 9..51x10-9 L/mol*s 1 600K - 1 500K E a = 1.76x10 5 J/mol = 176 kj/mol
Sample Problem The activation energy for the reaction CH 3 CO CH 3 + CO is 71 kj/mol. How many times greater is the rate constant for this reaction at 170 C than at 150 C? Answer: 2.5 12-26
Sample Problem The activation energy for the following first-order reaction is 102 kj/mol. N 2 O 5 (g) 2NO 2 (g) + (1/2)O 2 (g) The value of the rate constant (k) is 1.35 10 4 s 1 at 35 C. What is the value of k at 0 C? Answer: 8.2 10 7 s 1 12-27
Worksheet # 10-6 1. The rate constant for the gas phase decomposition of N2O5, N2O5 2NO2 + ½ O2 has the following temperature dependence: Temperature (K) Rate constant, k,(1/second) 338 0.0049 318 0.00050 298 0.000035 Determine the activation energy for this reaction. 2. A first order reaction has rate constants of 0.046 /second and 0.081 /second at 0 C and 20.0 C, respectively. What is the value of the activation energy? 3. The activation energy for the decomposition of HI to H2 and I2 is 186 kj/mol. The rate constant at 555 K is 3.52 x 10-7 L/mol sec. What is the rate constant of 645 K? 12-28
Worksheet Worksheet # 10 #- 6: 10 Answers - 6 1. The rate constant for the gas phase decomposition of N2O5, N2O5 2NO2 + ½ O2 has the following temperature dependence: Temperature (K) Rate constant, k,(1/second) 338 0.0049 318 0.00050 298 0.000035 Determine the activation energy for this reaction. Ea = 100 kj/mol 2. A first order reaction has rate constants of 0.046 /second and 0.081 /second at 0 C and 20.0 C, respectively. What is the value of the activation energy? Ea = 18.8 kj/mol 3. The activation energy for the decomposition of HI to H2 and I2 is 186 kj/mol. The rate constant at 555 K is 3.52 x 10-7 L/mol sec. What is the rate constant of 645 K? k = 9.5 x 10-5 L/mol sec 12-29 12-29
Effect of Surface Area Steel wool has more iron atoms exposed on the surface than the same mass of iron nail. Figure 12.4 Iron Nail Steel wool 12-30
Effect of Catalysts The catalyst called catalase in this piece of liver causes the decomposition of H 2 O 2 to occur faster. Figure 12.5 12-31
Effect of Catalysts Adding an appropriate catalyst increases the number of effective collisions by lowering the activation energy. This also increases the fraction of collisions that are effective. Figure 12.10 12-32
Effect of Catalysts Catalytic converters dramatically speed up the reactions of toxic gases to form harmless products: CO to CO 2 NO to N 2 and O 2 Figure 12.11 Catalyst is a palladium/platinum metal surface 12-33
Effect of Catalysts The thousands of enzymes in our bodies act to catalyze specific biological processes. Figure 12.12 12-34
Effect of Catalysts The enzyme sucrase catalyzes the decomposition of sucrose by making bondbreaking easier: Figure 12.13 12-35
Effect of Catalysts The metal-catalyzed hydrogenation of ethylene H 2 C CH 2 (g) + H 2 (g) H 3 C CH 3 (g)
Effect of Catalysts
Effect of Catalysts This catalyzed reaction has two transition states, and an intermediate that is lower in energy than the two transition states. 12-38
Effect of Catalysts Reaction energy diagram of a catalyzed and an uncatalyzed process.
Effect of Catalysts Each catalyst has its own specific way of functioning. In general a catalyst lowers the energy of activation. Lowering the E a increases the rate constant, k, and thereby increases the rate of the reaction
Effect of Catalysts A catalyst increases the rate of the forward AND the reverse reactions. A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction. A catalyst lowers E a by providing a different mechanism, for the reaction through a new, lower energy pathway. 12-41
Effect of Catalysts Chlorine atoms from CF 2 Cl 2 catalyze the decomposition of ozone in the stratosphere: O 3 (g) + Cl(g) ClO(g) + O 2 (g) ClO(g) + O 3 (g) Cl(g) + 2O 2 (g) The ClO(g) formed in step 1 is an intermediate that is formed temporarily. The Cl (g) is a catalyst as it is used in the first reaction and formed in the second reaction 12-42
Effect of Catalysts A catalyst is not a reactant or product. It interacts with the reactants, but is not permanently changed during the reaction. Since catalysts are recycled, small amounts are needed and last a long time. 12-43
Sample Problem The decomposition of HI is an exothermic reaction: 2 HI(g) H 2 (g) + I 2 (g) Draw an energy diagram for the uncatalyzed reaction. Label reactants and products. Sketch a possible activated complex. Use a dotted line to show the energy changes when platinum metal, a catalyst that increases the rate of the reaction, is added to the system. 12-44
Sample Problem 12-45
Sample Problem Ethene (H 2 C=CH 2 ) can be converted to ethanol (CH 3 CH 2 OH) by a three-step process. Identify any intermediates or catalysts. H 2 C=CH 2 + H 3 O + H 3 C-CH 2+ + H 2 O H 3 C-CH 2+ + H 2 O CH 3 CH 2 OH 2 + CH 3 CH 2 OH 2+ + H 2 O CH 3 CH 2 OH + H 3 O + 12-46
Sample Problem Identify any intermediates or catalysts. H 2 C=CH 2 + H 3 O + H 3 C-CH 2+ + H 2 O H 3 C-CH 2+ + H 2 O CH 3 CH 2 OH 2 + CH 3 CH 2 OH 2+ + H 2 O CH 3 CH 2 OH + H 3 O + Intermediates in this process include H 3 C-CH 2+ and CH 3 CH 2 OH 2+. H 3 C-CH 2+ is formed in step 1, then used in step 2. CH 3 CH 2 OH 2+ is formed in step 2, then used in step 3. H 3 O + is used in step 1, then regenerated in step 3, so it is a catalyst. 12-47
Effect of Catalysts
Worksheet # 10-5 1. One mechanism for the destruction of ozone in the upper atmosphere is: O3 (g) + NO (g) NO2 (g) + O2 (g) NO2 (g) + O (g) NO (g) + O2 (g) Overall reaction O3 (g) + O (g) 2O2 (g) a) Which species is the catalyst? b) Which species is the intermediate? c) The energy of activation of the overall reaction is 14.0 kj. Ea for the same reaction when catalyzed is 11.9 kj. Draw the energy diagram of both the catalyzed and uncatalyzed reaction. 2. Chlorine atoms (from the decomposition of freons in the upper atmosphere) can act as catalyst for the destruction of ozone. The activation energy for the reaction O3 + Cl ClO + O2 is 2.1 kj/mol. Which is the more effective catalyst for the destruction of ozone, Cl or NO? 12-49
Worksheet # 10 5: Answers 1. One mechanism for the destruction of ozone in the upper atmosphere is: O3 (g) + NO (g) NO2 (g) + O2 (g) NO2 (g) + O (g) NO (g) + O2 (g) Overall reaction O3 (g) + O (g) 2O2 (g) a) Which species is the catalyst? NO b) Which species is the intermediate? NO 2 c) The energy of activation of the overall reaction is 14.0 kj. Ea for the same reaction when catalyzed is 11.9 kj. Draw the energy diagram of both the catalyzed and uncatalyzed reaction. 2. Chlorine atoms (from the decomposition of freons in the upper atmosphere) can act as catalyst for the destruction of ozone. The activation energy for the reaction O3 + Cl ClO + O2 is 2.1 kj/mol. Which is the more effective catalyst for the destruction of ozone, Cl or NO? Cl 12-50 12-50
Chemical Instruments Spectrophotometric monitoring of a reaction.
Chemical Instruments Conductometric monitoring of a reaction Manometric monitoring of a reaction