1. 8–1 John A. Schreifels Chemistry 212 Chapter 14-1 Chapter 14 Rates of Reaction

2. 8–2 John A. Schreifels Chemistry 212 Chapter 14-2 Overview Reaction Rates –Definition of Reaction Rates –Experimental Determination of Rate –Dependence of Rate on Concentration –Change of Concentration with Time –Temperature and Rate; Collision and Transition-State Theories. –Arrhenius Equation Reaction Mechanisms –Elementary Reactions –Rate Law and the Mechanism –Catalysis

3. 8–3 John A. Schreifels Chemistry 212 Chapter 14-3 Reaction Rates Deal with the speed of a reaction and controlled by: –Proportional to concentrations of reactants –Proportional to catalyst concentration; catalyst = a substance that increases the rate of reaction without being consumed in the reaction. –Larger surface area of catalyst means higher reaction rate (more sites for reaction to take place). –Temperature: Higher temperature of reaction means faster.

4. 8–4 John A. Schreifels Chemistry 212 Chapter 14-4 Definition of Reaction Rate Reaction rate = increase in concentration of product of a reaction as a function of time or decrease in concentration of reaction as a function of time. Thus the rate of a reaction is: Rates are expressed as positive numbers. For the reaction in the graph we have:

5. 8–5 John A. Schreifels Chemistry 212 Chapter 14-5 Reaction Rates and Stoichiometry A + B  C; R C = R A = R B. A + 2B  3C; E.g.Calculate the rate of decomposition of HI in the reaction: 2HI(g)  H 2 (g) + I 2 (g). Given: After a reaction time of 100 secs. the concentration of HI decreased by 0.500 M. For the general reaction: aA + bB  cC + dD E.g. For the reaction 2A + 3B  4C + 2D; determine the rates of B, C and D if the rate of consumption of A is 0.100 M/s.

6. 8–6 John A. Schreifels Chemistry 212 Chapter 14-6 Rate Laws and Reaction Order Rate Law – an equation that tells how the reaction rate depends on the concentration of each reaction. Reaction order – the value of the exponents of concentration terms in the rate law. For the reaction: aA + bB  cC + dD, the initial rate of reaction is related to the concentration of reactants. Varying the initial concentration of one reactant at a time produces rates, which will lead to the order of each reactant. The rate law describes this dependence: R = k[A] m [B] n where k = rate constant and m and n are the orders of A and B respectively. –m = 1 (A varied, B held constant) gives R = k’[A]. Rate is directly proportional to [A]. Doubling A doubles R –m = 2 (A varied, B held constant) gives R = k’[A] 2. The rate is proportional to [A] 2. Doubling A quadruples R. E.g. Determine order of each reactant: HCOOH(aq) + Br 2 (aq)  2H + (aq) + 2Br  (aq) + CO 2 (g) R = k[Br 2 ] E.g. The formation of HI gas has the following rate law: R = k[H 2 ][I 2 ]. What is the order of each reactant?

7. 8–7 John A. Schreifels Chemistry 212 Chapter 14-7 Experimental Determination of a Rate Law: First Order Varying initial concentration of reactants changes the initial rate (usually all but one held constant) like one with two unknowns. Initial rate is the initial slope of the graph shown. As the initial concentration of that compound increases so does the rate. –Initial rate vs. [A] o plotted. –If straight line then reaction is first order and slope is rate constant. Second order rate law determined in like manner.

8. 8–8 John A. Schreifels Chemistry 212 Chapter 14-8 Rate Law for All Reactants Order for all components done same way. E.g. Determine the reaction order for each reactant from the table. (aq)+5Br  (aq)+6H + (aq)  3Br 2 (aq)+3H 2 O(l) Eg. 2: Determine the reaction orders for the reaction indicated from the data provided. A + 2B + C  Products.

9. 8–9 John A. Schreifels Chemistry 212 Chapter 14-9 Integrated Rate Law: First–Order Reaction For a first order reaction, Rate =  [A]/  t = k[A] or R A =  d[A]/dt = k[A]. Use of calculus leads to: or Allows one to calculate the [A] at any time after the start of the reaction. E.g. Calculate the concentration of N 2 O remaining after its decomposition according to 2N 2 O(g)  2N 2 (g) + O 2 (g) if it’s rate is first order and [N 2 O] o = 0.20M, k = 3.4 s  1 and T = 780°C. Find its concentration after 100 ms. Linearized forms: or Plot ln[A] vs t. Slope of straight line leads to rate constant, k. E.g. When cyclohexane(let's call it C) is heated to 500 o C, it changes into propene. Using the following data from one experiment, determine the first order rate constant.:

10. 8–10 John A. Schreifels Chemistry 212 Chapter 14-10 Half-Life: First Order Reaction Half-life of First order reaction, t 1/2 = 0.693/k. the time required for the concentration of the reactant to change to ½ of its initial value. i.e. at t 1/2, [A] = ½ [A] o E.g. For the decomposition of N 2 O 5 at 65 °C, the half-life was found to be 130 s. Determine the rate constant for this reaction. For n half-lives t = n*t 1/2 [A] = 2  n [A] o

11. 8–11 John A. Schreifels Chemistry 212 Chapter 14-11 Second–Order Reactions: Integrated Rate Law Rate law: R = k[A] 2 and the integrated rate equation is: Plot of vs. t gives a straight line with a slope of k. Half-life is: E.g. At 330°C, the rate constant for the decomposition of NO 2 is 0.775 L/(mol*s). If the reaction is second-order, what is the concentration of NO 2 after 2.5x10 2 s if the starting of concentration was 0.050 M?

12. 8–12 John A. Schreifels Chemistry 212 Chapter 14-12 Reaction Mechanisms Give insight into sequence of reaction events leading to product (reaction mechanism). Each of the steps leading to product is called an elementary reaction or elementary step. Consider the reaction of nitrogen dioxide with carbon dioxide which is second order on NO 2 : NO 2 (g) + CO(g)  NO(g) + CO 2 (g) Rate = k[NO 2 ] 2. Rate law suggests at least two steps. A proposed mechanism for this reaction involves two steps. –NO 3 is a reaction intermediate = a substance that is produced and consumed in the reaction so that none is detected when the reaction is finished. The elementary reactions are often described in terms of their molecularity. –Unimolecular One particle in elementary. –Bimolecular = 2 particles and –Termolecular = 3 particles

13. 8–13 John A. Schreifels Chemistry 212 Chapter 14-13 Rate Laws and Reaction Mechanisms Overall reaction order is often determined by the rate determining step. Use rate law of limiting step; No intermediates! 2NO 2 (g)  NO 3 (g) + NO(g), R 1 = k 1 [NO 2 ] 2 Slow NO 3 (g) +CO(g)  NO 2 (g) + CO 2 (g) R 2 = k 2 [NO 3 ][CO]Fast NO 2 + CO  NO + CO 2 R obs = k[NO 2 ] 2 E.g. Determine the rate law for the following mechanism:

14. 8–14 John A. Schreifels Chemistry 212 Chapter 14-14 Reaction Rates and Temperature: The Arrhenius Equation Rate (rate constant) increases exponentially with temperature. Collision theory indicates collisions every 10  9 s – 10  10 s at 25°C and 1 atm. i.e. only a small fraction of the colliding molecules actually react. Collision theory assumes: –Reaction can only occur if collision takes place. –Colliding molecules must have correct orientation and energy. –Collision rate is directing proportional to the concentration of colliding particles. A + B  Products; R c = Z[A][B] 2A + B  Products; R c = Z[A] 2 [B], etc. Only a fraction of the molecules, p (“steric factor”), have correct orientation; multiply collision rate by p.steric factor Particle must have enough energy. Fraction of those with correct energy follows Boltzmann equation where E a = activation energy, R = gas constant and T = temp. (Kelvin scale only please). This gives: k = Zpf

15. 8–15 John A. Schreifels Chemistry 212 Chapter 14-15 Transition State Theory Explains the reaction resulting from the collision of molecules to form an activated complex. Activated complex is unstable and can break to form product. Exothermic Reaction Endothermic Reaction

16. 8–16 John A. Schreifels Chemistry 212 Chapter 14-16 The Arrhenius Equation Summary: where A = frequency factor. Linear form:. Plot ln k vs. 1/T; the slope gives E a /R. E.g. determine the activation energy for the decomposition of N2O5 from the temperature dependence of the rate constant. Two point equation sometimes used also: E.g.2: Determine the rate constant at 35°C for the hydrolysis of sucrose, given that at 37°C it is 0.91mL/(mol*sec). The activation energy of this reaction is 108 kJ/mol. Rate constant increases when T 2 >T 1 k, s  1 Temp., °CTemp., K 4.8x10  4 45.0318.15 8.8x10  4 50.0323.15 1.6x10  3 55.0328.15 2.8x10  3 60.0333.15

17. 8–17 John A. Schreifels Chemistry 212 Chapter 14-17 Catalysis Catalysts a substance that increases the rate of a reaction without being consumed in the reaction. Catalyst provides an alternative pathway from reactant to product which has a rate determining step that has a lower activation energy than that of the original pathway. E.g. Hydrogen peroxide and bromine: 2H 2 O 2 (aq)  2H 2 O(l)+ O 2 (g). Mechanism is believed to be : 1. Br 2 redBr 2 (aq) + H 2 O 2 (aq)  2Br  (aq) +2H + (aq)+O 2 (g) 2. Br  oxid2H + (aq)+2Br  (aq)+H 2 O 2 (aq)  Br 2 (aq) + 2H 2 O(l). Overall2H 2 O 2 (aq)  2H 2 O(l)+O 2 (g) Notice that bromine is not consumed, even though it has participated in the reaction.

18. 8–18 John A. Schreifels Chemistry 212 Chapter 14-18 Homogeneous and Heterogeneous Catalysts Homogeneous catalyst: catalyst existing in the same phase as the reactants. Heterogeneous catalysis: catalyst existing in a different phase than the reactants. –The previous section gave an example of a homogeneous catalyst since the catalyst Br 2 was in the same phase as the hydrogen peroxide. The catalytic hydrogenation of ethylene is an example of a heterogeneous catalysis reaction: ENZYMES (biological catalysts) –They are proteins (large organic molecules that are composed of amino acids). –Slotlike active sites. The molecule fits into this slot and reaction proceeds. Poisons can block active site or reduce activity by distorting the active site.

19. 8–19 John A. Schreifels Chemistry 212 Chapter 14-19 “Steric Factor” Molecules must have the correct orientation before a reaction can take place. Figure 14.12 Importance of Molecular Orientation Return to p. 14-14