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

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

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 John A. Schreifels Chemistry 212 8–2 Chapter 14-2

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. 8–3 John A. Schreifels Chemistry 212 Chapter 14-3

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: Concentration vs Reaction Time A 2B -- 3C Concentration, M 0.090 conc change A [ A ] time change t [ A ] [ A ]1 2 t 2 t1 Rate A Init Rate 0.045 Inst. Rate Ave. Rate 0.000 0 250 500 Tim e, s Rates are expressed as positive numbers. For the reaction in the graph we have: R A [ A ] t John A. Schreifels Chemistry 212 RB [B] t RC [C] t Chapter 14-4 8–4

Reaction Rates and Stoichiometry A B C; RC RA RB. A 2B 3C; R 1 R 1 R A B C 2 3 E.g.Calculate the rate of decomposition of HI in the reaction: 2HI(g) H2(g) I2(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 a a a R A RB R C RD b c d 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. 8–5 John A. Schreifels Chemistry 212 Chapter 14-5

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) Br2(aq) 2H (aq) 2Br (aq) CO2(g) R k[Br2] E.g. The formation of HI gas has the following rate law: R k[H2][I2]. What is the order of each reactant? John A. Schreifels Chemistry 212 Chapter 14-6 8–6

Experimental Determination of a Rate Law: First Order Concentration vs Reaction Time A 2B -- 3C 0.100 [A] 0, M 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. 0.000 0 250 500 Tim e, s Initial Rate vs. [A]o – Initial rate vs. [A]o plotted. – If straight line then reaction is first order and slope is rate constant. 0.0005 0.0004 0.0003 Ro Second order rate law determined in like manner. 0.050 0.0002 0.0001 0.0000 0.00 8–7 0.03 0.05 0.08 [A]o John A. Schreifels Chemistry 212 Chapter 14-7 0.10

Rate Law for All Reactants Order for all components done same way. E.g. Determine the reaction order for each reactant from the table. BrO (aq) 5Br (aq) 6H (aq) 3Br2(aq) 3H2O(l) 3 [BrO 3 ]o 0.10 0.20 0.10 0.20 [Br ]o 0.10 0.10 0.30 0.10 [H ]o 0.10 0.10 0.10 0.15 Ro 1.2 2.4 3.5 5.4 Eg. 2: Determine the reaction orders for the reaction indicated from the data provided. A 2B C Products. [A]o 2.06 0.87 0.50 1.00 John A. Schreifels Chemistry 212 [B]o 3.05 3.05 0.50 0.50 [C]o 4.00 4.00 0.50 1.00 Ro 3.7 0.66 0.013 0.072 8–8 Chapter 14-8

Integrated Rate Law: First–Order Reaction For a first order reaction, Rate [A]/ t k[A] or RA d[A]/dt k[A]. [A] k [A] log t Use of calculus leads to: ln kt or [ A ]o 2.303 [ A ]o Allows one to calculate the [A] at any time after the start of the reaction. E.g. Calculate the concentration of N2O remaining after its decomposition according to 2N2O(g) 2N2(g) O2(g) if it’s rate is first order and [N2O]o 0.20M, k 3.4 s 1 and T 780 C. Find its concentration after 100 ms. Linearized forms: ln[ A ] kt ln[ A ] oor log[ A ] k t log[ A ]o 2.303 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 oC, it changes into propene. Using the following data from one experiment, determine the first order rate constant.: t,min 0.00 5.00 10.00 15.00 [C],mM 1.50 1.24 1.00 0.83 John A. Schreifels Chemistry 212 Chapter 14-9 8–9

Half-Life: First Order Reaction Half-life of First order reaction, t1/2 0.693/k. the time required for the concentration of the reactant to change to ½ of its initial value. i.e. at t1/2 , [A] ½ [A]o E.g. For the decomposition of N2O5 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*t1/2 [A] 2 n [A]o John A. Schreifels Chemistry 212 1 / 2[ A]o ln k t1/ 2 [ A]o 1 ln k t1/ 2 2 t1/ 2 0.693 / k 1 n ln k n t1 / 2 2 1 ln k tn 2n 1 [ A ]t n [A] 2 o 8–10 Chapter 14-10

Second–Order Reactions: Integrated Rate Law Rate law: R k[A]2 and the integrated rate equation is: 1 1 kt [ A ]t [ A ]o 1 Plot of [ A ] vs. t gives a straight line with a slope of k. t 1 Half-life is: t1/ 2 k [ A ] o E.g. At 330 C, the rate constant for the decomposition of NO2 is 0.775 L/(mol*s). If the reaction is second-order, what is the concentration of NO2 after 2.5x102 s if the 8–11 starting of concentration was 0.050 M? John A. Schreifels Chemistry 212 Chapter 14-11

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 NO2: NO2(g) CO(g) NO(g) CO2(g) Rate k[NO2]2. Rate law suggests at least two steps. A proposed mechanism for this reaction involves two steps. Step 1 Step 2 Overall 2NO2(g) NO3(g) NO(g) NO3(g) CO(g) NO2(g) CO2(g) NO2 CO NO CO2 – NO3 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 John A. Schreifels Chemistry 212 8–12 Chapter 14-12

Rate Laws and Reaction Mechanisms Overall reaction order is often determined by the rate determining step. Use rate law of limiting step; No intermediates! 2NO2(g) NO3(g) NO(g), R1 k1[NO2]2 Slow NO3(g) CO(g) NO2(g) CO2(g) R2 k2[NO3][CO] Fast NO CO2 Robs k[NO2]2 NO2 CO E.g. Determine the rate law for the following mechanism: 2*[N2O5(g) NO2 (g) NO3 ( g) ] Fast k2 NO3(g) NO2(g) NO 2 (g) NO(g) O 2 (g) Slow k1 k 1 k3 NO3 NO 2NO 2 (g) Fast k obs 2N2O5(g) 4NO 2 (g) O 2 (g) Use steady state approximation. at “equilibrium” rates of forward and reverse reactions are same. Use to eliminate intermediates from rate law equations. R1 R 1 k 1 [N2O5 ] k 1[NO 2 ][NO3 ] John A. Schreifels Chemistry 212 or [NO3 ] k 1 [N2O5 ] k 1 [NO2 ] 8–13 Chapter 14-13

Reaction Rates and Temperature: The Arrhenius Equation Rate (rate constant) increases exponentially with temperature. Collision theory indicates collisions every 10 9s – 10 10s 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; Rc Z[A][B] 2A B Products; Rc Z[A]2[B], etc. Only a fraction of the molecules, p (“steric factor”), have correct orientation; multiply collision rate by p. Particle must have enough energy. Fraction of those with correct energy follows Boltzmann equation where Ea activation energy, R gas constant and T temp. (Kelvin scale Eonly please). f e a / RT This gives: k Zpf John A. Schreifels Chemistry 212 8–14 Chapter 14-14

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 John A. Schreifels Chemistry 212 Endothermic Reaction 8–15 Chapter 14-15

The Arrhenius Equation ln k E where A frequency factor. k A exp a RT Arrhenius Plot Linear form: . ln k ln A E a -5 RT Plot ln k vs. 1/T; the slope gives E a/R. -6 E.g. determine the activation energy for the decomposition of N2O5 from the temperature dependence of the rate constant. -7 k, s 1 Temp., C Temp., K Summary: 4.8x10 4 45.0 318.15 8.8x10 4 50.0 323.15 1.6x10 3 55.0 328.15 2.8x10 3 60.0 -8 0.00300 0.00305 0.00310 0.00315 1/T, K 333.15 E 1 k 1 ln 2 a k R T T 1 1 2 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 T2 T1 John A. Schreifels Chemistry 212 8–16 Chapter 14-16

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: 2H2O2(aq) 2H2O(l) O2(g). Mechanism is believed to be : 1. Br2 red 2. Br oxid Overall Br2(aq) H2O2(aq) 2Br (aq) 2H (aq) O2(g) 2H (aq) 2Br (aq) H2O2(aq) Br2(aq) 2H2O(l). 2H2O2(aq) 2H2O(l) O2(g) Notice that bromine is not consumed, even though it has participated in the reaction. 8–17 John A. Schreifels Chemistry 212 Chapter 14-17

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 Br2 was in the same phase as the hydrogen peroxide. The catalytic hydrogenation of ethylene is an example of a heterogeneous catalysis reaction: Pt H2C CH2 ( g) H2 (g) H3C CH3 (g) 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. John A. Schreifels Chemistry 212 8–18 Chapter 14-18

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