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Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemical Kinetics The area of chemistry that concerns reaction rates.

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Presentation on theme: "Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemical Kinetics The area of chemistry that concerns reaction rates."— Presentation transcript:

1 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemical Kinetics The area of chemistry that concerns reaction rates.

2 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Factors affecting reactions The commercial use of a chemical reaction requires knowledge of three factors: Stoichiometry Thermodynamics Kinetics

3 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 Kinetics Kinetics describes how fast reactants are used up and products are formed. Some reactions occur very fast (combustion) while others proceed very slowly (rusting). To be useful, reactions musty occur at a reasonable rate.

4 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Reaction Rate Change in concentration (conc) of a reactant or product per unit time.

5 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 Example 1 Consider the following reaction: CO + NO 2 → CO 2 + NO At t 1 (0 seconds) [CO] = 0.0223 mol/L At t 2 (12.5 seconds) [CO] = 0.0119 mol/L (the brackets [ ] indicate concentration) Average rate = -∆[CO] = - [CO] 2 – [CO] 1 ∆t t 2 – t 1

6 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Example 1 = - 0.0119 mol/L – 0.0223 mol/L 12.5 s = 0.000832 mol/L∙s

7 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 General form For A + B → C + D Average reaction rate = - ∆[A] = - ∆[B] = ∆[C] = ∆[D] ∆t ∆t ∆t ∆t Expressed in terms of each reactant.

8 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Collision theory Particles must collide Particles must collide with the correct orientation. Particles must collide with sufficient energy.

9 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Factors affecting reaction rates Surface area Concentration Temperature Presence of a catalyst

10 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Rate laws Must always be obeyed or you will be arrested by the kinetics police A mathematical expression that relates reaction rates to the concentration of reactants. The reaction order mathematically defines the extent to which the reaction rate depends on the concentrations of the reactants.

11 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Rate Laws Rate = k[NO 2 ] n k = rate constant n = rate order

12 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 Rate law expression For aA + bB → cC + dD Rate = k[A] m [B] n m ≠ a and n ≠ b The rate constant k must be determined experimentally. As the concentration of reactants change, the rate also changes.

13 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Rate law expressions The rate constant = k [A] = concentration of A [B] = concentration of B m = order of the reaction with respect to A n = order of the reaction with respect to B m + n = overall order of the reaction

14 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 Rate constant k Its value is for a specific reaction Its units depend on the overall order of the reaction Its value does not change with concentration Its value does not change over time Changes with temperature. Must be experimentally determined Depends upon the presence of a catalyst

15 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Order of reaction For A + B → C + D Rate = k[A] m [B] n If the rate of the reaction is directly proportional to a change in [A] then the reaction is 1 st order (with respect to A) and m = 1 If the rate of the reaction does not change with [A], then the reaction is 0 order with respect to A. m = 0 The overall rate of the reaction is m + n

16 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 example 2NO + O 2 → 2NO 2 Rate = k[NO] 2 [O 2 ] This reaction is 2 nd order with respect to [NO] and 1 st order with respect to [O 2 ] The overall order of the reaction is 3 If you double the concentration of NO, the rate will quadruple (2 2 = 4) If you double the concentration of O 2 the rate of the reaction also doubles.

17 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 Determining rate laws 2NO + H 2 → N 2 + 2H 2 O Trial #Initial [NO] Mol/L Initial [H 2 ] Mol/L Rate mol/L∙S 14.0 × 10 -3 2.0 × 10 -3 1.2 × 10 -5 28.0 × 10 -3 2.0 × 10 -3 4.8 × 10 -5 34.0 × 10 -3 2.4 × 10 -5

18 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Determining rate laws Rate = k[NO] m [H 2 ] n We want to determine m and n m = 2 n = 1 The overall order of the reaction is 3 The rate law expression is k[NO] 2 [H 2 ]

19 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 Types of Rate Laws Differential Rate Law: expresses how rate depends on concentration. Integrated Rate Law: expresses how concentration depends on time.

20 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 Example CH 3 CHO → CH 4 + CO trial[CH 3 CHO]Rate Mol/L∙S 10.100.020 20.200.080 30.300.180 40.400.320

21 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 Example A + 2B → AB 2 trial[A][B]Rate mol/L∙S 10.01M 1.5 × 10 -4 20.01M0.02M1.5 × 10 -4 30.02M0.03M6.0 × 10 -4

22 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 Example 2A + B 2 +C → D Trial #[A][B 2 ][C]Rate mol/L∙s 1.01 1.25×10 3 2.02.01 5.00×10 3 3.03.01.051.125× 10 4 4.04.02.018.00×10 4

23 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Example (cont.) Rate = k[A] x [B 2 ] y [C] z By inspection x = 2 Solve for y Use trial 1 and 4 since [C] does not change Rate 4 = k[A] 2 4 [B 2 ] y 4 [C] z 4 Rate 1 k[A] 2 1 [B 2 ] y 1 [C] z 1

24 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 Example (continued) 8.00 ×10 4 = (.04/.01) 2 × (.02/.01) y 1.25×10 3 64 = 16 × 2 y 2 y = 64/16 = 4 y = 2 Now solve for z

25 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Example (continued) Use trial 1 and 3 because [B] does not change Rate 3 = k[A] 2 3 [C] z 3 Rate 1 k[A] 2 1 [C] z 1 1.125×10 4 = (.03/.01) 2 × (.05/.01) z 1.250×10 3 9 = 9(5) z 5 z = 1 Z = 0

26 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 Example continued Rate = k[A] 2 [B 2 ] 2 k = rate/[A] 2 [B 2 ] 2 k = 1.25×10 3 / (.01) 2 (.01) 2 k = 1.25×10 11 L 3 /mol 3 ∙s Rate = 1.25×10 11 [A] 2 [B 2 ] 2

27 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27 Method of Initial Rates Initial Rate: the “instantaneous rate” just after the reaction begins. The initial rate is determined in several experiments using different initial concentrations.

28 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 Overall Reaction Order Sum of the order of each component in the rate law. rate = k[H 2 SeO 3 ][H + ] 2 [I  ] 3 The overall reaction order is 1 + 2 + 3 = 6.

29 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29 First-Order Rate Law Integrated first-order rate law is ln[A] =  kt + ln[A] o For aA  Products in a 1st-order reaction,

30 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 Half-Life of a First-Order Reaction t 1/2 = half-life of the reaction k = rate constant For a first-order reaction, the half-life does not depend on concentration.

31 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 Second-Order Rate Law For aA  products in a second-order reaction, Integrated rate law is

32 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 Half-Life of a Second-Order Reaction t 1/2 = half-life of the reaction k = rate constant A o = initial concentration of A The half-life is dependent upon the initial concentration.

33 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33 A Summary 1.Simplification: Conditions are set such that only forward reaction is important. 2.Two types: differential rate law integrated rate law 3.Which type? Depends on the type of data collected - differential and integrated forms can be interconverted.

34 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 A Summary (continued) 4.Most common: method of initial rates. 5.Concentration v. time: used to determine integrated rate law, often graphically. 6.For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).

35 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 Reaction Mechanism 4 The series of steps by which a chemical reaction occurs. 4 A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

36 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 Reaction Mechanism (continued) 4 The reaction has many steps in the reaction mechanism.

37 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Often Used Terms Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. Molecularity: the number of species that must collide to produce the reaction indicated by that step. Elementary Step: A reaction whose rate law can be written from its molecularity. uni, bi and termolecular

38 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 Rate-Determining Step In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction.

39 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39 Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why? Arrhenius: An activation energy must be overcome.

40 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 40 Transition state theory In a chemical reaction, the reactants must pass through a short-lived, high energy intermediate state called a transition state, before any products are formed. Activation energy is energy which must be absorbed by the reactants in the ground state to allow them to reach the transition state.

41 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 Transition state theory continued If the energy of the products (E p ) is greater than the energy of the reactants (E r ), then the process is endothermic. If E p < E r then the reaction is exothermic.

42 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 42 Arrhenius Equation 4 Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). 4 Orientation of reactants must allow formation of new bonds.

43 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43 Arrhenius Equation (continued) k = rate constant A = frequency factor E a = activation energy T = temperature R = gas constant

44 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 44 Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

45 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 45 Heterogeneous Catalysis 1. Adsorption and activation of the reactants. 2. Migration of the adsorbed reactants on the surface. 3. Reaction of the adsorbed substances. 4. Escape, or desorption, of the products. Steps:


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