Chemical Kinetics

Slide 2 of 55 Contents 15-1The Rate of a Chemical Reaction 15-2Measuring Reaction Rates 15-3Effect of Concentration on Reaction Rates: The Rate Law 15-4Zero-Order Reactions 15-5First-Order Reactions 15-6Second-Order Reactions 15-7Reaction Kinetics: A Summary

Slide 3 of 55 Contents 15-8Theoretical Models for Chemical Kinetics 15-9The Effect of Temperature on Reaction Rates 15-10Reaction Mechanisms 15-11Catalysis Focus On Combustion and Explosions

4 kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. experimentally it is shown that there are 4 factors that influence the speed of a reaction: –nature of the reactants, –temperature, –catalysts, –concentration Kinetics

5 Defining Rate rate is how much a quantity changes in a given period of time the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) –so the rate of your car has units of mi/hr

6 Defining Reaction Rate the rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time –or product concentration increases for reactants, a negative sign is placed in front of the definition

7 Reaction Rate Changes Over Time as time goes on, the rate of a reaction generally slows down –because the concentration of the reactants decreases. at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium.

8 at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1

9 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2

10 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3

11 Hypothetical Reaction Red  Blue Time (sec) Number Red Number Blue in this reaction, one molecule of Red turns into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time

12 Hypothetical Reaction Red  Blue

13 Hypothetical Reaction Red  Blue

What if we need to make a (simple) sandwich? We need to use the equation below. Reaction Rate and Stoichiometry So each sandwich consumes 2 pieces of toast (T) and 1 piece of cheese (C) The rate depends on which reactant we choose … this cannot be right To get this right we have to divide by the coefficients in the balanced equation × √

15 Reaction Rate and Stoichiometry in most reactions, the coefficients of the balanced equation are not all the same H 2 (g) + I 2 (g)  2 HI (g) for these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another –for the above reaction, for every 1 mole of H 2 used, 1 mole of I 2 will also be used and 2 moles of HI made –therefore the rate of change will be different in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient

We can express the rate of any reaction as follows: In terms of reactants (negative) In terms of Products (positive) General Rate of Reaction

Slide 17 of 55 General Rate of Reaction a A + b B → c C + d D Rate of reaction = rate of disappearance of reactants = Δ[C] ΔtΔt 1 c = Δ[D] ΔtΔt 1 d Δ[A] ΔtΔt 1 a = - Δ[B] ΔtΔt 1 b = - = rate of appearance of products

Examples

Slide 19 of The Rate of a Chemical Reaction Rate of reaction: is the change of concentration with time. 2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) t = 38.5 s [Fe 2+ ] = M Δt = 38.5 sΔ[Fe 2+ ] = ( – 0) M Rate of formation of Fe 2+ = ½ = = 1.29x10 -5 M s -1 Δ[Fe 2+ ] ΔtΔt M 38.5 s

Slide 20 of 55 Rates of Chemical Reaction Δ[Sn 4+ ] ΔtΔt 2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) Δ[Fe 2+ ] ΔtΔt = 1 2 Δ[Fe 3+ ] ΔtΔt = - 1 2

Slide 21 of Measuring Reaction Rates H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) 2 MnO 4 - (aq) + 5 H 2 O 2 (aq) + 6 H + → 2 Mn H 2 O(l) + 5 O 2 (g)

Slide 22 of 55 H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) Example (-1.7 M / 2600 s) = 6 x M s -1 -(-2.32 M / 1360 s) = 1.7 x M s -1 Determining and Using an Initial Rate of Reaction. Rate = -Δ[H 2 O 2 ] ΔtΔt

Slide 23 of 55 Example Δ[H 2 O 2 ] = -([H 2 O 2 ] f - [H 2 O 2 ] i ) = 1.7 x M s -1  Δt Rate = 1.7x M s -1 ΔtΔt = - Δ[H 2 O 2 ] [H 2 O 2 ] 100 s – 2.32 M = -1.7 x M s -1  100 s = 2.17 M = 2.32 M M [H 2 O 2 ] 100 s What is the concentration at 100s? [H 2 O 2 ] i = 2.32 M

Slide 24 of Effect of Concentration on Reaction Rates: The Rate Law a A + b B …. → g G + h H …. Rate of reaction = k [A] m [B] n …. Rate constant = k Overall order of reaction = m + n + ….

Slide 25 of 55 Example 15-3 Method of Initial Rates Establishing the Order of a reaction by the Method of Initial Rates. Use the data provided establish the order of the reaction with respect to HgCl 2 and C 2 O 2 2- and also the overall order of the reaction.

Slide 26 of 55 Example 15-3 Notice that concentration changes between reactants are by a factor of 2. Write and take ratios of rate laws taking this into account.

Slide 27 of 55 Example 15-3 R 2 = k[HgCl 2 ] 2 m [C 2 O 4 2- ] 2 n R 3 = k[HgCl 2 ] 3 m [C 2 O 4 2- ] 3 n R2R2 R3R3 k(2[HgCl 2 ] 3 ) m [C 2 O 4 2- ] 3 n k[HgCl 2 ] 3 m [C 2 O 4 2- ] 3 n = 2 m = 2.0 therefore m = 1.0 R2R2 R3R3 k2 m [HgCl 2 ] 3 m [C 2 O 4 2- ] 3 n k[HgCl 2 ] 3 m [C 2 O 4 2- ] 3 n == 2.0= 2mR32mR3 R3R3 = k(2[HgCl 2 ] 3 ) m [C 2 O 4 2- ] 3 n

Slide 28 of 55 Example 15-3 R 2 = k[HgCl 2 ] 2 1 [C 2 O 4 2- ] 2 n = k(0.105)(0.30) n R 1 = k[HgCl 2 ] 1 1 [C 2 O 4 2- ] 1 n = k(0.105)(0.15) n R2R2 R1R1 k(0.105)(0.30) n k(0.105)(0.15) n = 7.1x x10 -5 = 3.94 R2R2 R1R1 (0.30) n (0.15) n = = 2 n = 2 n = 3.98 therefore n = 2.0

Slide 29 of 55 + = Third Order R 2 = k[HgCl 2 ] 2 [C 2 O 4 2- ] 2 First order Example Second order 2

Slide 30 of Zero-Order Reactions A → products R rxn = k [A] 0 R rxn = k [k] = mol L -1 s -1

Slide 31 of 55 Integrated Rate Law -[A] t + [A] 0 = kt [A] t = [A] 0 - kt ΔtΔt -Δ[A] dt = k -d[A] Move to the infinitesimal = k And integrate from 0 to time t

Slide 32 of First-Order Reactions H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) = -k [H 2 O 2 ] d[H 2 O 2 ] dt = -kt ln [A] t [A] 0 ln[A] t = -kt + ln[A] 0 [k] = s -1

Slide 33 of 55 First-Order Reactions

Slide 34 of 55 Half-Life t ½ is the time taken for one-half of a reactant to be consumed. = -kt ln [A] t [A] 0 = -kt ½ ln ½[A] 0 [A] 0 - ln 2 = -kt ½ t ½ = ln 2 k k =

Slide 35 of 55 Half-Life Bu t OOBu t (g) → 2 CH 3 CO(g) + C 2 H 4 (g)

Slide 36 of 55 Some Typical First-Order Processes

Slide 37 of Second-Order Reactions Rate law where sum of exponents m + n + … = 2. A → products = kt + 1 [A] 0 [A] t 1 dt = -k[A] 2 d[A] [k] = M -1 s -1 = L mol -1 s -1

Slide 38 of 55 Second-Order Reaction

Slide 39 of 55 Pseudo First-Order Reactions Simplify the kinetics of complex reactions Rate laws become easier to work with. CH 3 CO 2 C 2 H 5 + H 2 O → CH 3 CO 2 H + C 2 H 5 OH If the concentration of water does not change appreciably during the reaction. –Rate law appears to be first order. Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.

Slide 40 of 55 Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot 1/[A] vs t.

Slide 41 of Reaction Kinetics: A Summary Calculate the rate of a reaction from a known rate law using: Determine the instantaneous rate of the reaction by: Rate of reaction = k [A] m [B] n …. Finding the slope of the tangent line of [A] vs t or, Evaluate –Δ[A]/Δt, with a short Δt interval.

Slide 42 of 55 Summary of Kinetics Determine the order of reaction by: Using the method of initial rates. Find the graph that yields a straight line. Test for the half-life to find first order reactions. Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.

Slide 43 of 55 Summary of Kinetics Find the rate constant k by: Find reactant concentrations or times for certain conditions using the integrated rate law after determining k. Determining the slope of a straight line graph. Evaluating k with the integrated rate law. Measuring the half life of first-order reactions.

Slide 44 of Theoretical Models for Chemical Kinetics Kinetic-Molecular theory can be used to calculate the collision frequency. –In gases collisions per second. –If each collision produced a reaction, the rate would be about 10 6 M s -1. –Actual rates are on the order of 10 4 M s -1. Still a very rapid rate. –Only a fraction of collisions yield a reaction. Collision Theory

Slide 45 of 55 Activation Energy For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s). Activation Energy is: –The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 46 of 55 Activation Energy

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 47 of 55 Kinetic Energy

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 48 of 55 Collision Theory If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower. As temperature increases, reaction rate increases. Orientation of molecules may be important.

Collision Theory

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 50 of 55 Transition State Theory The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

15-9 Effect of Temperature on Reaction Rates Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = Ae -E a /RT ln k = + ln A R -Ea T 1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 52 of 55 Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) = -1.2  10 4 K R -Ea-Ea -E a = 1.0  10 2 kJ mol -1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 53 of 55 Arrhenius Equation k = Ae -E a /RT ln k = + ln A R -Ea-Ea T 1 ln k 2 – ln k 1 = + ln A - - ln A R -Ea-Ea T2T2 1 R -Ea-Ea T1T1 1 ln = - R -Ea-Ea T2T2 1 k2k2 k1k1 T1T1 1

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 54 of Catalysis Alternative reaction pathway of lower energy. Homogeneous catalysis. –All species in the reaction are in solution. Heterogeneous catalysis. –The catalyst is in the solid state. –Reactants from gas or solution phase are adsorbed. –Active sites on the catalytic surface are important.

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 55 of Catalysis

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 56 of 55 Catalysis on a Surface

Prentice-Hall © 2002General Chemistry: Chapter 15Slide 57 of 55 Enzyme Catalysis E + S  ES k1k1 k -1 ES → E + P k2k2