General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

فصل چهاردهم: سینتیک واکنشهای شیمیایی

Contents 14-1The Rate of a Chemical Reaction 14-2Measuring Reaction Rates 14-3Effect of Concentration on Reaction Rates: The Rate Law 14-4Zero-Order Reactions 14-5First-Order Reactions 14-6Second-Order Reactions 14-7Reaction Kinetics: A Summary

Contents 14-8Theoretical Models for Chemical Kinetics 14-9The Effect of Temperature on Reaction Rates 14-10Reaction Mechanisms 14-11Catalysis Focus On Combustion and Explosions

سینتیک یعنی مطالعه سرعت واکنشهای شیمیایی و راههای کنترل سرعت آنها. واکنشهای شیمیایی به صورت همگن و ناهمگن طبقه بندی می شوند. واکنشهای همگن تنها در یک فاز صورت می گیرند و واکنشهای ناهمگن در فصل مشترک فازها. مقدمه

معادله سرعت سرعت واکنش با غلظت مواد آن مرتبط است. به لحاظ ریاضی سرعت واکنش، سرعت از بین رفتن مواد اولیه یا سرعت تولید مواد حاصل در واحد زمان است. سرعت واکنش را با R و غلظت مولار را با [ ] نمایش می دهند. مطابق تعریف بالا می توان نوشت:

14-1 The Rate of a Chemical Reaction Rate of 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+ = = Δ[Fe 2+ ] ΔtΔt M 38.5 s = 2.6 x M s -1

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

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

14-2 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) Experimental set-up for determining the rate of decomposition of H 2 O 2. Oxygen gas given off by the reaction mixture is trapped, and its volume is measured in the gas buret. The amount of H 2 O 2 consumed and the remaining concentration of H 2 O 2 can be calculated from the measured volume of O 2 (g).

H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) Example: Initial rate: -(-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

Example: -Δ[H 2 O 2 ] = -([H 2 O 2 ] f - [H 2 O 2 ] i )= 1.7 x M s -1 x Δt Rate = 1.7 x M s -1 ΔtΔt = - Δ[H 2 O 2 ] [H 2 O 2 ] 100 s – 2.32 M =-1.7 x M s -1 x 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

14-3 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 + ….

مرتبه واکنش واکنشهای مرتبه صفر واکنشهای مرتبه اول واکنشهای مرتبه دوم واکنشهای مرتبه سوم A B

واکنشهای مرتبه اول نمودار لگاریتمی نسبت غلظتها برحسب زمان خطی است.

Example: 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.

Example: Notice that concentration changes between reactions are by a factor of 2. Write and take ratios of rate laws taking this into account.

Example: 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 (0.105) m [C 2 O 4 2- ] 2 n k (0.052) m [C 2 O 4 2- ] 3 n = 2 m = 2.0 therefore m = 1.0 R2R2 R3R3 = 2m2m 7.1 x x =

Example: 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 = 7.1 x x = 3.94 R2R2 R1R1 (0.30) n (0.15) n = = 2 n = 2 n = 3.98 therefore n = 2.0

+ = Third OrderFirst order Example: Second order R = k [HgCl 2 ] [C 2 O 4 2- ] 2

15-4 Zero-Order Reactions A → products R rxn = k [A] 0 R rxn = k [k] = mol L -1 s -1

Integrated Rate Law -∫-∫ dt= k d[A] ∫ [A] 0 [A] t 0 t -[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

15-5 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 = - k dt [H 2 O 2 ] d[H 2 O 2 ] ∫ [A] 0 [A] t ∫ 0 t = -kt ln [A] t [A] 0 ln[A] t = -kt + ln[A] 0 [k] = s -1

First-Order Reactions

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 = For a first order reaction:

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

Some Typical First-Order Processes Some typical first-order processes

15-6 Second-Order Reactions Rate law where sum of exponents m + n + … = 2 A → products dt= - k d[A] [A] 2 ∫ [A] 0 [A] t ∫ 0 t = kt + 1 [A] 0 [A] t 1 dt = -k[A] 2 ; d[A] [k] = M -1 s -1 = L mol -1 s -1

Second-Order Reaction = kt + 1 [A] 0 [A] t 1

Pseudo First-Order Reactions Simplify the kinetics of complex reactions Rate laws become easier to work with 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. CH 3 CO 2 C 2 H 5 + H 2 O → CH 3 CO 2 H + C 2 H 5 OH

Testing for a Rate Law Plot [A] vs t. Plot ln[A] vs t. Plot 1/[A] vs t. 2nd order

15-7 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.

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.

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.

15-8 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:

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.

Activation Energy

Kinetic Energy

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

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

Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) = -1.2 x 10 4 K R -Ea-Ea E a = 1.0 x 10 2 kJ mol -1

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 EaEa T1T1 1 k1k1 k2k2 T2T2 1 log = R EaEa T1T1 1 k1k1 k2k2 T2T2 1

15-10 Reaction Mechanisms A step-by-step description of a chemical reaction. Each step is called an elementary process. –Any molecular event that significantly alters a molecules energy of geometry or produces a new molecule. Reaction mechanism must be consistent with: –Stoichiometry for the overall reaction. –The experimentally determined rate law.

Elementary Processes Unimolecular or bimolecular. Exponents for concentration terms are the same as the stoichiometric factors for the elementary process. Elementary processes are reversible. Intermediates are produced in one elementary process and consumed in another. One elementary step is usually slower than all the others and is known as the rate determining step.

A Rate Determining Step

Slow Step Followed by a Fast Step H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) dt = k[H 2 ][ICl] d[P] Postulate a mechanism: H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) slow H 2 (g) + ICl(g) HI(g) + HCl(g) fast HI(g) + ICl(g) I 2 (g) + HCl(g) dt = k[H 2 ][ICl] d[HI] dt = k[HI][ICl] d[I 2 ] dt = k[H 2 ][ICl] d[P] (Process)

Slow Step Followed by a Fast Step

Fast Reversible Step Followed by a Slow Step 2NO(g) + O 2 (g) → 2 NO 2 (g) dt = -k obs [NO] 2 [O 2 ] d[P] Postulate a mechanism: dt = k 2 [N 2 O 2 ][O 2 ] d[NO 2 ] slow N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k2k2 dt = k 2 [NO] 2 [O 2 ] d[NO 2 ] k -1 k1k1 2NO(g) + O 2 (g) → 2 NO 2 (g) K = k -1 k1k1 = [NO] 2 [N 2 O 2 ] = K [NO] 2 k -1 k1k1 = [NO] 2 [N 2 O 2 ] fast 2NO(g) N 2 O 2 (g) k1k1 k -1

The Steady State Approximation dt = k 1 [NO] 2 – k 2 [N 2 O 2 ] – k 3 [N 2 O 2 ][O 2 ] = 0 d[N 2 O 2 ] N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k3k3 2NO(g) N 2 O 2 (g) k -1 k1k1 2NO(g) N 2 O 2 (g) N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k3k3 N 2 O 2 (g) 2NO(g) k2k2 k1k1 2NO(g) N 2 O 2 (g) dt = k 3 [N 2 O 2 ][O 2 ] d[NO 2 ]

The Steady State Approximation dt = k 1 [NO] 2 – k 2 [N 2 O 2 ] – k 3 [N 2 O 2 ][O 2 ] = 0 d[N 2 O 2 ] k 1 [NO] 2 = [N 2 O 2 ](k 2 + k 3 [O 2 ]) k 1 [NO] 2 [N 2 O 2 ] = (k 2 + k 3 [O 2 ]) dt = k 3 [N 2 O 2 ][O 2 ] d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = (k 2 + k 3 [O 2 ])

Kinetic Consequences of Assumptions dt d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = (k 2 + k 3 [O 2 ]) N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k3k3 N 2 O 2 (g) 2NO(g) k2k2 k1k1 2NO(g) N 2 O 2 (g) dt d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = ( k 3 [O 2 ]) k 1 [NO] 2 = dt d[NO 2 ] k 1 k 3 [NO] 2 [O 2 ] = ( k 2 ) [NO] 2 [O 2 ] = k1k3k1k3 k2k2 Let k 2 << k 3 Let k 2 >> k 3 Or

11-5 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.

11-5 Catalysis

Catalysis on a Surface

Enzyme Catalysis E + S ES k1k1 k -1 ES E + P k2k2

Saturation Kinetics E + S ES k1k1 k -1 → E + P k2k2 dt = k 1 [E][S] – k -1 [ES] – k 2 [ES]= 0 d[P] dt = k 2 [ES] d[P] k 1 [E][S] = (k -1 +k 2 )[ES] [E] = [E] 0 – [ES] k 1 [S]([E] 0 –[ES]) = (k -1 +k 2 )[ES] (k -1 +k 2 ) + k 1 [S] k 1 [E] 0 [S] [ES] =

Michaelis-Menten dt = d[P] (k -1 +k 2 ) + k 1 [S] k 1 k 2 [E] 0 [S] dt = d[P] (k -1 +k 2 ) + [S] k 2 [E] 0 [S] k1k1 dt = d[P] K M + [S] k 2 [E] 0 [S] dt = d[P] k 2 [E] 0 dt = d[P] KMKM k2k2 [E] 0 [S]

Chapter 15 Questions Part 1: 2, 4, 6, 10, 18, 20 Part 2: 24, 29, 32, 37, 38 Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding. Choose a variety of problems from the text as examples. Practice good techniques and get coaching from people who have been here before.