CHE1102, Chapter 13 Learn, 1 Chapter 13 Chemical Kinetics Practice Exercises 13.1,6,8- 12,16,19,22-3,26,30 Examples: ,10-11, 13 In Text Problems

CHE1102, Chapter 13 Learn, 2 – The speed (or rate) at which a reaction takes place 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) 2 ways to measure the rate of any reaction 1. Measure the Increase in [Products] over time 2. Measure the Decrease in [Reactants] over time How do you measure this ? Measure a change in concentration per change in time Kinetics

CHE1102, Chapter 13 Learn, 3 slope = yxyx Rate =   [ ]   t slope = Rate C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq)

CHE1102, Chapter 13 Learn, 4 * By convention, all rates are reported as positive values Rate =   [ Products ]   t when   [ Products ]   t slope = + # when   [ Reactants ]   t slope = - # Rate = -   [ Reactants ]   t

CHE1102, Chapter 13 Learn, 5 Instantaneous Rate – rate of rxn at a particular instant of time = slope of tangent to the line at that instant of time

CHE1102, Chapter 13 Learn, 6

CHE1102, Chapter 13 Learn, 7 Initial Rate – instantaneous rate at t = 0 (very beginning) – All reactions slow down over time. – The best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. – At t = 0, what is the rate?

CHE1102, Chapter 13 Learn, 8 In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. Rate =  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Reaction Rates and Stoichiometry

CHE1102, Chapter 13 Learn, 9 If slope of tangent line at t = 0 is x M/sec Rate = -  [H 2 O 2 ]  t = - ( x M/sec) Rate =2.0 x M/sec 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) Consider Then,

CHE1102, Chapter 13 Learn, 10 Can the rate of H 2 O 2 being consumed be related to the rate of products being generated ? 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) [H 2 O] increases at a rate of 2.0 x M/sec as [H 2 O 2 ] decreases at a rate of 2.0 x M/sec [O 2 ] increases at a rate of 1.0 x M/sec as [H 2 O 2 ] decreases at a rate of 2.0 x M/sec What is the rate that O 2 forms when the [H 2 O 2 ] is decreasing at a rate of 2.0 x M/sec ?

CHE1102, Chapter 13 Learn, 11 Reaction Rates and Stoichiometry To generalize, for the reaction a A + b Bc C + d D Rate =  1a1a  [A]  t =  1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t =

CHE1102, Chapter 13 Learn, 12 Reaction Rates and Stoichiometry How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 (g) 3 O 2 (g)? If the rate at which O 2 appears,  [O 2 ]/  t, is 6.0 X M/s at a particular instant, at what rate is O 3 disappearing at this time, -  [O 3 ]/  t?

CHE1102, Chapter 13 Learn, 13 Factors that Affect Reaction Rates 1)Physical state of the reactants (usually means surface area) 2)Concentration of the reactants 3)Temperature of the reactants 4)Presence of a catalyst

CHE1102, Chapter 13 Learn, 14 Factors That Affect Reaction Rates Physical state of the reactants –In order to react, molecules must come in contact with each other. –The more homogeneous the mixture of reactants, the faster the molecules can react.

CHE1102, Chapter 13 Learn, 15 Factors That Affect Reaction Rates Concentration of reactants –Increasing reactant concentration generally increases reaction rate. –Since there are more molecules, more collisions occur.

CHE1102, Chapter 13 Learn, 16 Factors That Affect Reaction Rates Temperature –Reaction rate generally increases with increased temperature. –Kinetic energy of molecules is related to temperature. –At higher temperatures, molecules move more quickly, increasing numbers of collisions and the energy the molecules possess during the collisions.

CHE1102, Chapter 13 Learn, 17 Factors That Affect Reaction Rates Presence of a catalyst –Catalysts speed up reactions by changing the mechanism of the reaction. –Catalysts are not consumed during the course of the reaction. Do not appear in balanced equation.

CHE1102, Chapter 13 Learn, 18 Reaction Rate vs. Concentration 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) Rate = k [H 2 O 2 ] x Rate law expression takes the form: Rate = rate of the reaction k = rate constant [H 2 O 2 ] = concentration of reactant x = order of the reactant (usually an integer)

CHE1102, Chapter 13 Learn, 19 What if the rxn has multiple reactants ? 2 HgCl 2 + C 2 O Hg 2 Cl CO Cl - Rate = k [HgCl 2 ] x [C 2 O ] y k = rate constant indicative of this rxn x = order wrt [HgCl 2 ] y = order wrt [C 2 O ] x and y must be determined experimentally

CHE1102, Chapter 13 Learn, 20 Trial[HgCl 2 ][C 2 O 4 2- ]initial rate, M/sec M0.30 M 7.65 x M0.30 M 1.53 x M0.10 M1.70 x To determine the numerical value for x and y, examine the data and find 2 trials where [reactant] of interest is varied, while all other reactants are held constant. Compare rate law expressions for these two trials to determine the order of that reactant. Method of Initial Rates

CHE1102, Chapter 13 Learn, 21 Trial[HgCl 2 ][C 2 O 4 2- ]initial rate, M/sec M0.30 M 7.65 x M0.30 M 1.53 x M0.10 M1.70 x Use trials 1 and 2 to obtain the numerical value for the order wrt [HgCl 2 ] (which is x) Examine trials 1 and 2 [HgCl 2 ] varies while the [C 2 O ] is held constant Rate = k [HgCl 2 ] x [C 2 O ] y x = 1 unitless

CHE1102, Chapter 13 Learn, 22 Trial[HgCl 2 ][C 2 O 4 2- ]initial rate, M/sec M0.30 M 7.65 x M0.30 M 1.53 x M0.10 M1.70 x Use trials 2 and 3 to obtain the numerical value for the order wrt [C 2 O ] (which is y) Examine trials 2 and 3 Rate = k [HgCl 2 ] x [C 2 O ] y [C 2 O ] varies while the [HgCl 2 ] is held constant y = 2 unitless

CHE1102, Chapter 13 Learn, 23 Rate = k [HgCl 2 ] x [C 2 O ] y x = 1 y = 2 Rate = k [HgCl 2 ] 1 [C 2 O ] 2 x and y do NOT vary under any circumstances ! or Rate = k [HgCl 2 ] [C 2 O ] 2 rxn is 1 st order wrt HgCl 2 rxn is 2 nd order wrt C 2 O 4 2 -

CHE1102, Chapter 13 Learn, 24 The experimentally determined rate law for the reaction 2 NO(g) + 2 H 2 (g) N 2 (g) + 2 H 2 O(g) is rate = k[NO] 2 [H 2 ]. What are the reaction orders in this rate law? What is the effect of doubling the concentration of H 2 on the rate? What is the effect of doubling the concentration of NO? Rate Laws

CHE1102, Chapter 13 Learn, 25 Rate = k [HgCl 2 ] [C 2 O ] 2 Determine the Numerical Value for the Rate Constant, k = k Rate [HgCl 2 ] [C 2 O ] 2 1 M 2 ·sec 8.5 x k = k is constant and does NOT vary with [ ] units are very important k does vary with temperature

CHE1102, Chapter 13 Learn, 26 Rate constants can be used to compare the relative rates of reactions. –the larger the value of k, the faster the reaction. –the smaller the value of k, the slower the reaction. The units of the rate constant depend on the overall reaction order of the rate law. For a second order rxn: For a first order rxn: Units of rate constant = M -1 s -1 Units of rate constant = s -1 eg. Rate = k[N 2 O 5 ] eg. Rate = k[H 2 ][I 2 ] Rate Constants

CHE1102, Chapter 13 Learn, 27 Overall Reaction Order x + y = =3 reaction is overall 3 rd order – sum of the individual reaction orders Rate = k [HgCl 2 ] [C 2 O ] 2

CHE1102, Chapter 13 Learn, 28 [ ] vs. time Δ aAaAbB + jJ if Rate = k[A] 1 rxn is overall 1 st order ln [A] o [A] t = kt [A] o = initial [A] at t = 0 [A] t = [A] at some time later, t k = rate constant t = time from t = 0

CHE1102, Chapter 13 Learn, 29 SO 2 Cl 2 decomposes as follows: SO 2 Cl 2 SO 2 + Cl 2 The rate constant, k = 2.2 x s -1 at 300  C. Suppose 0.30 M SO 2 Cl 2 is placed in a container at 300  C and allowed to decompose. What is the [SO 2 Cl 2 ] after 75 minutes ? Whenever the rate constant, k has units of 1/time ( or time - 1 ), the reaction is overall 1 st order ! 1 t

CHE1102, Chapter 13 Learn, 30 Half-life Abbreviation for half-life is t 1/2 For rxn is overall 1 st order t 1/2 = k – time required to convert exactly ½ reactants into products

CHE1102, Chapter 13 Learn, 31

CHE1102, Chapter 13 Learn, 32

CHE1102, Chapter 13 Learn, 33 Prozac is a prescription drug which is metabolically eliminated from the body by overall first order kinetics. Prozac has a body clearance half-life of x 10 5 sec. If a patient is given a 20.0 mg dose, how many hours until only 13.8 mg Prozac remains in the body ?

CHE1102, Chapter 13 Learn, 34 [ ] vs. time Δ aAaAuU + kK if Rate = k[A] 2 rxn is overall 2 nd order 1 [A] t = kt 1 [A] o t 1/2 = 1 k [A] o

CHE1102, Chapter 13 Learn, 35 ln [ ] vs. time is linear [ ] vs. time is NOT linear only occurs when the rxn is overall 1 st order slope = - k CH 3 NC CH 3 CN

CHE1102, Chapter 13 Learn, 36 NO 2 NO + ½ O 2 This rxn is NOT 1 st order only occurs when the rxn is overall 2 nd order vs. time is linear 1 [ ] slope = k

CHE1102, Chapter 13 Learn, 37 Temperature vs. Rate An increase in temperature results in an increase in the rate of a reaction This is because k is temperature-dependent.

CHE1102, Chapter 13 Learn, 38

CHE1102, Chapter 13 Learn, 39 Collision Theory – For a reaction to occur, reacting species must first collide 1. collision must possess enough energy for molecules to react 2. collision must have reactants in the proper orientation to react

CHE1102, Chapter 13 Learn, 40 Collision Theory There is a minimum amount of energy required for reaction: the activation energy, E a. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation-energy barrier. E a is usually the most important factor in determining whether a particular collision results in a reaction.

CHE1102, Chapter 13 Learn, 41 The Collision Model Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Cl + NOCl NO + Cl 2

CHE1102, Chapter 13 Learn, 42 Activation Energy, E a

CHE1102, Chapter 13 Learn, 43 Relating E a to Speeds of Rxn Consider these energy profiles: Rank the reactions according to rate.

CHE1102, Chapter 13 Learn, 44 How is Activation Energy, E a, Determined? In lab, keep [reactants] constant and vary the temperature. Calculate values of the rate constant, k at each temperature. Svante Arrhenius 1859 – 1927

CHE1102, Chapter 13 Learn, 45 Arrhenius Equation k = (A) (e (−Ea/RT) ) y = mx + b ln k = + ln A - E a RT take ln of both sides slope = - Ea R a plot of ln k vs. gives a straight line 1T1T

CHE1102, Chapter 13 Learn, 46 1T1T K -1 slope = - 19,000 K ln k vs. 1T1T K -1

CHE1102, Chapter 13 Learn, 47 slope = -EaR-EaR E a = - (slope) (R) slope = - 19,000 K E a = - ( - 19,000 K) (8.314 J/mole·K) E a = 157,966 J/mole E a = 158 kJ/mole E a must be positive

CHE1102, Chapter 13 Learn, 48 Arrhenius Equation ln k =  ( ) + ln A 1T1T EaREaR E a can also be obtained non-graphically if we know the rate constant of a reaction at two or more temperatures. ln k 1 =  ( ) + ln A 1T11T1 EaREaR ln k 2 =  ( ) + ln A 1T21T2 EaREaR ln k 1 - ln k 2 = ln = (1/T 2 – 1/T 1 ) EaREaR k1k1 k2k2

CHE1102, Chapter 13 Learn, 49 Arrhenius equation (again) = ln k1k2k1k2 1T11T1 1T21T2 EaREaR Trialk, 1/M·secTemp x  C x  C 2 HI H 2 + I 2 Given the data above, determine the E a of the rxn

CHE1102, Chapter 13 Learn, 50 Catalysts 1. Decrease the E a by providing a different pathway to products 2. Provide the correct orientation for the reactants to react – increase the rate of rxn without being consumed

CHE1102, Chapter 13 Learn, 51 uncatalyzed rxncatalyzed rxn

CHE1102, Chapter 13 Learn, 52

CHE1102, Chapter 13 Learn, 53 Fritz Haber 1868 – 1934 Karl Bosch 1874 – 1940 N H 2 2 NH 3 Haber-Bosch Process

CHE1102, Chapter 13 Learn, 54 Catalyst is primarily Fe & Fe 2 O 3 H 2 and N 2 continuously added NH 3 has no affinity to absorb on surface H 2 and N 2 adsorb on the surface of the iron 3 H and N react on surface of the iron

CHE1102, Chapter 13 Learn, 55

CHE1102, Chapter 13 Learn, 56 Enzymes Enzymes are biological catalysts. They have a region where the reactants attach. That region is called the active site. The reactants are referred to as substrates.

CHE1102, Chapter 13 Learn, 57 Lock-and-Key Model In the enzyme–substrate model, the substrate fits into the active site of an enzyme, much like a key fits into a lock. They are specific.