Unformatted text preview: Chemical Kinetics Read & Study: Chapter 13 Read Chapter View the videos in the Science Learning View Center (SLC), N-604. One is entitled, “Catalysis: Technology for a Clean Environment” and the other is “Reaction Rates”. Rates”. Chapter 13 Chapter Chapter Overview
1. Introduction - Background and Key Definitions 1. Background 2. Factors Influencing the Rates of Reactions 2. Rates 3. Rate Laws 4. Activation Energies 5. Reaction Mechanisms - The “Paragraphs” of the 5. The Chemical Language Chemical 6. Effects of Catalysts
12/04/09 2 Introduction
1. Chemical Kinetics - The study of how fast chemi1. The cal reactions take place; a study of the “RATE” of a cal “RATE” reaction. 2. Chemical Reaction Rate - A change in the amount 2. amount or concentration of a reactant or product in a given or concentration period of time. Rate = - ∆ [Reactant] = + ∆ [Product] Rate ∆ time ∆ time time
12/04/09 3 3. Kinetics vs. thermodynamics - Factors to be con3. Factors sidered when predicting whether or not a change will take place: 1) Energy (Enthalpy) Change Thermodynamics 2) The Change in disorder (Entropy) 3) The RATE of the change 3) RATE Kinetics 4. Reaction Mechanism - A detailed molecular-, 4. atomic-, or ionic-level “picture” or model of how how a reaction takes place.
12/04/09 4 How does NO (g) react with O2 (g) in the atmosphere to form NO2 (g), a major contributor to smog? 2 NO (g) + O2 (g) NO Step 1: NO (g) + O2 (g) Step NO Step 2: NO3 (g) Step NO Step 3: NO3 (g) + NO (g) Step NO 2 NO2 (g) NO3 (g) Fast NO F ast NO (g) + O2 (g) Fast NO2 (g) Sl ow 12/04/09 5 By measuring the rates of reactions under various By rates conditions, rate laws can be deduced. The chemist, conditions, rate then, designs reaction mechanisms to “explain” the then, reaction observed rate laws. observed 5. Factors affecting the rates of reactions 5. Memorize!! 1) The Nature of the Reactants 1) Memorize!! 2) The Reactants’ Concentrations Memorize!! 3) The Reactants Total Surface Area Memorize!! 4) The Temperature Memorize!! 5) The presence or absence of catalysts
12/04/09 6 FACTORS AFFECTING REACTION RATES FACTORS 1) The Nature of the Reactants • Ions in solution tend to react very quickly. Ions • Covalent molecules tend to react more slowly. Covalent • Large covalent molecules tend to react more Large slowly than small covalent molecules. • Molecules with strong covalent bonds tend to Molecules react more slowly than those with weak bonds.
12/04/09 7 2) The Concentrations of the Reactants For a reaction to occur, three things must happen • The reacting molecules, atoms, or ions, must collide. collide • They must be properly oriented with respect to one oriented another. • The must have sufficient energy. energy The probability of these three factors being present is increased as the concentration of the reactants is concentration increased.
12/04/09 8 Proper Orientation is Necessary 12/04/09 9 Suffcient Energy is Necessary Bounces Off! Reacts! 12/04/09 10 Homogeneous Reactions - Reactions that take place Homogeneous Reactions in a single solution (phase). Cu2+ + 4 NH3 (aq) NaOH (aq) + HCl (aq) Cu(NH3)42+ NaCl (aq) + H2O (l) 3) The Reactants’ State of Subdivision Heterogeneous Reactions - Reactions in which Heterogeneous reactants are in different phases. The reactions reactants take place at the interface between the phases. take interface The larger the surface area of the interface, the The surface higher the rate of reaction. higher rate
12/04/09 11 V = 1 m3 A = 6 m2 V = 1 m3 A = 8 m2 V = 1 m3 A = 12 m2 V = 1 m3 A = 36 m2 V = 1 m3 Powder A = ?? m2 V = 1 m3 A = 20 m2
12 12/04/09 http://blog.wired.com/wiredscience/2008/03/top-10-amazing.html http://pubs.acs.org/cen/government/86/8608gov1.html 12/04/09 13 Examples of Heterogenous Reactions C (s) + O (g)
2 CO (g)
2 Zn (s) + 2 HCl (aq) ZnCl (aq) + H (g)
2 2 4) The Temperature - Reaction Ag(NH ) increased 4) (s) + 2 NH (aq)Reaction rates are + Cl AgCl by increases in temperature.
3 3 2+ - Rule of Thumb: A 10oC rise in temperature will 10 double the rate of the reaction. the
12/04/09 14 Practice Exercise: The temperature at which a reaction Practice The is run is raised from 0 C to 40 C. How much is the
o o rate of the reaction increased? toC Rate 0 R1 R =2 R 10 2R1 20 4R1 30 8R1 If you want to slow down the reaction that causes your If slow 40 16R1 = R2 milk to sour, you put the milk in the refrigerator to lower its temperature.
12/04/09 15 What about the Thermite Reaction and temperature?
http://blog.wired.com/wiredscience/2008/03/top-10-amazing.html 12/04/09 16 5) Catalysis - How can a catalyst change the rate of a 5) How reaction and yet NOT be consumed? Thermodynamic Answer: By changing the energy Thermodynamic By pathway. Activation Energy Energy Reactants Products With a catalyst, the “energy hump” is lowered -the energy pathway is changed. 12/04/09 17 Kinetic Answer: The catalyst is a reactant in one Kinetic reactant step of the reaction mechanism and a product in a product subsequent step. subsequent Homogeneous Catalysis - Reactions wherein the Homogeneous catalyst is present in the same phase as the reactants. catalyst 2 SO (g) + O (g) Catalyst NO (g) NO (g) 2 SO (g) Heterogeneous Catalysis - Reactions wherein the catHeterogeneous Reactions alyst is present in a separate phase from the reactants. Pt (s) 2 H (g) + O (g) 2 H O (g) Catalyst
12/04/09 18 Heterogeneous Catalysis 12/04/09 19 Hydrogen reacts very exothermically with oxygen. Even so, they can be held in the same container together for years without reacting unless a catalyst or source of years catalyst energy, such as a spark, are introduced into the mixture. energy, Rat e L aws
Rate Law - A mathematical equation that shows how the Rate mathematical rate of a reaction is affected by the concentration of reactants or products in a reaction. Measuring the Rate of a Reaction - Follow the disapMeasuring Follow pearance of a reactant or the appearance of a product.
12/04/09 20 2 N2O5 (g) 2 N2O4 (g) + O2 (g) Rate = +∆[O2]/∆ t = +(1/2)(∆[N2O4]/∆ t) t) = - (1/2)(∆ [N2O5]/∆ t) 12/04/09 21 Decomposition of N2O5 (g) (g) 12/04/09 22 Reaction of Ethylene w/ O3 (g) (g)
Another Case! 12/04/09 23 Measuring the Rate of a Reaction
Method of Initial Rates – A method that involves running Method method A reaction several times starting with different initial reaction different Concentrations of reactants. 1. 2. 3. Hold the temperature constant. If Present, keep the solvent constant. Vary (change) only one concentration at a time. 12/04/09 24 Method of Initial Rates
2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)
Run Number Reactant Concentration [NO] [H2] [NO] 0.10 0.10 0.30 0.010 0.040 0.010 Rate of decrease in [NO] 0.062 0.246 0.558
25 1 2 3
12/04/09 Method of Initial Rates Effect of [H ] on the Rate:
2 0.040 M 0.246 M/s 0.246 As the [H2Mis raised by a factor.062 M/s rate of 0.010 ] 0 of 4.0, the decrease in [NO] (the rate of the reaction) is raised by a factor of 4.0. ∴ Rate α [H2] Effect of [NO] on the Rate: 0.30 M = 3.0 3.0 0.10 M
12/04/09 = 4.0 4.0 = 4.0 4.0 ∴ Rate α [NO]2 0.558 M/s = 9.0 0.558 9.0 0.062 M/s
26 Method of Initial Rates where k is called the “Reaction Rate Constant”! where is k is independent of reactant concentrations but IS is dependent on temperature. The Rate Law exponents are usually NOT the same as The the coefficients in the balanced chemical equation. Therefore, the rate law cannot be predicted. It MUST be Therefore, experimentally determined.
12/04/09 27 aA + bB cC + dD aA k Rate = k [A] [B]
Run Number Number 1 2 3
12/04/09 Reactant Concentration [A] [B] [A] 0.030 0.060 0.030 0.010 0.010 0.020 Initial Initial Rate 1.7 x 10-8 6.8 x 10-8 4.9 x 10-8
28 rate 1 = k(0.030)m(0.010)n = 1.7 x 10-8 M/s rate 2 = k(0.060)m(0.010)n = 6.8 x 10-8 M/s
-8 rate 1 = 1.7 x 10-8 = (0.030)m = 0.030 m 1.7 (0.030) 0.030 rate 2 6.8 x 10-8 (0.060)m 0.060 rate 0.25 = (0.50) 12/04/09 29 rate 1 = k(0.030)m(0.010)n = 1.7 x 10-8 M/s rate 3 = k(0.030)m(0.020)n = 4.9 x 10-8 M/s
-8 rate 1 = 1.7 x 10-8 = (0.010)n = 0.010 n 1.7 (0.010) 0.010 rate 3 4.9 x 10-8 (0.020)n 0.020 rate 0.35 = (0.50) Rate = k [A]2[B]3/2
12/04/09 30 2 I- + S2O82Run Run [I ]
- I2 (aq) + 2 SO42[S O ]
2 82- Initial Rate 1 2 3 0.15 0.15 0.50 0.45 0.25 0.45 2.6 x 10 - 4 2.6 1.4 x 10 - 4 8.6 x 10 - 4 Find the Rate Law: Rate = k [I-] [S2O82- ]
12/04/09 31 m n rate 1 = k(0.15) (0.45) = 2.6 x 10 -4 M/s -4
m n m m = = rate 3 = k(0.50) (0.45) = 8.6 x = -4 M/s = 0.30 10 -4 rate 1
-4 2.6 xm10-4 2.6 (0.15) (0.15)
m 0.15 0.15 0.50 0.303= (0.30) 10-4 rate 8.6 x (0.50)
m 12/04/09 Log (0.30) = Log (0.30) = m Log (0.30) 32 m n rate 1 = k(0.15) (0.45) = 2.6 x 10 -4 M/s -4
m n n n = = rate 2 = k(0.15) (0.25) = 1.4 x = -4 M/s = 1.86 10 -4 rate 1 2.6 x 10-4 2.6 n -4 (0.45) (0.45)
n 0.45 0.45 0.25 1.86 2 (1.80)x 10-4 rate = (1.8 1.4 (0.25)
n ∴ Rate = k [I-][S2O82- ]
33 12/04/09 Log (1.86) = Log (1.80) = n Log (1.80) Rate Laws
The Order of a Reaction – The sum of the exponents in the The Order The Rate Law for the reaction Rate for 866 C
o 2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g) Rate = k1 [H2][NO]2 [H k First Order with respect to [H2] Second Order with respect to [NO] Third order OVERALL Third OVERALL
12/04/09 1 Determined by Experiment 34 Rate Laws aA + bB cC + dD
2 Rate = k2 [A]2[B]3/2 [A] k Determined by Experiment Second Order with respect to [A] 1.5 Order with respect to [B] 3.5 order OVERALL 3.5 OVERALL 2 I- + S O Rate = k [I-][S O ]
12/04/09 I (aq) + 2 SO
2 42- 2 822 82- First Order with respect to both First [I-] and [S2O82-] and [I second order OVERALL second OVERALL 35 Rate Laws aA + bB + cC dD + eE
4 Subtances involved in a reaction that have NO concentration Subtances NO effect on the rate of the reaction are said to have a “zeroth order” effect effect. The reaction above is zeroth order with respect to [C]. effect. zeroth k REMEMBER: The rate law for a reaction CANNOT be predicted from the chemical equation! It MUST be predicted MUST experimentally determined!
12/04/09 36 Rate Laws
Class Exercise: The following reaction is second Class order with respect to acetaldehyde. Write the Rate Law for this reaction. Law
CH3-C –H (g) CH4 (g) + CO (g) -C ||| k | O Rate = k [CH -C –H]
3 2 Rate Constants: The value of the rate constant, k, for a Rate given reaction depends on the NATURE of the reactants and the given NATURE TEMPERATURE, but NOT on the concentration of reactants. TEMPERATURE but NOT
12/04/09 37 ||| | O Rate Constants
Class Exercise: A + 2 B C Class Rate = k [B] = 0.012 mol·L ·s
When [B]0 = 0.60 M and [A]0 = 0.020 M, Find k. k = rate/[B]0 = (0.012 mol·L-1·s-1)/(0.60 mol·L-1) Find the rate of the reaction when [A]0 = [B]0 = 0.010 M Find [A] 12/04/09 38 Rate Law Characteristics
Zero Order Reactions: Reactions with rates that are Zero not dependent on the concentration of reactants. not Reaction Rate [Reactant]
12/04/09 Time Time The rate of metabolism of alcohol in the body is independent of [alcohol]. The RATE is constant! dent RATE
39 Rate Law Characteristics
First Order Reactions: First 2 N2O5 (g) 4 NO (g) + O2 (g) In General: Rate = k[A] In Rate Rate = k[N2O5] Rate Time
12/04/09 ln [A] [A] Time
40 Rate Law Characteristics [A] = [A]o e-kt
Integrated Rate Law (1st Order) Integrated ln [A] = ln [A]o + (-kt) ln [A] = - kt + ln [A]o y = mx + b
Half-Life: The time required for one-half of a reactant Half-Life: The to disappear in a reaction. For FIRST ORDER reactions, to FIRST The half-life is INDEPENDENT of reactant concentration. The INDEPENDENT of O.K. Let’s Prove That!
12/04/09 41 Rate Law Characteristics
ln [A] = ln [A]o + (-kt) [A] [A] = -kt ln [A]o [A]
ln [A]o [A] 2[A]o 2[A]
1/2 1/2 1/2 1/2 But…[A] = ½ [A]o at time t = ln (1/2) = -kt kt = - ln (1/2) = ln 2 = 0.693 ln
12/04/09 1/2 42 1/2 The Half-Life (t ) of a First Order process is inversely The Half-Life of proportional to the rate constant, k, and is independent proportional independent 2 ofOrder Reactions: the reactant concentration.
nd Rate = k[A]2 Simplest Type Rate Simplest Rate = k[A][B] Also 2nd Order k 2 NO2 (g) 2 NO (g) + O2 (g) ntegrated Rate Law (2nd Order) 1/[A] = kt + 1/[A]o y = mx + b
12/04/09 43 Rate = k[NO2]2 1/2 1/2 At time t , [A] = 1/2[A] . Therefore At 2/[A]o = kt + 1/[A]o 1/2 1 1/2 [A]o = 1/[A] ∴ kt k 2/[A]o -yout tello 1= ot1/2 = 1/k[A]ofrom 2nd order? How can st rder reactions 1t 1sst ln [A] 1/[A] 12/04/09 time time 2 nd d Or de r rr de de Or Or 44 Temperature Effects on Reaction Rates Arrhenius Plot Take log of both sides:
-Ea/RT ln k = ln A + ln e -Ea/RT ln k = Ae-Ea/RT ln k = ln A - Ea/RT ln k = -(Ea/R)(1/T) y = mx
12/04/09 + + ln A b E = -R(Slope)
45 Slope = - Ea/R Temperature Effects on Reaction Rates If you know the k at one temperature, you can find k for a different temperature with the Arrhenius Equation. ln k1 = -(Ea/R)(1/T1) + ln A ln k2 = -(Ea/R)(1/T2) + ln A ln k1 + Ea/RT1 = ln A ln k2 + Ea/RT2 = ln A ln k1 + Ea/RT1 = ln k2 + Ea/RT2
12/04/09 46 Temperature Effects on Reaction Rates ln k1 + Ea/RT1 = ln k2 + Ea/RT2
ln k1 - ln k2 = [Ea/R][(1/T2)-(1/T1)] = ln (k1/k2) Class Exercise: ln k2 = ln k1 - [Ea/R][(1/T2)-(1/T1)] ln k2 = -2.538
12/04/09 k2 = 0.0798 s-1
47 Activation Energy: The minimum energy necessary Activation The for a reaction to occur. 12/04/09 48 Temperature Effects on Reaction Rates Arrhenius Plot Ea/RT k = Ae12/04/09 49 Reaction Mechanisms Reaction Mechanism: A series of consecutive molecularReaction series level events or steps that lead from reactants to products. They can be disproven but never proven. They disproven proven For a mechanism to be valid, it must account for everything that is known about the reaction, including the rate law for the reaction and its stoichiometry. Elementary Process: A single reaction step in a reaction Elementary single mechanism. 12/04/09 50 ...
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