This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Overview
Chapter 14 Kinetics (Rates of Reactions) Chapter 15 Equilibrium Chapter 19 Thermodynamics Later chapters are applications of these topics We will look at these topics at both the macroscopic level and the microscopic level.
1 Chapter 14 Chemical Kinetics
Kinetics tells us how fast the reaction will go. Kinetics is the study of rates of reactions factors that affect these rates: concentration, temperature, pressure, catalysis, surface area, solvent, environment... 2 12.1 Reaction Rates
Reaction rate is a measure of how fast a reaction occurs. Some reactions are inherently fast and some are slow: Effect of Concentration
Changing the concentration of a reactant can change the reaction rate:
NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) HCl(aq) NaCl(aq) O(l 3 4 Effect of Surface Area Effect of Temperature
A light stick glows brighter at higher temps, but it will run out sooner. Milk sours more quickly when left at room temperature. Food cooks more quickly at higher temps.
5 6 Iron Nail Steel wool Rusting Fast Rusting Slow 1
1 Catalysis
The catalyst called catalase in this piece of liver causes the decomposition of H2O2 to occur faster. 14.1 Reaction Rates
Must be able to measure rates to determine how to control them with concentration or temperature. H2 O 2 decomposition
7 8 14.1 Reaction Rates
Rate = conc/time or conc/time so it conc/ conc/ will have a positive value For A B A
Rate =  A B = time time Rates and Stoichiometry
Use reaction stoichiometry to specify reference points for rates. When reacting in ratios not equal to 1:1, one substance disappears or appears faster than the other.
5Br (aq) + BrO3 (aq) + 6H+ (aq) 3Br2(aq) + 3H2O(l) aq) aq) aq) Rate Measurement
9 Br disappears 5 times as fast as BrO3 [BrO3] 1 [Br] 1 [H+] +1 [Br2 ]  =   =   =   t 5 t 6 t 3 t Can't measure change in H2O because its Can' concentration is so high, it stays constant.
10 0.085
5Br (aq) + BrO3 (aq) + 6H+ (aq) 3Br2(aq) + 3H2O(l) Reaction Rates
Write equations for the rate of this reaction in terms of each reactant and product. 2N2O5(g) 4NO2(g) + O2(g) [Br] [BrO3]
0.017 [Br2]
0.051
11 12 2
2 Calculation of Rates
2N2O5(g) 4NO2(g) + O2(g) Start with 2.330 M solution at measure the concentration periodically. 1 ([N2O5]2  [N2O5 ]1) 1 [N2O5] Rate =   =   2 t 2 (t 2  t 1) Rate during time period 52,400 to 72,000 sec, where the concentrations are measured to be 1.350 M and 1.110 M.
13 2N2O5 4NO2 + O2(g)
Time (s) [N2O5] (M) 45oC; 0 11100 19200 31600 *52400 *72000 112700 138900 188600 2.33 2.08 1.91 1.67 1.35 1.11 0.72 0.55 0.34 14 Calculation of Rates
Rate = 1/2( [N2O5] / t) 1/2( t = 72,000 52,400 = 19,600 sec [N2O5] = (1.110 M 1.350 M) = 0.240 M Rate = 1/2 (240 M/19,600 s) = 6.12 106 M/s
15 Rates
The rate changes as the reaction proceeds. Usually it gets smaller with time. The rates calculated are average rates for the particular time interval. 16 Rates
Can get average rate and instantaneous rate from a graph of concentration vs. time Initial Rates
To get around the problem of the rate changing with time, we sometimes measure initial rates. rates. 17 18 3
3 Initial Rates
If the reaction is not too fast, then C/t is a C/ good approximation to the rate. We know concentration values much better at the beginning of the reaction than we do later.r different concentrations. What is the later. relationship between initial rate and concentration?
19 14.2 The Dependence of Rate on Concentration
Rates usually vary with initial concentrations of some or all reactants, possibly products. Eg. 2 g Na2CO3 + 50 mL HCl in a flask Eg. enclosed with a balloon. Which reaction is fastest? Effect of Conc Effect of SA 1M 3M 6M
20 Rate and Concentration
Why does rate generally decrease as the reaction proceeds?
concentration of reactants decreases fewer molecules to collide, so collision frequency decreases Rate Laws A + B C + D
We describe the dependence of rate on concentration by an equation called the rate law: Rate = k[A]x[B]y... where x & y are not stoichiometric coefficients k, x, y must be determined experimentally
k is the rate constant x,y are reaction orders for the individual species x+y+... = overall reaction order (n)
21 22 Rate Law
Rate = k[A]x[B]y... Orders (x, y, ...) are usually positive or negative integers Each order tells us how the rate depends on that concentration; this order must be determined experimentally; only occasionally experimentally; does it match the coefficients from the balanced reaction equation
23 How do we Determine Rate Laws?
Method 1: When rate is dependent upon only 1: one species. 2N2O5 4NO2 + O2(g) [N2O5] [NO2] [O2] Rate
2.21 2.21 2.21 2.21 2.00 1.00 1.00 0.00 1.00 1.00 2.00 0.00 1.14 x 105 1.13 x 105 1.12 x 105 1.13 x 105 The rate is constant when we change [NO2] and [O2], so orders are 0 (not in rate law)
24 4
4 How do we determine rate laws?
Thus, Rate = k[N2O5]x Determine Reaction Order
Rate = k[N2O5]x What is the order? (What is the value of x?)
[N2O5] Rate (M/s) Vary [N2O5] while measuring rate. Since k should have a constant value, calculate: k = Rate k = Rate/[N2O5] k = Rate/[N2O5]2 And see which is constant 2.00 1.79 1.51 1.23 0.92
25 1.05x10 6 9.67x10 6 7.83x10 6.31x106 6 4.81x10
26 5 How do we determine rate laws?
Method 2: If the rate depends on the concentration of more than one substance, design the experiment so only one concentration changes appreciably. Vary one concentration at a time and measure the rate. Compare one variable at a time. Group Work
Derive the rate law and value of k for the reaction, F2 + 2ClO2 2FClO2 [F2]o [ClO2]o Rate (M/s)
Exp 1 Exp 2 Exp 3 0.100 0.100 0.200 0.0100 0.0400 0.0100 0.0012 0.0048 0.0024 R2/R1 = 4 and C2/C1 = 4, so 1st order in ClO2 R3/R1 = 2 and C3/C1 = 2, so 1st order in F2 Rate = k[F2][ClO2] and k = 1.2 M1s1
27 28 Units on k
The units on k depend on the overall order for the reaction. Determine the units using dimensional analysis. What are the units on k for:
Zero Order 1st Order 2nd Order
29 Group Work
Determine the rate law and the value of the rate constant for the following reaction: 2I(aq) + 2VO2+(aq) + 4H+(aq) I2(aq) + 2VO2+(aq) + 2H2O(l) [I] 0.00200 0.00400 0.00200 0.00200 [VO2+] 0.0100 0.0100 0.0200 0.0100 [H+] 0.100 0.100 0.100 0.0500 Rate 26.0 52.1 51.9 6.50
30 Exp 1 Exp 2 Exp 3 Exp 4 5
5 Group Work
Determine the rate law and the value of the rate constant for the following reaction: 2I(aq) + 2VO2+(aq) + 4H+(aq) I2(aq) + 2VO2+(aq) + 2H2O(l) [I] [VO2+] [H+] Rate 0.00200 0.0100 0.100 26.0 Exp 1 0.00400 0.0100 0.100 52.1 Exp 2 0.00200 0.0200 0.100 51.9 Exp 3 0.00200 0.0100 0.0500 6.50 Exp 4 0.00200 0.0100 0.150 ? Exp 5
31 Rate Laws
Rate Laws in the form Rate = k[A]x[B]y tell us about the rate of a reaction at a specific instant, and about the dependence of initial concentration on intitial rate. This form of the rate law does not tell us information about time, such as "how long it will take for a specific concentration to react completely, or half way, ..." Integrated rate laws will give us this information.
32 14.4 The Change of Concentration with Time How do we determine rate laws? The dependence of concentration on reaction rate for each reactant is either zero order, 1st order, or 2nd order (usually). Zero Order: Rate = [A]/t = k[A]0 = k [A]/ 1st Order: Rate = [A]/t = k[A]1 [A]/ 2nd order: Rate = [A]/t = k[A]2 [A]/ Converting these equations to linear equations as a function of time will allow us to graph the three possibilities (Method 3)
33 14.3 The Change of Concentration with Time How do we determine rate laws? Zero Order in A: (n=0) Rate = [A]/t = k[A]0 = k [A]/ [A]/t = k [A]/ Rearranging and taking the integral of both sides gives the integrated rate law: law: [A] = [A]o  kt Example: Mg + 2HCl MgCl2 + H2
34 How do we determine rate laws?
If zero order ([H2]/t = k), a graph of [H2] ( ]/ vs. time should be linear. [H2] = [H2]o + kt
Why +kt in +kt this case? How do we Determine Rate Laws?
1st Order in A (n=1) Rate = [A]/t = k[A] [A]/ Rearranging and taking the integral of both sides gives the integrated rate law: law: ln [A] = ln [A]o  kt 35 36 6
6 How do we Determine Rate Laws?
1st Order in A (n=1) ln [A] = ln [A]o  kt 2N2O5 4NO2 + O2(g) If n = 1, a graph of ln [N2O5] vs. time should be linear. How do we determine rate laws?
2nd Order in A (n=2) Rate = [A]/t = k[A]2 [A]/ Rearranging and taking the integral of both sides gives the integrated rate law: law: 1/[A] = 1/[A]o + kt 37 38 How do we determine rate laws?
2nd Order in A (n=2) 1/[A] = 1/[A]o + kt 2UO2+ + 4H+ U4+ + UO22+ + 2H2O If n = 2, a graph of 1/[UO2+] vs. time should be linear. How do we determine rate laws?
Summary: n = 0 [A]/t = k [A]/ n = 1 [A]/t = k[A] [A]/ n = 2 [A]/t = k[A]2 [A]/ [A] = [A]o  kt ln [A] = ln [A]o  kt 1/[A] = 1/[A]o + kt Method 3 for determining rate law: Graph [A] vs. t, ln [A] vs. t, and 1/[A] vs. t and see which graph is linear.
39 40 What is the order? What is the order? 41 42 7
7 How do we determine rate laws?
Method 4: Measure halflife (t1/2)  the time 4: halfto consume half of the current reactant  reduce its concentration to 1/2 How do we determine rate laws? Half Lives
Zero Order Half Life n = 0, [A] = [A]o  kt [A]  [A]o = kt [A]o  [A] = kt At t1/2, [A] = 1/2[A]o [A]o  1/2[A]o = k t1/2 1/2[A]o = k t1/2 t1/2 = [A]o/2k
t1/2 decreases as the reaction proceeds 43 44 How do we determine rate laws?
1st Order Half Life n = 1, ln[A] = ln[A]o  kt ln[A] ln[A]  ln[A]o = kt ln[A] ln[A]o  ln[A] = kt ln[A] ln ([A]o / [A]) = kt At t1/2, [A] = 1/2[A]0 ln ([A]o / 1/2[A]o) = k t1/2 ln2 = k t1/2 t1/2 = ln 2/k = 0.693/k
t1/2 is constant throughout the reaction
45 How do we determine rate laws?
2nd Order Half Life n = 2, 1/[A] = 1/[A]o + kt Rearrange and substitute [A] = 1/2[A]0 at t1/2 1/[A]o = kt1/2 n = 2, t1/2 = 1/k[A]o
t1/2 increases as the reaction proceeds
46 Summary of HalfLife Equations
n = 0, t1/2 = [A]o/2k
t1/2 decreases as the reaction proceeds Group work: 2N2O5 4NO2 + O2 1st order Approximate the k Value n = 1, t1/2 = ln 2/k = 0.693/k
t1/2 is constant throughout the reaction n = 2, t1/2 = 1/k[A]o
t1/2 increases as the reaction proceeds Use these equations to calculate k or t1/2 from one another. Often used for radioactivity  Chapter 21
47 48 8
8 FirstOrder Problem Group Work
The rate constant (k) for a 1st order reaction was determined to be 0.0462 s1. Predict how long it should take for the reaction to be
a) 50% complete (1/2 unreacted) unreacted) b) 75% complete (1/4 unreacted) unreacted) c) 87.5% complete (1/8 unreacted) unreacted) Types of Problems
Use integrated rate laws to calculate:
half life from k k from half life Conc., after time t, given initial conc. time it takes for a certain amount to react 49 50 Group Work
Consider the decomposition of 3.00 M H2O2, which follows first order kinetics with a rate constant of 0.042 min1 at 70C. 70
1. What is the half life? 2. What conc. of H2O2 is left after 40.0 min? 3. How long will it take for the conc. of H2O2 to decrease to 0.100 M? 14.4 Temperature and Rate
Examples N2 and O2 coexist in air, but combine at high temperatures (in auto engines) to form NOx Rates generally increase with increasing T 2Cr3+ + 6Mn3+ + 7H2O CrO42 + 6Mn2+ + 14H+ 51 cold hot cold hot 52 Rates and Temperature
k increases with temperature, causing the increase in rate 2NO2 2NO + O2 Rate = k[NO2]2 Often k
increases ~ x 2 for 10oC T increase
53 Rates and Temperature
What must the reactants do in order for a reaction to occur? Consider the following reaction as an example: 2O3(g) 3O2(g) 54 9
9 Collision Theory explains why k increases with T
Collision Theory:
For reaction to occur, 1. molecules must collide; more collisions higher collide; rate 2. colliding molecules must be oriented properly to react 3. properly oriented colliding molecules must have sufficient energy (which explains the T dependence) (The average kinetic energy of molecules increases with temperature.) Gas simulation Collision Theory
Not all collisions lead to products. 55 56 Group Work
NO + O3 O2 + NO2 Average Kinetic Energy
The number of molecules with sufficient kinetic energy to react increases as the temperature increases. Which collisions are likely to be effective?
Not all molecules have the same kinetic energy.
57 58 Activation Energy
Why don't all properlyoriented collisions lead don' properlyto products? There must be sufficient energy to break reactant bonds. The activation energy, Ea, is the minimum energy required to initiate a chemical reaction. Ea is often referred to as the energy barrier for a reaction.
59 Energy Barrier 60 10
10 EnergyReaction Coordinate Activation Energy
CH3NC CH3CN Ea Explains Methane Combustion 61 62 Activation Energy
The energy barrier is called the activation energy, Ea. This is the difference between the transition state energy and the average reactant energy. Related to the energy change of the reaction: H = Ea,f  Ea,r O3 + NO O2 + NO2 H = 199.8 kJ/mol, Ea,f = 10.7 kJ/mol What is Ea,r ? Ea,r = 10.7  (199.8) = 210.5 kJ/mol
63 O3 + NO O2 + NO2
250 Ea,f=10.7 kJ 200 150 E
100 Ea,r=210.5 kJ
50 0 Reaction Coordinate 64 O3 + NO
250 O2 + NO2
250 O3 + NO O2 + NO2 200 200 150 150 E
100 E
100 50 50 0 0 Reaction Coordinate 65 Reaction Coordinate 66 11
11 O3 + NO O2 + NO2
250 250 O3 + NO O2 + NO2 200 200 150 150 E
100 E
100 50 50 0 0 Reaction Coordinate 67 Reaction Coordinate 68 O3 + NO O2 + NO2
250 250 O3 + NO O2 + NO2 200 200 150 150 E
100 E
100 50 50 0 0 Reaction Coordinate 69 Reaction Coordinate 70 Activation Energy
Exothermic reaction Activation Energy
Endothermic reaction 71 72 12
12 Arrhenius Equation
Relates k to T and Ea and is used to calculate Ea k = A eEa/RT
A is frequency factor, T in Kelvins, R = 8.314 J/mol K 14.5 Reaction Mechanisms
Mechanism is a detailed pathway for a reaction Proposed mechanism must be consistent with the rate law and other experimental evidence Consider two possible mechanisms for the reaction: SN1 or SN2
73 Measured experimentally by measuring k at different temperatures ln k = lnA  Ea/RT Graph ln k vs. 1/T CH3Cl + OH CH3OH + Cl 74 Mechanisms
Usually have several mechanisms consistent with a given rate law Mechanism identifies the elementary reactions (onestep reactions) that combine to (onemake up the overall reaction. Elementary steps must add up to give the balanced chemical equation. Molecularity: number of reactant molecules Molecularity: in an elementary reaction
unimolecular, bimolecular, termolecular (more is rare)
75 Mechanisms
Overall reaction may be elementary: CH3NC CH3CN Rate = k[CH3NC] For an elementary reaction
orders = coefficients O3 + NO O2 + NO2 Rate = k[O3][NO] This reaction may be elementary. 76 Multistep Mechanisms
Many rate laws are more complex, so the overall reaction cannot be an elementary reaction. These have multistep mechanisms. One step in multistep mechanism is slower than the others called the ratedetermining ratestep. Multistep Mechanisms
Tl3+ + 2Fe2+ Tl+ + 2Fe3+ Rate = k[Tl3+][Fe2+]2/[Fe3+]
Tl3+ + Fe2+ Tl2+ + Fe3+ fast Tl2+ + Fe2+ Tl+ + Fe3+ slow (rate determining) Intermediate: substance produced in one step and consumed in a later step (Tl2+)
Ozone mechanism Hydrazine mechanism
77 78 13
13 Multistep Mechanisms
Group Work
S2O82 + I 2SO42 + I+ I+ + I I2 Which step is rate determining? What is the intermediate? 1st step; I+
79 Multistep Mechanisms
Group Work slow fast S2O82 + I 2SO42 + I+ I+ + I I2 slow fast S2O82 + 2I 2SO42 + I2 (overall or net equation) 80 Group Work
O3(g) + Cl(g) ClO(g) + O2(g) Cl(g ClO(g ClO(g) + O3(g) Cl(g) + 2O2(g) Cl(g ClO(g
Identify any intermediates. Write equation for the overall (net) reaction. 14.6 Catalysis
Catalyst: substance that changes the rate of reaction but is not permanently changed itself Metals catalyze many reactions of gases; called heterogeneous catalysis Mechanism: absorption, reaction, desorption H2 O 2 decomposition
81 82 Heterogeneous Catalysis Catalysis
Catalyst changes the reaction mechanism and lowers the activation energy C2 H4 + H 2 C2 H6 Catalytic Converter: NOx to N2 and O2 CO to CO2 and O2
CO + O2 83 84 14
14 Homogeneous Catalysis
In homogeneous catalysis, everything is in solution (or in gas phase) Identify the intermediate(s) and catalyst(s) in the following mechanism NO + O2 NO3 NO3 + NO 2NO2 NO2 + SO2 NO + SO3 NO2 + SO2 NO + SO3 2SO2 + O2 2SO3 Homogeneous Catalysis
In homogeneous catalysis, everything is in solution (or in gas phase) Identify the intermediate(s) and catalyst(s) in the following mechanism NO + O2 NO3 NO3 + NO 2NO2 NO2 + SO2 NO + SO3 NO2 + SO2 NO + SO3 2SO2 + O2 2SO3 85 86 Enzymes as Catalysts
Enzymes have cavities that just fit the reacting molecules and act to bring the reactants together in the proper orientation. anim
87 15
15 ...
View
Full
Document
This note was uploaded on 04/16/2008 for the course CHM 116 taught by Professor Unknown during the Spring '08 term at ASU.
 Spring '08
 Unknown
 Chemistry, Equilibrium, Kinetics

Click to edit the document details