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Chemical Kinetics
Honors Unit 11
Chemical Kinetics Chemical equations do not give us information
on how fast a reaction goes from reactants to products.
We can use thermodynamics to tell if a reaction is product – or reactant – favored but how do we know about the speed?
Chemical Kinetics KINETICS =
The study of reaction rates and the mechanism (the way the reaction proceeds)
***Only kinetics will tell us how fast the reaction happens!
Reaction Rate A rate is any change per interval of time.
Example: speed (distance/time) is a rate!
Reaction rate = change in concentration of a reactant or product over time Units: , M time-1, or
Rate of Reaction Reaction rate = change in concentration of a
reactant or product over time
Units:
o
o M time-1
Decomposition of N2O5
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0 20 40 60 80 100
Time (sec)
Mo
lar
Co
nce
ntr
atio
n o
f N
2O5
(mo
l/L
)
1) Initial rate = rate at “time zero”
2) Average rate = the rate over a given time interval
3) Instantaneous rate = the slope of the tangent line at a given point
Types of Rates
Types of Rates
The general reaction rate is calculated by dividing rate expressions by stoichiometric coefficients.
For the reaction aA + bB cC +dD
o Disappearance of A =
o Appearance of D =
Expressing a Rate
Collision Theory of Reactants Reactions occur when molecules collide to
exchange or rearrange atoms Effective collisions occur when molecules have
correct energy and orientation
Factors Affecting Rates1. Concentration and physical state of reactants
and productso A Rate Law relates the rate of the reaction to
the concentration of the reactants
2. Temperature
3. Catalystso Catalysts are substances that speed up a
reaction but are unchanged by the reaction
Writing Rate LawsFor aA + bB cC + dD
The rate law is: Rate = k[A]m[B]n
k is the rate constant
The exponents must be determined by performing an experiment.
They are NOT derived from the stoichiometry coefficients in an overall chemical equation!!
Rate Laws & Orders of Reactions
Rate Law for a reaction: Rate = k[A]m[B]n[C]p
The exponents m, n, and p: Are the reaction order Can be 0, 1, 2, or fractions (may be other whole
numbers in fictional examples) Must be determined by experimentation
Overall Order = sum of m, n, and p
Interpreting Rate Laws If m = 1 (1st order)
Rate = k [A]1
If [A] doubles, then the rate doubles (goes up by a factor of 2)
If m = 2 (2nd order) Rate = k [A]2
If [A] doubles, then rate quadruples (increases rate by a factor of 4)
If m = 0 (zero order) Rate = k [A]0 If [A] doubles, rate does not change!
For reaction aA products Rate = k[A]m
Rate Constant, k
Relates rate and concentration at a given temperature.
General formula for units of k: M(1- overall order) time-1
Overall Order Units of k0 M time-1
1 Time-1
2 M-1 Time-1
3 M-2 Time-1
Example #1 (part a): Rate Law ProblemThe initial rate of decomposition of acetaldehyde, CH3CHO, was measured at a series of different concentrations and at a constant temperature.
CH3CHO(g) CH4(g) + CO(g)
(a) Using the data, determine the order of the reaction; that is, determine the value of m in the equation
Rate = k[CH3CHO]m
[CH3CHO] (mol/L) 0.162 0.195 0.273 0.410 0.518
Rate (mol/L*min) 3.15 4.56 8.94 20.2 35.2
Strategy
You are looking at how the concentration affects the rate so compare the two in a proportion!
Use the equation:
Pick any two points from the given data!
mm
2 2m
1 1
A ARate 2Rate 1 AA
Example #1:
CH3CHO (mol/L) 0.162 0.195 0.273 0.410 0.518
Rate (mol/L*min) 3.15 4.56 8.94 20.2 35.2
CH3CHO(g) CH4(g) + CO(g)
Rate = k[CH3CHO]m
Significance of Rate Laws
Rate of rxn = k[CH3CHO]2
Here the rate goes up by FOUR when the initial concentration doubles.
Therefore, the reaction is SECOND order overall.
Example #1 (part b & c):Using the same set of data from part a, and knowing the order of the reaction, determine:
b) the value of the rate constant, k (w/ units!)c) the rate of the reaction when
[CH3CHO] = 0.452 mol/L
Strategy: Use any set of data to find k. Solve for rate using k, rate order equation, and given
concentration.
Example #1 (part b & c):
Example #2:The data below is for the reaction of nitrogen (II) oxide with hydrogen at 800oC.
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
(a) Determine the order of the reaction with respect to both reactants and write the rate law (b) calculate the value of the rate constant, and (c) determine the rate of formation of product when [NO]=0.0024 M and [H2]=0.0042 M.
Strategy: Choose two experiments where the concentration of one
reactant is constant and the other is changing! Solve for m and n separately
Initial
Concentration ()
Initial Concentration ()
Rate of Formation of N2
()
Experiment [NO] [H2] 1 0.0010 0.0040 0.122 0.0020 0.0040 0.483 0.0030 0.0040 1.084 0.0040 0.0010 0.485 0.0040 0.0020 0.966 0.0040 0.0030 1.44
Example #2:
Initial
Concentration ()
Initial Concentration ()
Rate of Formation of N2
()
Experiment [NO] [H2] 1 0.0010 0.0040 0.122 0.0020 0.0040 0.484 0.0040 0.0010 0.485 0.0040 0.0020 0.96
Example #2:
Initial
Concentration ()
Initial Concentration ()
Rate of Formation of N2
()
Experiment [NO] [H2] 1 0.0010 0.0040 0.122 0.0020 0.0040 0.484 0.0040 0.0010 0.485 0.0040 0.0020 0.96
Example #2:
Example #3:The initial rate of a reaction A + B C was measured with the results below. Write the rate law, the value of the rate constant (with units), and the rate of reaction when [A] = 0.050 M and [B] = 0.100 M.
Experiment [A] (M) [B] (M) Initial Rate (M/s)1 0.1 0.1 4.0x10-5
2 0.1 0.2 4.0x10-5
3 0.2 0.1 16.0x10-5
Example #3:
Potential Energy Diagrams Molecules need a minimum amount of energy for a
reaction to take place. Activation energy (Ea) – the minimum amount of energy
that the reacting species must possess to undergo a specific reaction
Activated complex - a short-lived molecule formed when reactants collide; it can return to reactants or form products. Formation depends on the activation energy & the correct
geometry (orientation)
Potential Energy Diagrams
Potential Energy Diagrams
Potential Energy Diagrams
Catalyzed Pathway
Catalysts lower activation energy!!!
Reaction MechanismsMechanism – how reactants are converted to products at the molecular level
Most reactions DO NOT occur in a single step! They occur as a series of elementary steps
(a single step in a reaction).
Rate Determining StepRate determining step –
the slowest step in a reaction
COCl2 (g) COCl (g) + Cl (g) fast
Cl (g) + COCl2 (g) COCl (g) + Cl2 (g) slow
2 COCl (g) 2 CO (g) + 2 Cl (g) fast 2 Cl (g) Cl2 (g) fast
Rate Determining Step COCl2 (g) COCl (g) + Cl (g) fast
Cl (g) + COCl2 (g) COCl (g) + Cl2 (g) slow
2 COCl (g) 2 CO (g) + 2 Cl (g) fast 2 Cl (g) Cl2 (g) fast
2 COCl2 (g) 2 Cl2 (g) + 2 CO (g)
Adding elementary steps gives the net (or overall) reaction!
Intermediates Intermediates are produced in one elementary
step but reacted in another
NO (g) + O3 (g) NO2 (g) + O2 (g)
NO2 (g) + O (g) NO (g) + O2 (g)
O3 (g) + O (g) 2 O2 (g)
Catalysts Catalyst – a reactant in an elementary step but
unchanged at the end of the reactionA substance that speeds up the reaction but is not
permanently changed by the reactionBoth an original reactant and a final product
NO (g) + O3 (g) NO2 (g) + O2 (g)
NO2 (g) + O (g) NO (g) + O2 (g)
O3 (g) + O (g) 2 O2 (g)
Example #4
Cl2 (g) 2 Cl (g) Fast
Cl (g) + CHCl3 (g) CCl3 (g) + HCl (g) Slow
CCl3 (g) + Cl (g) CCl4 (g) Fast
Identify: The rate determining step The overall (net) reaction The identity of any intermediates The identity of any catalysts
Example #5
H2O2(aq) + I1-(aq) H2O(l) + IO1-(aq) Slow
H2O2(aq) + IO1-(aq) H2O(l) + O2(g) + I1- (aq) Fast
Identify: The rate determining step
The overall (net) reaction
The identity of any intermediates
The identity of any catalysts
Example #6
O3 (g) + Cl (g) O2 (g) + ClO (g) Slow
ClO (g) + O (g) Cl (g) + O2 (g) Fast
Identify: The rate determining step The overall (net) reaction The identity of any intermediates The identity of any catalysts
Chemical Equilibrium
UNIT 11 (PART 2)
Pb2+ (aq) + 2 Cl- (aq) PbCl2 (s)
What is equilibrium?
Definition (dictionary.com): a state of rest or balance due to the equal action of opposing forces
Chemical Equilibrium: A process where a forward and reverse reaction occur at equal rates
***Not all chemical reactions are reversible!!!
General Characteristics of Equilibrium
1. DYNAMIC (in constant motion)
2. REVERSIBLE
3. Can be approached from either direction (reaction can run in the forward direction or the reverse direction)
1. DYNAMIC (in constant motion)
2. REVERSIBLE
3. Can be approached from either direction (reaction can run in the forward direction or the reverse direction)
Pink to blueCo(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O
Blue to pinkCo(H2O)4Cl2 + 2 H2O Co(H2O)6Cl2
Video Clip - Cobalt Chloride Complex Equilibrium
Characteristics of Dynamic Equilibrium
Characteristics of Dynamic Equilibrium
After a period of time, the concentrations of reactants and products are constant.
The forward and reverse reactions continue after equilibrium is attained.
+
Fe3+ + SCN- FeSCN2+
Examples of Chemical Equilibria
Phase changes such as:
H2O(s) H2O(liq)
Graphing Dynamic Equilibrium
Equilibrium achieved when
product and reactant
concentrations remain
constant!!
The Equilibrium Expression, Keq
Keq = equilibrium constant (for a given T)
Brackets "[ ]" = concentration (molarity)
"a, b, c, and d" = coefficients from balanced equation
The "c" in Kc = concentration
(Kc = a special Keq based on concentration)
dDcCbB aA
c d
c a b
C DK
A B
There are two cases when a species is not shown in the equilibrium expression:
#1:
SOLIDS – (s) after the formula
#2:
pure LIQUIDS – (l) after the formula
Example #7Write the equilibrium expression for the oxidation-reduction reaction occurring between iron(III) chloride and tin(II) chloride:
2 FeCl3 (aq) + SnCl2 (aq) 2 FeCl2 (aq) + SnCl4 (aq)
Example #8Write the equilibrium expression for the replacement of silver ions by copper:
Cu (s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag (s)
“If a system at equilibrium is stressed, the system tends to shift its equilibrium position to
counter the effect of the stress.”
Le Chatelier’s Principle
How does a “stress” influence equilibrium?
The impact of addition of reactants on reaction rate
The Seesaw Analogy
“Stresses” to a System
1) Changes in amount of species2) Changes in pressure or volume3) Changes in temperature4) Adding an inert substance5) Adding a catalyst
“Stresses” (factors) that can cause changes at equilibrium
1) Changes in amount of specieso Add reactant; system shifts to the RIGHT!
(produces more products)
o Add product; system shifts to the LEFT! (produces more reactants)
o Remove reactant; system shifts to LEFT!
o Remove product; system shifts to RIGHT!
Example #9: Predict the direction of shift based on the following concentration changes for the reaction:
CH4(g) + 2S2(g) CS2(g) + 2H2S(g)
(A) Some S2(g) is added.
(B) Some CS2(g) is added.
(C) Some H2S(g) is removed.
(D) Some argon gas (an inert gas) is added.
“Stresses” (factors) that can cause changes at equilibrium
2) Changes in pressure or volumeo If P goes down (same as V goes up), system shifts
to increased # of moles of gas
o If P goes up (same as V goes down), system shifts to decreased # of moles of gas
Example #10: Predict the effect of increasing pressure (decreasing volume) on each of the following reactions.
(a) CH4(g) + 2S2(g) CS2(g) + 2H2S(g)
(b) H2(g) + Br2(g) 2HBr(g)
(c) CO2(g) + C(s) 2CO(g)
(d) PCl5(g) PCl3(g) + Cl2(g)
“Stresses” (factors) that can cause changes at equilibrium
3) Changes in temperatureo Write heat as a product (exothermic) or reactant
(endothermic)
o System shifts to get rid of added heat:o System shifts LEFT for exo. reactions as T goes upo System shifts RIGHT for endo. reactions as T goes up
Example #11: Predict the effect of increasing temperature on each of the following reactions:
(a) CO (g) + 3 H2 (g) CH4 (g) + H2O (g) ΔH < 0
(b) CO2(g) + C (s) 2 CO (g) ΔH > 0
(c) 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g) EXO.
(d) 2 H2O (g) 2 H2(g) + O2(g) ENDO.
“Stresses” (factors) that can cause changes at equilibrium
4) Adding an inert substanceo If a substance is NOT in the reaction (or in the Keq
expression) any changes will have NO EFFECT on equilibrium!
Ex.) If CO2 is added to the system below, there would be no effect on the equilibrium.
H2 (g) + Br2 (g) 2 HBr (g)
“Stresses” (factors) that can cause changes at equilibrium
5) Adding a catalysto What do catalysts do? They increase the rate of the
reaction!
o Adding a catalyst will not affect equilibrium. It only changes the rate at which you reach equilibrium.
Example #12:How can the reaction below be shifted to the right? List all possibilities!
CO(g) + 2 H2 (g) <---> CH3OH (g)
Example #13:How can the reaction below be shifted to the right? This process (the Haber process) is used in the industry to produce ammonia. List all possibilities!
N2(g) + 3H2(g) 2NH3(g) ∆H = negative
N2(g) + 3 H2(g) 2 NH3(g) + heat
K = 3.5 x 108 at 298 K
Haber-Bosch Process for NH3
