ch17 hrw rxn kinetics
TRANSCRIPT
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Chapter 17 pp
Reaction Kinetics
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17.1 The Reaction Process Explain the concept of Explain the concept of reaction mechanismreaction mechanism.. Use Use collision theorycollision theory to interpret reactions. to interpret reactions. Define Define activated complexactivated complex.. RelateRelate activation energy to heat of reaction. activation energy to heat of reaction.
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Reaction Mechanisms The The series of stepsseries of steps that actually occur that actually occur
in a chemical reaction.in a chemical reaction. A balanced equation does not tell us A balanced equation does not tell us
howhow the reactants become products. the reactants become products. KineticsKinetics can tell us something about the can tell us something about the
mechanism.mechanism.
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A Molecular Representation of the Elementary Steps in the
Reaction of NO2 and CO
Overall: NO2 + CO NO + CO2
Step 1: NO2 + NO2 NO3 + NO (k1)
Step 2: NO3 + CO NO2 + CO2 (k2)
NO3 is an intermediate
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2NO2 + F2 2NO2F Rate = k[NO2][F2] The proposed mechanism is: NO2 + F2 NO2F + F (slow) F + NO2 NO2F (fast) F is called an intermediate. It is formed
then consumed in the reaction
Reaction Mechanisms
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Each of the two reactions is called an Each of the two reactions is called an elementary stepelementary step..
The rate for each reaction can be The rate for each reaction can be written from its written from its molecularitymolecularity..
Molecularity is the number of pieces Molecularity is the number of pieces that must come together (that must come together (collidecollide) to ) to produce the reaction indicated by that produce the reaction indicated by that step.step.
Reaction Mechanisms
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Finding the Reaction Mechanism Must satisfy two requirements: 1. Sum of the elementary steps does
give the overall balanced equation for the reaction
2. The proposed mechanism must agree with the experimentally derived rate law.
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N2 + 3H2 2NH3 As the reaction
progresses the concentration H2 goes down
Concentration
Time
[H[H22]]
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As N2 + 3H2 2NH3 progresses the
concentration N2 goes down 1/3 as fast
Concentration
Time
[H[H22]]
[N[N22]]
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As N2 + 3H2 2NH3 progresses the
concentration NH3 goes up.
Concentration
Time
[H[H22]]
[N[N22]]
[NH[NH33]]
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Definition of Rate.
2NO2 2NO + O2
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Collision Theory In order to react molecules and atoms
must touch each other.They must hit each other hard enough to
react.Anything that increases collision force will
make the reaction faster.They must also hit with the correct
orientation or the collision won’t be “successful” & won’t cause a reaction.
LD 4: 17.35 Collisions & Rxn mechanisms
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Tr96 Fig. 17-9 p. 533 Possible Collision Patterns
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Several Possible Orientations for a Collision Between Two BrNO Molecules. (a) & (b) can lead to
a collision, (c) cannot.
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ON
Br
ON
Br
ON
Br
ON
Br
O N Br ONBr ONBr
O NBr
O N BrONBr No Reaction
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Activation Energy Two things must happen for a reaction
to occur. Bonds must break within the reactants
(requires energy). Bonds must form to make products
(gives off energy). This process makes a temporary
compound that last while new bonds are breaking and forming.
This is an activated complex.
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Activation Energy Once an exothermic reaction gets
started, the energy from bonds forming to make products is enough to sustain the reaction.
But, initially need energy to start the exothermic reaction.
Activation Energy - the minimum energy needed to transform the reactants into an activated complex.
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Ene
rgy
Reaction coordinate
Reactants
Products
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Ene
rgy
Reaction coordinate
Reactants
Products
Activation Energy - Minimum energy to make the reaction happen
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Ene
rgy
Reaction coordinate
Reactants
Products
Activated Complex or Transition State
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Ene
rgy
Reaction coordinate
Reactants
Products
Overall energy change
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Tr95A Ae Differences in Exothermic & Endothermic Reactions
•How much bigger is Ea(reverse) than Ea(forward)? . . .
•Ea(reverse) = Ea(forward) + ∆Hrxn
•What’s value of ∆H reverse rxn?
•∆H reverse rxn = (+) 393 kJ
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Tr95A Ae Differences in Exothermic & Endothermic Reactions
•What are the products of the reverse reaction? . . .
•2K(s) + 2H2O(l)
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Detour for STAR Test review The following slides will review the
important things you need to know for the STAR test.
You can also look at the chapters on equilibrium, nuclear chemistry, and organic chemistry.
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Le Châtelier’s Principle If something is changed in a system at If something is changed in a system at
equilibrium, the system will respond to equilibrium, the system will respond to relieve the stress.relieve the stress.
Three types of stress are applied:Three types of stress are applied:
– PressurePressure
– ConcentrationConcentration
– TemperatureTemperature
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Changes in PressureChanges in Pressure As the pressure increases the reaction As the pressure increases the reaction
will shift in the direction of the least will shift in the direction of the least gases.gases.
At At highhigh pressure pressure2H2H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(g)O(g)
At At lowlow pressure pressure 2H 2H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(g)O(g)
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Changing ConcentrationN2(g) + 3H2(g) 2NH3(g)
If you add reactants (i.e., increase their concentration) the forward reaction will speed up.
More product will form. Equilibrium “Shifts to the right” Reactants products
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Changing ConcentrationChanging ConcentrationNN2(g)2(g) + 3H + 3H2(g)2(g) 2NH 2NH3(g)3(g)
If you If you removeremove reactantsreactants (or (or decreasedecrease their concentration).their concentration).
The The reversereverse reaction will speed up. reaction will speed up. More More reactantreactant will form. will form. Equilibrium “Equilibrium “Shifts to the Shifts to the leftleft”.”. Reactants Reactants products products Used to control how much yield you get Used to control how much yield you get
from a chemical reaction.from a chemical reaction.
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Changing ConcentrationN2(g) + 3H2(g) 2NH3(g)
If you If you add productsadd products ( (i.e.i.e., increase , increase theirtheir concentration).concentration).
The The reversereverse reaction will speed up. reaction will speed up. More More reactantreactant will form. will form. Equilibrium “Equilibrium “Shifts to the leftShifts to the left”” Reactants Reactants products products
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Changing ConcentrationN2(g) + 3H2(g) 2NH3(g)
If you If you removeremove products (or products (or decreasedecrease their concentration).their concentration).
The The forwardforward reaction will speed up. reaction will speed up. More product will form.More product will form. Equilibrium “Equilibrium “Shifts to the Shifts to the rightright”” Reactants Reactants products products
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Changing Temperature N2(g) + 3H2(g) 2NH3(g) + 92 kJ
Reactions either require or release heat.Reactions either require or release heat. ExothermicExothermic go faster at go faster at lowerlower
temperatures.temperatures. EndothermicEndothermic reactions go faster at reactions go faster at higherhigher
temperature.temperature. All All reversiblereversible reactions will be reactions will be
endothermicendothermic one way and one way and exothermicexothermic the the reverse way.reverse way.
LD 3: 18.6 Reversible cobalt complexLD 3: 18.6 Reversible cobalt complex
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Exothermic Review N2(g) + 3H2(g) 2NH3(g) + 92 kJ
H < 0 Releases heat Think of heat as a product in exo rxns Raising temperature pushes toward
reactants. Shifts to left.
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Endothermic Review2H2O(g) + 484 kJ 2H2(g) + O2(g)
H > 0 Requires heat Think of heat as a reactant Raising temperature push toward
products. Shifts to right.
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Gas Phase Equilibrium catalyst
N2(g) + 3 H2(g) 2 NH3(g) + heathigh pressure and temperature
Endothermic or exothermic? . . . Exothermic. Adding heat shifts which way? Left Cooling the system shifts which way? Right
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Le Chatelier Review pppp
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H = 2820 kJ
Some CO2(g) is added and causes . . .
Shift to right so get more C6H12O6(s) . Temperature is raised . . . Shift to right, more C6H12O6(s) forms.
Volume is decreased There are equal moles of gaseous
reactants and products, so no change.
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Le Chatelier Review pppp
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H = 2820 kJ
Some O2(g) is removed . . .
Shift to right, more C6H12O6(s) forms.
Some C6H12O6(s) is removed . . . No change (glucose is a solid so is not in the
equilibrium expression). No more C6H12O6(s) . A catalyst is added . . . No effect on equilibrium, only allows it be attained
more quickly. No more C6H12O6(s) .
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Le Chatelier Review pppp
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ∆H = 2820 kJ
Some H2O is removed . . . No change, water is a liquid. No additional
C6H12O6(s) forms.
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Radiation
• Radiation comes from the nucleus of an atom.• Unstable nucleus emits a particle or energy
alpha
beta
gamma
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He42
A. Types of A. Types of RadiationRadiation pppp
A. Types of A. Types of RadiationRadiation pppp
Alpha particle () helium nucleus paper2+
Beta particle (-) electron
e0-1 1-
leadPositron (+)
positron e01
1+
Gamma () high-energy photon 0
concrete
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Radiation Protection
• Shielding
alpha – paper, clothing
beta – lab coat, gloves
gamma- lead, thick concrete
• Limit time exposed
• Keep distance from source
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Radiation Protection
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42B. Nuclear DecayB. Nuclear Decay ppppB. Nuclear DecayB. Nuclear Decay pppp
Alpha Emission
He Th U 42
23490
23892
parentnuclide
daughternuclide
alphaparticle
Atomic & Mass Numbers must balance!!
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43B. Nuclear DecayB. Nuclear Decay ppppB. Nuclear DecayB. Nuclear Decay pppp
Beta Emission
e Xe I 0-1
13154
13153
electronPositron Emission
e Ar K 01
3818
3819
positron
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44B. Nuclear DecayB. Nuclear Decay ppppB. Nuclear DecayB. Nuclear Decay pppp
Electron Capture
Pd e Ag 10646
0-1
10647
electron Gamma Emission
Usually follows other types of decay.
Transmutation One element becomes another.
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Gamma radiation pp
No change in atomic or mass number
11B 11B + 0
5 5 0
boron atom in a
high-energy state
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Learning Check NR1 pp
Write the nuclear equation for the beta emitter Cobalt-60. . .
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Solution NR1 pp
Write the nuclear equation for the
Beta emitter Cobalt-60.
60Co 60Ni + 0 e
27 28 -1
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Producing Radioactive Isotopes
Bombardment of atoms produces radioisotopes
= 60 = 6059Co + 1n 56Mn + 4H e
27 0 25 2
= 27 = 27
cobalt neutron manganese alpha atom radioisotope particle
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Learning Check NR2
What radioactive isotope is produced in the following bombardment of boron?
10B + 4He ? + 1n
5 2 0
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Solution NR2
What radioactive isotope is produced in the following bombardment of boron?
10B + 4He 13N + 1n
5 2 7 0
nitrogen
radioisotope
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Addition Polymers pp
Polymers are giant molecules, not small like the ones studied earlier in this chapter
Examples are plastics Polymer - large molecule formed by
the covalent bonding of smaller molecules called monomers
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Addition Polymers
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17.2 Reaction Rate
Define chemical kinetics & explain the 2 conditions necessary for chemical reactions to occur.
Discuss the 5 factors influencing rate. Define catalyst and discuss 2 types. Explain & write rate laws for chemical
reactions.
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Kinetics The study of reaction rates. Spontaneous reactions are reactions
that will happen - but we can’t tell how fast.
Diamond will spontaneously turn to graphite – eventually.
Reaction mechanism - the steps by which a reaction takes place.
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Things that Affect Rate#1 Nature of Reactants
Different substances have different reactivities.
Hydrogen reacts vigorously with chlorine, but only feebly with nitrogen.
The rate depends on the particular reactants and bonds involved.
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Things that Affect Rate#2 Surface Area
Particle size determines surface area.Particle size determines surface area. Molecules can only collide Molecules can only collide at the surfaceat the surface.. SmallerSmaller particles have particles have biggerbigger surface area. surface area. So, smaller particles have So, smaller particles have fasterfaster reaction. reaction. Smallest possible is molecules or ions.Smallest possible is molecules or ions. Dissolving speeds up reactions since the Dissolving speeds up reactions since the
particle size gets smaller.particle size gets smaller. Getting two solids to react with each other is Getting two solids to react with each other is
slow (big particle size).slow (big particle size). LD 4:17.42 LycopodiumLD 4:17.42 Lycopodium
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Things that Affect Rate#3 Temperature
Higher temperature - faster particles. Greater kinetic energy. More and stronger collisions. So, get faster Reactions. Decreasing temperature has opposite effect. Rule of thumb: Beginning at room
temperature (25 ºC), rate doubles with each 10 ºC rise in temperature.
Rules of thumb always have exceptions. LD 4: 17.45 Bleaching dye hot/cold water.
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Things that Affect Rate#4 Concentration
More concentrated means the molecules are closer together.
So, they collide more often.Get a faster reaction.LD 4:46 Mg + HCl, different
concentrations.
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Tr98 Fig. 17-13 p. 540 Concentration Affects Rate
•How many collisions can occur between 6 “A” particles & 4 “B” particles? . . .
•24 collisions.
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Things that Affect Rate#5 Catalysts
Substances that speed up a reaction Substances that speed up a reaction without without being used upbeing used up ( (e.g., e.g., enzyme).enzyme).
LD 4: 17.50 HLD 4: 17.50 H22OO2 2 catalytic decomposition.catalytic decomposition. Speeds up reaction by giving the reaction a Speeds up reaction by giving the reaction a new new
pathpath.. New path has a New path has a lowerlower activation energy. activation energy. Since more molecules have this energy the Since more molecules have this energy the
reaction goes faster.reaction goes faster. InhibitorInhibitor - a substance that blocks a catalyst. - a substance that blocks a catalyst.
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Ene
rgy
Reaction coordinate
Reactants
Products
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Catalysts Speed up a reaction without being used
up in the reaction. Enzymes are biological catalysts. Homogenous Catalysts are in the same
phase as the reactants. Heterogeneous Catalysts are in a
different phase as the reactants.
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Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction.
∆E is the same
in both cases!
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Tr97A Fig. 17-15 p. 541 Comparing Pathways for Decomposing H2O2 by Various Catalysts
•Why is H2O2 stored in brown bottles to protect it from UV light?
•UV light has enough energy to catalyze the reaction.
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Tr97A Fig. 17-15 p. 541 Comparing Pathways for Decomposing H2O2 by Various Catalysts
•How many activated complexes are created with the heterogenous catalyst, MnO2? . . .
•There are 3 activated complexes.
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Pt surface
HH
HH
HH
HH
Hydrogen bonds to surface of metal.
Break H-H bonds
Catalysts
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Pt surface
HH
HH
Catalysts
C HH C
HH
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Pt surface
HH
HH
Catalysts
C HH C
HH
The double bond breaks and bonds to the catalyst.
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Pt surface
HH
HH
Catalysts
C HH C
HH
The hydrogen atoms bond with the carbon
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Pt surface
H
Catalysts
C HH C
HH
H HH
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Heterogenous Catalysts Summary: usually in 4 steps: Adsorption & activation of reactants
onto the catalytic surface. Migration of the adsorbed reactants
across the surface. Reaction of the adsorbed reactants. Escape (desorption) of the products.
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Homogenous Catalysts Chlorofluorocarbons catalyze the
decomposition of ozone. Both are in same phase (gas) Another example: Enzymes regulating
the body processes. (Protein catalysts)
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Catalysts and rate Catalysts will speed up a reaction but
only to a certain point. Past a certain point adding more
reactants won’t change the rate. Becomes Zero Order (saturation
kinetics) See saturation kinetics with renal
disease & some drugs (aminoglycosides).
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Catalysts and rate.
Concentration of reactants
Rate
Rate increases until the active sites of catalyst are filled.
Then rate is independent of concentration
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Rate Laws: An Introduction
Reactions are reversible. As products accumulate they can begin
to turn back into reactants. Early on the rate will depend on only the
amount of reactants present. So, we want to measure the reactants
as soon as they are mixed. This is called the Initial rate method.
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Rate LawsRate Laws Two key points The concentration of the products do
not appear in the rate law because this is an initial rate (only the [reactants]).
The order must be determined experimentally.
Can’t be obtained from the equation’s coefficients (unless it is single step - more later on this).
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Rate Laws N2 + 3H2 2NH3
When forward & reverse reaction rates are equal then there is no change in concentration of either.
It’s at equilibrium. Rate = k[N2]n[H2]m
Note: “m” and “n” are not coefficients of the balanced equation (unless single step), must be determined experimentally.
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You will find that the rate will only depend on the concentration of the reactants.
Rate = k[NO2]n
This is called a rate law expression. k is called the rate constant. n is the order of the reactant - usually a
positive integer (1st, 2nd, etc.).
2 NO2 2 NO + O2
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Definition of Rate.
2NO2 2NO + O2
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Determining Rate Laws The first step is to determine the form of
the rate law (especially its order). Must be determined from experimental
data (not from the coefficients!!). You’ll do this type of lab in AP Chem.
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The method of Initial Rates This method requires that a reaction be
run several times (do in AP Chem). The initial concentrations of the
reactants are varied. The reaction rate is measured just after
the reactants are mixed. Eliminates the effect of the reverse
reaction and makes easier to calculate.
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Rate Law Example pp
Write the general form of rate law for:BrO3
- + 5 Br- + 6H+ 3Br2 + 3 H2O
Answer is . . . Rate = k[BrO3
-]n[Br-]m[H+]p
Determine n, m, p by comparing rates Add n, m, p values to get overall order
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Example Continued pp
So, for the reactionSo, for the reaction
BrOBrO33-- + 5 Br + 5 Br-- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22O The O The
general form of the Rate Law is general form of the Rate Law is Rate = Rate =
k[BrOk[BrO33--]]nn[Br[Br--]]mm[H[H++]]pp
Be able to write this as the the answer to Be able to write this as the the answer to a question, given the above reaction).a question, given the above reaction).
We use We use experimental dataexperimental data to determine to determine the values of n,m, and pthe values of n,m, and p
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Initial concentrations (M) pp
Rate (M/s)
BrOBrO33-- BrBr-- HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
Now we have to see how the rate changes with concentration
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Initial concentrations (M) pp
Rate (M/s)
BrOBrO33-- BrBr-- HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
Find n by ratio of Rate 2/Rate 1, in which only [BrO3
-] changes
n = 1 because (next slide)
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The math pp
Rate 2 = 1.6 x 10-3 mol/Ls = k(0.20 mol/L)n(0.10 mol/l)m(0.10 mol/L)p Rate 1 8.0 x 10-4 mol/Ls k(0.10 mol/L)n(0.10 mol/l)m(0.10 mol/L)p
= 2.0 = (0.20 mol/L / 0.10 mol/L)n = (2.0)n So, n = 1 Easy way: when the concentration was doubled
the rate was doubled so n = 1. This is the “take-home” message for you.
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Initial concentrations (M) pp
Rate (M/s)
BrOBrO33-- BrBr-- HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
Find m by ratio of Rate 3/Rate 2, in which only [Br-] changes
m = 1 (ditto on math)
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Initial concentrations (M) pp
Rate (M/s)
BrOBrO33-- BrBr-- HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
Find p by ratio of Rate 4/Rate 1, in which only [H+] changes
p = 2 (next slide)
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Initial concentrations (M) pp
Rate (M/s)
BrOBrO33-- BrBr-- HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
When double [H1+] the rate quadruples so p = 2
Double : quadruple = 2 : 4 = 2 : 22
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Example continued pp
So, rate of reaction is first order in BrO3-
and Br- and second order in H+
Overall rate = 1 + 1 + 2 = 4 Rate law can be written as:
Rate = k[BrO3-][Br-][H+]2
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Rate Laws & Reaction Pathways A chemical reaction usually takes place in A chemical reaction usually takes place in
several stepsseveral steps as the molecules collide. as the molecules collide. Elementary reactionElementary reaction - a reaction that happens - a reaction that happens
in a single step in a single step Reaction mechanismReaction mechanism is a description of how is a description of how
the chemical reaction really happens.the chemical reaction really happens. It is a It is a series of elementary reactionsseries of elementary reactions.. The The productproduct of an elementary reaction is an of an elementary reaction is an
intermediateintermediate.. An intermediate is a product that An intermediate is a product that immediately immediately
gets usedgets used in the next reaction. in the next reaction.
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Reaction Mechanisms The series of steps The series of steps that actually occurthat actually occur
in a chemical reaction.in a chemical reaction. A balanced equation does not tell us A balanced equation does not tell us
howhow the reactants become products. the reactants become products. Kinetics can tell us something about the Kinetics can tell us something about the
mechanism.mechanism.
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A Molecular Representation of the Elementary Steps in the
Reaction of NO2 and CO
Overall: NO2 + CO NO + CO2
Step 1: NO2 + NO2 NO3 + NO (k1)
Step 2: NO3 + CO NO2 + CO2 (k2)
NO3 is an intermediate
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2NO2NO22 + F + F22 2NO2NO22F F The The proposedproposed mechanism is: mechanism is: NONO22 + F + F22 NO NO22F + F + FF (slow)(slow) FF + NO + NO22 NO NO22FF (fast) (fast) F is called an F is called an intermediate.intermediate. It is formed It is formed
thenthen consumed in the reaction. consumed in the reaction. Both steps are “elementary” steps.Both steps are “elementary” steps.
Reaction Mechanisms pp
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Each of the two reactions in the last Each of the two reactions in the last slide is called an slide is called an elementary stepelementary step..
The rate for each reaction can be The rate for each reaction can be written from its written from its molecularitymolecularity..
Molecularity is the number of pieces Molecularity is the number of pieces that must come together (collide) to that must come together (collide) to produce the reaction indicated by that produce the reaction indicated by that step.step.
Reaction Mechanisms
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2NO2NO22 + F + F22 2NO2NO22F F The The
proposedproposed mechanism is: mechanism is: NO NO22 + F + F22
NO NO22F + F F + F (slow) F + NO(slow) F + NO22
NO NO22FF (fast) F is called (fast) F is called
an an intermediate.intermediate. It is formed It is formed thenthen consumed in the reaction.consumed in the reaction. Both Both steps are “elementary” steps.steps are “elementary” steps.
Since the first one is Since the first one is slow & elementaryslow & elementary it is rate-determining.it is rate-determining.
So, Rate = k[So, Rate = k[NONO22][][FF22]]
Reaction Mechanisms
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UnimolecularUnimolecular step involves one molecule - step involves one molecule - Rate is Rate is firstfirst order. order.
BimolecularBimolecular step - requires two molecules - step - requires two molecules - Rate is Rate is secondsecond order order
TermolecularTermolecular step - requires three molecules - step - requires three molecules - Rate is Rate is thirdthird order order
TermolecularTermolecular steps are almost never heard of steps are almost never heard of because the chances of three molecules because the chances of three molecules coming into contact at the same time are coming into contact at the same time are miniscule.miniscule.
With molecularity the coefficients With molecularity the coefficients cancan become become the exponents.the exponents.
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A A productsproducts A+A productsA+A products 2A 2A products products A+B productsA+B products A+A+B ProductsA+A+B Products 2A+B Products2A+B Products A+B+C ProductsA+B+C Products
Rate = k[A] Rate = k[A] Rate= k[A]Rate= k[A]22
Rate= k[A]Rate= k[A]22
Rate= k[A][B]Rate= k[A][B] Rate= k[A]Rate= k[A]22[B][B] Rate= k[A]Rate= k[A]22[B][B] Rate= k[A][B][C]Rate= k[A][B][C]
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101
Finding the Reaction Mechanism Must satisfy Must satisfy twotwo requirements: requirements: Sum of the Sum of the elementaryelementary steps does give steps does give
the the overalloverall balanced equation for the balanced equation for the reactionreaction
The proposed The proposed mechanismmechanism must must agreeagree with the with the experimentally derivedexperimentally derived rate law. rate law.
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Review Problems pp A reaction of X & Y occurs by a 1-step A reaction of X & Y occurs by a 1-step
mechanism: mechanism: X + 2Y X + 2Y XYXY22 Write the rate law . . .Write the rate law . . . Rate = k[X][Y]Rate = k[X][Y]22 What happens to rate if double [X]? . . .What happens to rate if double [X]? . . . Rate is doubled (1st order in X)Rate is doubled (1st order in X) If double [Y]? . . .If double [Y]? . . . Rate is quadrupled (2nd order in Y)Rate is quadrupled (2nd order in Y)
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103
Review Problems pp A reaction of X & Y occurs by a 1-step A reaction of X & Y occurs by a 1-step
mechanism: mechanism: X + 2Y X + 2Y XYXY22
Rate = k[X][Y]Rate = k[X][Y]22 What happens to the rate if What happens to the rate if
only use 1/3 of Y? . . .only use 1/3 of Y? . . . Rate is 1/9 the original because . . .Rate is 1/9 the original because . . . 22nd order in Y so (1/3)nd order in Y so (1/3)22 = 1/9 = 1/9
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Review Problems pp The rate of X reacting with Y doubles when The rate of X reacting with Y doubles when
[X] is doubled, but increases 8 times when [X] is doubled, but increases 8 times when [Y] is doubled. Write the rate law . . .[Y] is doubled. Write the rate law . . .
Rate = k[X][Y]Rate = k[X][Y]33 because . . . because . . . Double:double = 2:2Double:double = 2:21 1 so “n” = so “n” = 11, 1st order, 1st order Double:8 times = 2:2Double:8 times = 2:233 so “m”= so “m”= 33, 3rd order., 3rd order.
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Review Problems pp The rate law for a single-step reaction
that forms only one product, C, is Rate = k[A]2[B] Write the balanced equation . . .
2A + B C
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This reaction takes place in three steps
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Ea
First step is fast
Low activation energy
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Second step is slowHigh activation energy
Ea
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Ea
Third step is fastLow activation energy
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Second step is rate determining
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Intermediates are present
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Activated Complexes or Transition States
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Mechanisms and rates There is an activation energy for There is an activation energy for eacheach
elementary step.elementary step. Slowest stepSlowest step (rate determining) must (rate determining) must
have the have the highesthighest activation energy. activation energy.
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Nuclear Review for Star Test
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He42
Types of RadiationTypes of RadiationTypes of RadiationTypes of Radiation
Alpha particle () helium nucleus paper2+
Beta particle (-) electron
e0-1 1-
leadPositron (+)
positron e01
1+
Gamma () high-energy photon 0
concrete
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116 Nuclear DecayNuclear DecayNuclear DecayNuclear Decay
Alpha Emission
He Th U 42
23490
23892
parentnuclide
daughternuclide
alphaparticle
Atomic & Mass Numbers must balance!!
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117 Nuclear DecayNuclear DecayNuclear DecayNuclear Decay
Beta Emission
e Xe I 0-1
13154
13153
electronPositron Emission
e Ar K 01
3818
3819
positron
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118 Nuclear DecayNuclear DecayNuclear DecayNuclear Decay
Electron Capture (of inner orbital electrons)
Pd e Ag 10646
0-1
10647
electron Gamma Emission
Usually follows other types of decay.
Transmutation One element becomes another.
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119
Learning Check
Write the nuclear equation for the beta emitter Cobalt-60. . .
60Co 60Ni + 0 e 27 28 -1
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Learning Check NR2
What radioactive isotope is produced in the following bombardment of boron?
10B + 4He ? + 1n
5 2 0
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