Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
TP For CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
whatistheorderin‒OCH3?
25%
25%
25%
25%
1.
2.
3.
4.
1
2
Neitheroftheabove
Moreinfoneeded
Monday, May 1, 2017
‒OCH
3
Rate M/s
0.0001
0.0001
2.8x10‒5
0.0002
0.0001
5.6x10‒5
0.0001
0.0002
2.8x10‒5
CH3 3CBr
Response
Counter
5
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9:05 am)
• Rate versus … fromexperiment
• Makingsenseofrateversus … :Reactionmechanism
• Makingsenseofrateconstants:theArrheniusrelation
Finallecture:Rateversustemperature:Ea andA.Puttingitalltogether:
Firstlaw,secondlaw,equilibrium,andkinetics.
1
Copyright © 2017 Dan Dill [email protected]
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
Differential rate laws: rate vs […]
Rateversus … fromexperiment
Inliquidmethanol,CH3OH,thefollowingreactionoccurs:
CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
Weknowhowtodefinerate.
Wealsoknowthatrate
Writearatelawforthisreaction.
concentration
So,weneedtolearnjusthowratedependsonconcentrationandwhat
determinestherateconstant, .
We’llusedatatolearnhowratedependsonconcentration,theso‐called
differentialratelaw.
We’llseethateachreactionwillhaveitownparticulardependence
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Copyright © 2017 Dan Dill [email protected]
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1
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
(CH3)3CBr + ‒OCH3  (CH3)3COCH3 + Br‒
Differential rate laws: rate vs […]
Inliquidmethanol,CH3OH,thefollowingreactionoccurs:
‒OCH
3
Rate M/s
0.0001
0.0001
2.8x10‒5
0.0002
0.0001
5.6x10‒5
0.0001
0.0002
2.8x10‒5
CH3 3CBr
CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
Howcanweknowwhethertheratelawiscorrect?
ratefor CH3 3CBr a,ratefor ‒OCH3
1.
2.
3.
4.
b
Whatistheorder, ,in CH3 3CBr?
Whatistheorder, ,in‒OCH3?
Whatisthefulldifferentialratelaw?
Whatisthevalueofkfor?
10
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
TP For CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
whatistheorderin CH3 3CBr?
25%
25%
25%
25%
1.
2.
3.
4.
1
2
Neitheroftheabove
Moreinfoneeded
Response
Counter
Copyright © 2017 Dan Dill [email protected]
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
TP For CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
whatistheorderin‒OCH3?
‒OCH
3
Rate M/s
0.0001
0.0001
2.8x10‒5
0.0002
0.0001
5.6x10‒5
0.0001
0.0002
2.8x10‒5
CH3 3CBr
11
5
12
25%
25%
25%
25%
1.
2.
3.
4.
1
2
Neitheroftheabove
Moreinfoneeded
Response
Counter
CH3 3CBr
‒OCH
3
Rate M/s
0.0001
0.0001
2.8x10‒5
0.0002
0.0001
5.6x10‒5
0.0001
0.0002
2.8x10‒5
5
13
2
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
TP For CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
whatisthefulldifferentialratelaw?
25%
25%
25%
25%
1.
2.
3.
4.
rate k CH3 3CBr ‒OCH3
rate k CH3 3CBr
rate k ‒OCH3
Neitheroftheabove
Response
Counter
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Rate M/s
0.0001
2.8x10‒5
0.0002
0.0001
5.6x10‒5
0.0001
0.0002
2.8x10‒5
0.0001
Copyright © 2017 Dan Dill [email protected]
TP For CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
whatisthevalueofk?
‒OCH
3
CH3 3CBr
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5
14
Copyright © 2017 Dan Dill [email protected]
25%
25%
25%
25%
1.
2.
3.
4.
2.8x10‒5Ms‒1
2.8x10‒1s‒1
2.8x10‒1Ms‒1
Noneoftheabove
CH3 3CBr
‒OCH
3
Rate M/s
0.0001
0.0001
2.8x10‒5
0.0002
0.0001
5.6x10‒5
0.0001
0.0002
2.8x10‒5
Response
Counter
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5
15
Copyright © 2017 Dan Dill [email protected]
Making sense of rate versus […]
Makingsenseofrateversus … :Reactionmechanism
Fromdata,wehavefoundthatthedifferentialratelawfor
CH3 3CBr ‒OCH3  CH3 3COCH3 Br‒
israte k CH3 3CBr
Shouldn’ttheratedependon ‒OCH3 too?
16
Copyright © 2017 Dan Dill [email protected]
17
3
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
(CH3)3CBr + ‒OCH3  (CH3)3COCH3 + Br‒
Apossiblemechanismthataccountsforratenotdepending on
1 CH3 3CBr CH3 3C
2 CH3 3C
‒OCH
3
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
Making sense of rate constants
is…
Themechanism
1 CH3 3CBr CH3 3C
Br‒ ,slow
‒OCH3  CH3 3COCH3 ,fast
2 CH3 3C
Showthatthisaccountsfortheobservedratelaw.
Br‒ ,slow
‒OCH3  CH3 3COCH3 ,fast
presumesreaction1ismuchslowerthanreaction2.
Let’sseeifwecanunderstandhowcouldthisbe.
19
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
20
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
Making sense of rate constants
Makingsenseofrateconstants:Arrheniusequation
Themechanism
1 CH3 3CBr CH3 3C
2 CH3 3C
‒OCH
3
Br‒ ,slow
 CH3 3COCH3 ,fast
Reaction1ismuchslowerthanreaction2because…
k1,for ismuchsmallerthank2,for
Whatdeterminestherelativesizeofrateconstants?
28
Copyright © 2017 Dan Dill [email protected]
29
4
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
Making sense of rate constants
NO2(g) + CO(g)  NO(g) + CO2(g)
Tounderstandwhatdeterminestherelativesizeofrateconstants,let’sbegin
byimagininghowenthalpychangesasanexothermicreaction,R P,suchas
Becausereactionsinvolvebothbondbreakinginreactantsandbondmakingin
products,energymustbesupplied forbothR PandP R.
NO2 g CO g  NO g CO2 g
evolvesasreactants,R,aretransformedintoproducts,P.
Sketchadiagramshowinghowenthalpychanges verticalaxis astimepasses
horizontalaxis ,startingwithallreactants,R,andendingwithallproducts,P.
30
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
31
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
NO2(g) + CO(g)  NO(g) + CO2(g)
NO2(g) + CO(g)  NO(g) + CO2(g)
WheredoestheenergycomefromforR P?
Kineticenergyofreactantmolecules.
Onlymoleculeswithkineticenergygreaterthantheactivationenergy,Ea,can
react.
32
Copyright © 2017 Dan Dill [email protected]
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Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
NO2(g) + CO(g)  NO(g) + CO2(g)
NO2(g) + CO(g)  NO(g) + CO2(g)
OnlymoleculeswithkineticenergygreaterthanEa canreact
RaisingtemperatureincreasesthenumberwithatleastkineticenergyEa
Onlymoleculeswithkineticenergygreaterthantheactivationenergy,Ea,can
react.
exp
Ea/ RT
…
Fraction ofmoleculeswithminimumkineticenergyEa forreaction always
lessthan1 34
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Copyright © 2017 Dan Dill [email protected]
35
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
NO2(g) + CO(g)  NO(g) + CO2(g)
NO2(g) + CO(g)  NO(g) + CO2(g)
Whatroledoestherelativeorientationofthecollidingmolecules
collisiongeometry play?
Collisionsmusthavethecorrectgeometry:A
37
Copyright © 2017 Dan Dill [email protected]
Copyright © 2017 Dan Dill [email protected]
38
6
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
Lecture 37 CH102 A1 (MWF 9 am) Spring 2017
5/2/2017 2:07 PM
Copyright © 2017 Dan Dill [email protected]
Putting geometry and temperature together
k A
exp
Ea/ RT
A …
frequencyofcollisions …
probabilitythatcollisionshaveanappropriatecollisiongeometry
exp
Ea/ RT …
FractionofmoleculeswithminimumkineticenergyEa forreaction always
lessthan1 39
Copyright © 2017 Dan Dill [email protected]
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