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![Page 1: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/1.jpg)
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Chemical kinetics or dynamics3 lectures leading to one exam
question Texts: “Elements of Physical Chemistry”
Atkins & de Paula
Specialist text in Hardiman Library– “Reaction Kinetics” by Pilling & Seakins, 1995These notes available On NUI Galway web pages at
http://www.nuigalway.ie/chem/degrees.htm
What is kinetics all about?
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Academic? Ozone chemistry Ozone; natural formation ( 185240 nm)– O2 + h2O– O + O2 O3
Ozone; natural destruction ( 280320 nm) Thomas Midgely
– O3 + hO + O2 1922 TEL; 1930 CFCs
– O + O3 2O2
‘Man-made’ CCl2F2 + hCl + CClF2
– Cl + O3 ClO + O2
– ClO + O Cl + O2
– -----------------------------– Net result is: O + O3 2 O2
1995 Nobel for chemistry: Crutzen, Molina & Rowland 1996 CFCs phased out by Montreal protocol of 1987
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Chemical kinetics
Thermodynamics– Direction of
change Kinetics
– Rate of change– Key variable: time
What times?– 1018 s age of
universe– 10-15 s atomic nuclei– 108 to 10-14 s
Ideal theory of kinetics?– structure, energy– calculate fate
Now?– compute rates of
elementary reactions
– most rxns not elementary
– reduce observed rxn. to series of elementary rxns.
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Thermodynamics vs kinetics
Reaction Timescale K
H2O (aq) = H+ + OH– seconds 10–14
H+ + OH– = H2O (aq) microseconds 1014
2 H2 (g) + O2 (g) = 2 H2O (l) years 1041
4 Al (s) + 3 O2 (g) = 2 Al2O3 (s) decades 10277
Kinetics determines the rate at which change occurs
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Rate of reaction symbol: R, v, Stoichiometric equation
m A + n B = p X + q Y
Rate = (1/m) d[A]/dt = (1/n) d[B]/dt = (1/p) d[X]/dt = (1/q) d[Y]/dt– Units: (concentration/time)– in SI mol/m3/s, more practically mol dm–3 s–1
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Rate Law
How does the rate depend upon [ ]s? Find out by experiment
The Rate Law equation R = kn [A] [B] … (for many reactions)
– order, n = + + … (dimensionless)
– rate constant, kn (units depend on n)
– Rate = kn when each [conc] = unity
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Experimental rate laws?CO + Cl2 COCl2
Rate = k [CO][Cl2]1/2
– Order = 1.5 or one-and-a-half order
H2 + I2 2HI Rate = k [H2][I2]
– Order = 2 or second order
H2 + Br2 2HBr Rate = k [H2][Br2] / (1 + k’ {[HBr]/[Br2]} )
– Order = undefined or none
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Determining the Rate Law Integration
– Trial & error approach– Not suitable for multi-reactant systems– Most accurate
Initial rates– Best for multi-reactant reactions– Lower accuracy
Flooding or Isolation– Composite technique– Uses integration or initial rates methods
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Integration of rate laws
Order of reactionFor a reaction aA products, the rate law is:
nA
A
n
n
Akdt
Adr
akkdefining
Aakdt
Ad
Akdt
Ad
ar
][][
][][
][][1
rate of change in theconcentration of A
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First-order reaction
)()][]ln([
][
][
][
][
][][
00
][
][ 0
1
0
ttkAA
dtkA
Ad
dtkA
Ad
Akdt
Adr
At
A
A
t
A
A
A
t
![Page 11: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/11.jpg)
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First-order reaction
tkAA
ttkAA
At
At
0
00
]ln[]ln[
)(]ln[]ln[
A plot of ln[A] versus t gives a straightline of slope -kA if r = kA[A]1
![Page 12: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/12.jpg)
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First-order reaction
tkt
tkt
At
At
A
A
eAA
eA
A
tkA
A
ttkAA
0
0
0
00
][][
][
][
][
][ln
)(]ln[]ln[
![Page 13: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/13.jpg)
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A Passume that -(d[A]/dt) = k [A]1
0 5 10 151
2
3
4
5
6
7
8
[H2O
2] / m
ol d
m-3
Time / ms
![Page 14: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/14.jpg)
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Integrated rate equationln [A] = -k t + ln [A]0
0 5 10 15
0.2
0.4
0.6
0.8
1.0
ln [H 2
O2]
/ m
ol d
m-3
Time / ms
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Half life: first-order reaction
2/10
0
0
][
][21
ln
][
][ln
tkA
A
tkA
A
A
At
The time taken for [A] to drop to half its original value is called the reaction’s half-life, t1/2. Setting [A] = ½[A]0
and t = t1/2 in:
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Half life: first-order reaction
2/12/1
2/1
693.0693.0
693.02
1ln
tkor
kt
tk
AA
A
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When is a reaction over? [A] = [A]0 exp{-kt}
Technically [A]=0 only after infinite time
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Second-order reaction
tA
A
t
A
A
A
dtkA
Ad
dtkA
Ad
Akdt
Adr
][
][ 02
2
2
0][
][
][
][
][][
![Page 19: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/19.jpg)
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Second-order reaction
tkAA
ttkAA
At
At
0
00
][
1
][
1
)(][
1
][
1
A plot of 1/[A] versus t gives a straightline of slope kA if r = kA[A]2
![Page 20: 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.](https://reader030.fdocuments.us/reader030/viewer/2022032803/56649e295503460f94b16ed8/html5/thumbnails/20.jpg)
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Second order test: A + A P
2 4 6 8 1010
12
14
16
18
20
22
24
(1 / [A]0)
1 / [A
]
Time / ms
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Half-life: second-order reaction
2/10
2/10
2/10
0
][
1
][
1
][
1
][
2
][
1
][
1
tAk
ortkA
tkAA
tkAA
AA
Ao
At
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Initial Rate Method5 Br- + BrO3
- + 6 H+ 3 Br2 + 3 H2O General example: A + B +… P + Q + … Rate law: rate = k [A] [B] …??log R0 = log[A]0 + (log k+ log[B]0 +…) y = mx + c Do series of expts. in which all [B]0, etc are
constant and only [A]0 is varied; measure R0 Plot log R0 (Y-axis) versus log [A]0 (X-axis) Slope
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Example: R0 = k [NO][H2]
2 NO + 2 H2 N2 + 2 H2O Expt. [NO]0 [H2]0 R0
– 1 25 10 2.410-3
– 2 25 5 1.210-3
– 3 12.5 10 0.610-3
Deduce orders wrt NO and H2 and calculate k. Compare experiments #1 and #2 Compare experiments #1 and #3 Now, solve for k from k = R0 / ([NO][H2])
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How to measure initial rate? Key: - (d[A]/dt) -([A]/t) ([P]/dt)
A + B + … P + Q + … t=0 100 100 0 0 mol m3
s 99 99 1 1 ditto Rate? (100-99)/10 = -0.10 mol m3 s1
+(0-1)/10 = -0.10 mol m3 s1
Conclusion? Use product analysis for best accuracy.
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Isolation / floodingIO3
+ 8 I + 6 H 3 I3 + 3 H2O Rate = k [IO3
-] [I-] [H+]…– Add excess iodate to reaction mix– Hence [IO3
-] is effectively constant– Rate = k [I-] [H+]…– Add excess acid– Therefore [H+] is effectively constant
Rate k [I-]
Use integral or initial rate methods as desired
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Rate law for elementary reaction
Law of Mass Action applies:– rate of rxn product of active masses of
reactants
– “active mass” molar concentration raised to power of number of species
Examples:– A P + Q rate = k1 [A]1
– A + B C + D rate = k2 [A]1 [B]1
– 2A + B E + F + G rate = k3 [A]2 [B]1
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Molecularity of elementary reactions? Unimolecular (decay) A P
- (d[A]/dt) = k1 [A]
Bimolecular (collision) A + B P
- (d[A]/dt) = k2 [A] [B]
Termolecular (collision) A + B + C P
- (d[A]/dt) = k3 [A] [B] [C]
No other are feasible! Statistically highly unlikely.
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CO + Cl2 COCl2 Exptal rate law: - (d[CO]/dt) = k [CO] [Cl2]1/2
– Conclusion?: reaction does not proceed as written
– “Elementary” reactions; rxns. that proceed as written at the molecular level.
Cl2 Cl + Cl (1) Cl + CO COCl (2) COCl + Cl2 COCl2 + Cl (3) Cl + Cl Cl2 (4)
– Steps 1 thru 4 comprise the “mechanism” of the reaction.
decay collisional collisional collisional
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- (d[CO]/dt) = k2 [Cl] [CO]
If steps 2 & 3 are slow in comparison to 1 & 4
then, Cl2 ⇌2Cl or K = [Cl]2 / [Cl2]
So [Cl] = K × [Cl2]1/2
Hence:
- (d[CO] / dt) = k2 × K × [CO][Cl2]1/2
Predict that: observed k = k2 × K Therefore mechanism confirmed (?)
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H2 + I2 2 HI Predict: + (1/2) (d[HI]/dt) = k [H2] [I2] But if via:
– I22 I– I + I + H2 2 HI rate = k2 [I]2 [H2]– I + II2
Assume, as before, that 1 & 3 are fast cf. to 2Then: I2 ⇌2 I or K = [I]2 / [I2] Rate = k2 [I]2 [H2] = k2 K [I2] [H2] (identical)
Check? I2 + h 2 I (light of 578 nm)
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Problem In the decomposition of azomethane, A, at
a pressure of 21.8 kPa & a temperature of 576 K the following concentrations were recorded as a function of time, t:
Time, t /mins 0 30 60 90 120[A] / mmol dm3 8.706.524.893.672.75 Show that the reaction is 1st order in
azomethane & determine the rate constant at this temperature.
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Recognise that this is a rate law question dealing with the integral method.
- (d[A]/dt) = k [A]? = k [A]1
Re-arrange & integrate (bookwork) Test: ln [A] = - k t + ln [A]0
Complete table:Time, t /mins 0 30 60 90 120ln [A] 2.161.881.591.301.01 Plot ln [A] along y-axis; t along x-axis Is it linear? Yes. Conclusion follows.Calc. slope as: -0.00959 so k = + 9.610-3 min-
1
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More recent questions …
Write down the rate of rxn for the rxn:C3H8 + 5 O2 = 3 CO2 + 4 H2O
for both products & reactants [8 marks]For a 2nd order rxn the rate law can be written:
- (d[A]/dt) = k [A]2
What are the units of k ? [5 marks] Why is the elementary rxn NO2 + NO2 N2O4
referred to as a bimolecular rxn? [3 marks]
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Rate constant expression
RT
EAk aexp
RT
EE
k
k
RT
ERT
E
A
A
k
k
RT
EAk
aa
a
a
a
)(exp
)(
)(
exp
exp
21
2
1
2
1
2
1
11