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Introduction to Kinetics

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Thermodynamics tells us if a reaction can proceed and gives equilibrium concentration. ... Where Mo is the molecular number density at STP. ... – PowerPoint PPT presentation

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Title: Introduction to Kinetics


1
Introduction to Kinetics
  • AOSC 620
  • Fall 2009
  • R. Dickerson
  • See Finlayson-Pitts Chapter 5
  • Seinfeld and Pandis Chapters 4 9

2
Kinetics A. Rates, rate constants, and reaction
order. Thermodynamics tells us if a reaction
can proceed and gives equilibrium concentration.
Kinetics tells us how fast reactions proceed. If
thermodynamics alone controlled the atmosphere it
would be dissolved in the oceans as nitrates - we
would be warm puddles of carbonated water. 1.
First Order Reactions A ? PRODUCTS
EXAMPLES 222Rn -gt 218Po ? N2O5 ? NO2 NO3
3
The red line describes first order loss with a
rate constant of 1 min-1 The blue line is the
rate of formation of the product.
minutes
4
Radon is important source of indoor air
pollution, and N2O5 is nitric acid anhydride,
important in air pollution nighttime chemistry.
The rate equations take the form dProd./dt
kA -dReact./dt For example dPo/dt
kRn Rn -dRn/dt Where k is the first
order rate constant and k has units of time-1
such as s-1, min-1, yr-1. We usually express
concentration, Rn, in molecules cm-3 and k in
s-1.
5
Another example, nitric acid anhydride. dNO2
/dt k N2O5 and dN2O5/dt -k N2O5
Integrating 1/N2O5 dN2O5 -k dt ln (
N2O5 t / N2O5 o) -k ?t If we define
the starting time as zero N2O5t / N2O5o
exp(-kt)
6
The rate constants for these first order
reactions are kRn 0.182 days-1 kN2O5
0.26 s-1 (at room temperature)
7
2. Second Order Reactions A B ? PRODUCTS
EXAMPLES NO O3 ? NO2 O2 HCl OH ? H2O
Cl Examples of the rate equations are as
follows dNO/dt -kNOO3 dCl/dt
kOHHCl Units of k are conc-1 time-1.
1/(molecules/cm3) (s-1) cm3 s-1
8
For our second order kinetics examples Units of
k are conc-1 time-1. 1/(molecules/cm3) (s-1)
cm3 s-1 Rate constants have the following
values kNO-O3 1.8x10-14 cm3 s-1 kHCl-OH
8.0x10-13 cm3 s-1
9
3. Third Order Reactions A B C ?
PRODUCTS dA/dt -kABC Examples 2NO
O2 ? 2NO2 O O2 M ? O3 M
10
2NO O2 ? 2NO2 O O2 M ? O3 M M is any
third body (usually N2) needed to dissipate
excess energy. From the ideal gas law and Avg's
number M Mo P/Po To /T 2.69x10-19
(P/1) (273/T) MoP Where Mo is the molecular
number density at STP. Third order rate constants
have units of conc-2 time-1. These are usually
(cm-3)-2 s-1. kNO-O2 2.0 x 10-38 cm6 s-1
kO-O2 4.8 x 10-33 cm6 s-1
11
USEFUL IDEA For the following reaction A B
? C D dC/dt kf AB - kr CD At
steady state dC/dt 0, by definition. Thus
kf / kr CD/AB Keq
12
Halflife and Lifetime Definition Halflife, t1/2
the time such that At1/2 / A0 1/2
Definition of lifetime or residence time, ? ,
comes from kinetics, where k is the first order
rate constant with units of time-1. We know
that At/A0 exp(-k t) The lifetime, ?, is
when t 1/k so ? ? 1/k We can link half-life
and lifetime t1/2 ln(2)/k ? 0.69/k
13
For radon 222 (222Rn) the lifetime is 5.5 days,
but the half-life is only 3.8 days. For second
order reactions we need pseudo first order
conditions ? (kA bar)-1 For example NO
O3 ? NO2 O2 k 1.8 x 10-14 cm3 s-1
ASSUME O3 gtgt NO and dO3/dt 0.0 LET
O3 bar 50 ppb (a reasonable value for air
near the surface). ?NO 1/ kO3 bar
1/1.8 x 10-14 x 50x10-9 x 2.5x1019 44 s
CONCLUSION any NO injected into such an
atmosphere (by a car for example) will quickly
turn into NO2 , if there are no other reactions
that play a role. We will call kO3 the pseudo
first order rate constant. For third order
reactions we must assume that two components are
constant.
14
? (kA barB bar)-1 For example O O2
M ? O3 M k 4.8x10-33 cm6s-1 ASSUME
dO2/dt dM/dt 0.0 We know that O2
0.21 and that M O2 N2 1.00. At RTP
P02 0.21 atm and PM 1.0 atm. Therefore the
lifetime of O atoms is ? 4.8x 10-33 x 0.21 x
1.0 x (2.5x1019) 2-1 1.6x10-6 s
VERY SHORT!
15
Example 2 Same reaction at stratospheric
temperature and pressure. P30km P0
exp(-30/7) 0.014 atm ? 4.8x10-33 x 0.21
x 1.0 x (0.014 x 2.5x1019) 2-1 2.1x10-3 s
The lifetime of an O atom is is still short, but
it is a thousand times longer than in the
troposphere! The pressure dependence has a major
impact on the formation and destruction of
tropospheric and stratospheric ozone.
16
If two of the reactants in a third order reaction
are the same, we can derive a useful expression
for the rate of loss of the reactant. A A B
? PROD For a great excess of B dA/dt
-(2kB)A 2 A -2 dA -(2kB)dt Integrat
ing from 0 to time t A-2 dA
-(2kB)dt -At-1 A0-1 -(2kB)t
At-1 2kBt A0-1 Now we can calculate
the concentration at any time t in terms of the
initial concentration and the rate constant k.
17
Transport and Evolution of a Pollution Plume from
Northern ChinaA Satellite-based Case Study
  • Can Li1, Nick Krotkov2, 3, Russ Dickerson1,
    Zhanqing Li1, 4, and Mian Chin21AOSC, UMD 2GSFC,
    NASA 3GEST, UMBC 4ESSIC, UMD

Why China? A lot of emissions Local, regional,
and global effects Why transport and evolution?
Key factors determining the large-scale impact of
air pollution (e.g., conversion from SO2 to
sulfate aerosols, adding CCN to the system) Why
satellites? Transport episodes associated with
synoptic weather system, and are of regional
scale new satellite sensors provide great
spatial coverage, daily observation, and sensors
of different strengths can be combined
18
On April 5,2005 on the aircraft Lots of
SO2 Lots of dust
1 hr later and from space Lots of SO2 Lots of
dust
19
Satellite Snapshots
Trajectory model projects movement of the plume,
satellite sensors take snapshots every day
AMF Correction to operational product,
combination of satellites and models, and MODIS
AOD data
Trajectory Projection
SO2 lifetime 1-4 day SO2 to sulfate 0.1-0.2
increase in AOD near plume core in one day
Uncertainties? Details? Suggestion? Please come
to see our poster.
20
The comparison demonstrates that operational OMI
algorithm can distinguish between heavy pollution
( April 5 ahead of cold front )
SO2
OMI SO2
AQUA- MODIS RGB
21
and background SO2 conditions (on April 7 behind
cold front)
SO2
OMI SO2
MODIS RGB
22
SO2 lifetime estimate
By following the path of the air mass, and
correcting for the altitude effect, the amount of
SO2 lost to dry deposition and chemical reactions
was approximated. Ln (C/C0) -kt
The mass decreases from the 5th to the 7th looks
to be first order with a lifetime of 4 d. The
mass on April 8 shows no change from April 7
because the air mass was very disperse (large
box) and mixed with other plumes.
23
SO2 OH ? prod (H2SO4) If attack by OH is the
only sink for SO2, and if the lifetime is 4 d
3x105 s. Lifetime t (OHMk)-1 k 2x10-12
cm-3 s-1 Molecular number density of OH 1.4x106
cm-3
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