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FScN 8334
Topic 2
Reaction OrderDetermination
and reaction kinetics
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Topic 2 Or der Det er m in a t ion Page # 2
I. General definitions
A. general equation for equilibrium
kaA bB cC dD
k
KC D
A B
k
k
f
b
eq
c d
a b
f
b
+ +
= [ ] [ ][ ] [ ]
=
[X] = concentration of each species
a,b,c etc = stoichiometry # of species to give mass balance
kf = forward rate constant (units depend on stoichiometry)
kb = backward rate constant
Keq= equilibrium constant
B. Relationship to thermodynamics
G H T S RT K eq= = ln
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Topic 2 Or der Det er m in a t ion Page # 3
C. molecularity vs stoichiometry
molecularity - actual # of reacting species
stoichiometry - # to mass balance equation
molecularity stoichiometry
(1) A --------> C 1 A --->C
(2) A +A-->A + A*---->A + C 2 A ---->C
D. Water
In general neglect water as reactant as concentration is 55 M/L
and does not change with reaction extent.
? of low moisture systems
? of pH - if H+ or OH- catalyzed - especially as water content
changes
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Topic 2 Or der Det er m in a t ion Page # 4
II. Reaction order
A. Gener alized ra te equa tion
= [ ] [ ] =dAdt
k A Bn na b rate of gain or loss per unit time
= amount / time
k = rate constant
[A] = concentration
[B] = concentrationn = order with repect to A or Bx
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Topic 2 Or der Det er m in a t ion Page # 5
B. Order definitions
overall order = sum of exponents = a+b = n
dAdt
k A Ba b= [ ] [ ]
specific order - may not be the stoichiometric parameters,
rather it is the curve fitting parameters
GEORGE BOX :
ALL MODELS ARE WRONG BUT SOME ARE USEFUL
for A order = a
for B order b
for A -> C overall order = 1
for A + A -> A + C order = 2
order can be fractional - complex reaction
order can be zero -
- change in A negligible over time eg drug in
suspension
- not true stoichiometry
pseudo order - mechanism unknown - curve fitting
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Topic 2 Or der Det er m in a t ion Page # 6
C. Units convention for rate constant
mass balance
A B C + >
2 3
moles = 0
dA
dtk A B
dB
dtk A B k k
dC
dtk A B k k
dA
dt
dB
dt
dC
dt
A
B b A
C C A
= [ ][ ]
= [ ][ ] =
= [ ][ ] =
+
+
+ =
2
2
2
2
3
0
thus always look at how the rate constant was measured
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Topic 2 Or der Det er m in a t ion Page # 7
D. Reaction Rate for simple two component equilibrium reaction
1. reaction (example: mutarotation of reducing sugars in
solution)
k
A B
k
KB
A
k
k
f
b
eq
f
b
= [ ][ ]
=
2. boundary conditions for equilibrium reactions
at time = 0 A = Ao
at time = t A=Ao - x B = x
at time = teq
B = Xe A=Ao - Xe
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Topic 2 Or der Det er m in a t ion Page # 8
3. solution
+ = = [ ] [ ]
= [ ] [ ]
= =
=
= [ ] [ ] = [ ] [ ][ ]
= [ ]
dB
dt
dA
dt
k A k B
k A x k x
Kk
k
x
A xk k
A x
x
dx
dtk A x k x k A x
k
x A x x
k
x A x xx
f b
f o b
eq
f
b
e
o e
b fo e
e
f o b f o
f
e
o e
f
e
o e e [ ][ ]
= [ ]
k
x A x x
k
x A x x
f
e
o e
f
e
o e
ln lnA A
A A
B
B B
k
BA t k k t o e
t e
e
e
f
e
o f b
=
= = +[ ]
thus plot either the A or B function vs time .
Need to know A or B as function of time and Ao and either Ae or Be
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Topic 2 Or der Det er m in a t ion Page # 9
E. Complex reactions
A--->B---->C
at time = t =0 Ao=Aat time = t Ao = A + B + C (mass balance)
dA
dtk A A A e
dB
dtk A k B
B A A C A e C
dC
dtk B k A e k C
dC
dt
k C k A e
A o
k t
A B
o o
k t
B B o
k t
B
B B o
k t
A
A
A
A
= =
=
= = [ ]
= = [ ]
+ = [ ]
1
1
1
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Topic 2 Or der Det er m in a t ion P age # 1 0
first order differential equation general solution
dy
dxP is function such that P'(x) = p(x)
+ =
( ) =
=
= +
p x y q x
P x p x
e
y q dx c
xP x
x x x
( ) ( )
( )
( )( )
( ) ( ) ( )
T h u s :
Bk A
k ke e
C k Ak
ek k
e e
A o
B A
k t k t
B o
B
k t
B A
k t k t
A B
B A B
=
[ ]
= ( ) ( )
1
11
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Topic 2 Or der Det er m in a t ion P age # 1 1
Stepwise solution for magnitudes of rate constants
Plot A vs time ---> kA from slope
for B and C
need to solve for constants by curve fitting techniques
no simple plots - use non-linear regression techniques
eg Excell Solver Function
JMP
Sigma Plot Curve Fitting Function
solve for kB first as know kA from ln A vs time plot
use Levenburg- Marquat or Rungga Katta techniques
then do the same for same for C as a function of time
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Topic 2 Or der Det er m in a t ion P age # 1 2
III. Order determination
= [ ]dA
dt
k AAn
A. Method of differentiation
1. plot concentration vs time. and draw smooth curve (fig 2-1)
2. slope is dA/dt.
determine slope at various points in time by:
a. drawing tangent to curve
b. taking A for each t (need lots of data points)
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Topic 2 Or der Det er m in a t ion P age # 1 3
3. plot ln dA/dt vs ln A as in Figure 2-2
ln ln ln
dA
dt k n AA= [ ] + [ ]slope = n the order
rate constant from algebraic substitution
Figure 2-1 Figure 2-2
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Topic 2 Or der Det er m in a t ion P age # 1 4
4. Class example
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Topic 2 Or der Det er m in a t ion P age # 1 5
` a. slope determinations
order close to 1
rate constant 0.06 units (time^ -0.68)
5. On-line computer instrument analyzers - depends on time
differential used and reaction rate.
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Topic 2 Or der Det er m in a t ion P age # 1 6
B. Method of integration
1. Order = 0 pseudo zero order
a. general solution
= [ ] =
=
[ ] =
=
dA
dtk A k
dA k dt
A A k t
A A k t
z
z
t
A
A
z
o z
0
0
0
0
plot A vs time ---> straight line
or plot Ao-A vs time --> straight line
slope = rate constant = kz
kA A
t tz =
2 1
2 1
units = amount per time (eg mg/L hr)
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Topic 2 Or der Det er m in a t ion P age # 1 7
b. data for amount vs time for zero order plot
c. draw line on graph and calculate slope (next page)
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Topic 2 Or der Det er m in a t ion P age # 1 8
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Topic 2 Or der Det er m in a t ion P age # 1 9
d. results
class value kavg= (95% CL) = t/n
my value kz=
e. linear regression generated results
kz = 0.7718 units/time 0.128 (95% CL)
kupper = 0.90 klower= 0.644 r2 = 0.954
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Topic 2 Or der Det er m in a t ion P age # 2 0
Figure 2-3 Example from nonenzymatic browning
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Topic 2 Or der Det er m in a t ion P age # 2 1
2. first order
a. derivation of equations
=
=
=
=
=
=
=
dAdt
k A
dA
Ak dt
A
Ak t
A
Ak t
A A e
A
A
kt
A A
f
A
A
f
t
f
f
k t
f
k
t
f
f
0 0
0
0
0
0
02 303
2 303
10
ln
ln
log.
.
[A]
time
log
slope * 2.3=k
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Topic 2 Or der Det er m in a t ion P age # 2 2
b. rate constant units = time-1
kf = slope of ln [A] vs t plot
= slope of ln [A/Ao] vs t plot
kf= 2.3 * slope on semi-log plot of [A] vs time
or of semi-log plot of [A/Ao] vs time
k
A
A
tf = 2 303
10
.
log[ ]
[ ]
c. class test dat a
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Topic 2 Or der Det er m in a t ion P age # 2 3
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Topic 2 Or der Det er m in a t ion P age # 2 4
d. results of calculations
instructor kf =
class kf= (95%CL) = t/n
linear regression slope = 0.00634 time-1(r2 = 0.989)
thus kf =2.3 * slope = 0.0146 0.0012 (95%CL)
kupper=0.0158 klower=0.0134
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Topic 2 Or der Det er m in a t ion P age # 2 5
Figure 2-4 Example of first order Ascorbic acid loss
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Topic 2 Or der Det er m in a t ion P age # 2 6
e. first order production of quantity B starting with quantity Ao
(1) mathematical derivation note assumes all A -> B
A Bat t A A B
at t A B B
at t t B A A A A B
A A e
A B A e B A e or B B e
kt
kt
kt kt
= = =
= = =
= = =
=
== =
0 0
0
1 1
0
0 0
0
0 0
0 ( ) ( )
non linear solution
solve for k
here Beq is value when all A-->B
can assume at large time B= Be = Ao
Note: usually only can measure A or B not both
B
time
Beq
Note: ln ln ln( ) B B ekt= +
1
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Topic 2 Or der Det er m in a t ion P age # 2 7
b. algebraic plot solution
algebraic solution
dB
dt kA A A B
dB
A
dB
A Bkdt
A B
A Bkt or
B B
B Bkt
if B then B A
A
A Bkt
o
B
B
oB
B
o
t
o
o o
= + =
=
=
[ ][ ]
=[ ][ ]
=
=
[ ][ ]
=
ln ln
ln
0 0
0
0
0
0
0
0
can plot ln(function) vs time to get slope = k
need good estimate of Beq when Ao value not known
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Topic 2 Or der Det er m in a t ion P age # 2 8
3. Other orders when n not equal to 1
a. derivation of rate equation (one reactant)
dAdt
k A
dA
Ak dt
A An k t
A
n k
nn
n
A
A
n
t
n n
f
n
f
= [ ]
=
= [ ]
[ ]
0 0
1
0
1
1
1 11
1
1
for n 1
plot vs t
slope =
positive or negative slope depending on loss vs gain
can also plot Y = A(1-n)*(1/1-n) vs time gives straight line with slope of k
basis of Macintosh computer program ( same plot for n = 2)
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Topic 2 Or der Det er m in a t ion P age # 2 9
b. class example for second order
A + A ----> B
n=2 therefore 1-n = -1
plot Y= [A]-1vs time
slope = k
k = (concentration time)-1eg mM-1L-1
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Topic 2 Or der Det er m in a t ion P age # 3 0
c second order plot
second order n=2 r2 = 0.949
slope = k = 0.000313 0.0001
kupper= 0.000413 klower= 0.000213
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Topic 2 Or der Det er m in a t ion P age # 3 1
d. Two reactants (each 1st order)
A + B ----> C
= [ ][ ] = =
= =
[ ] =
=
=
+ [ ]
= [ ]
dA
dtk A B
dC
dt
dx
dt
A A x
B B x
C x
A x
B x
A
B
A
B A B kt
A
A
B
B A B kt
o
o
[ ] [ ]
[ ] [ ]
ln ln ln
ln
0
0
0
0
0 0
0
00 0
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Topic 2 Or der Det er m in a t ion P age # 3 2
(1) if can measure A and B independently then
plot ln [A/B] vs time
slope = [Ao-Bo] k
(2) if know Ao and Bo and can measure C thenplot ln[(Ao-C)/(Bo-C)] vs time.
same slope = [Ao-Bo] k
(3) if all three measurable can make both plots and
compare k values - allows for ? of accuracy of
measurement
(4) can also assume initial rate holds
measure loss of A kA= k [Bo]
measure loss of B kB = k[ Ao]
find k separately either way
compare to plot for both simultaneously
(5) same as (4) but also measure C
if smaller suggests that C---> D
(6) examples ascorbic acid + oxygen in packageamino acid + reducing sugar (NEB)
(7) error based on error of measurement of concentrations
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Topic 2 Or der Det er m in a t ion P age # 3 3
C. Method of half lives
1. first order n=1
ln
ln ln . .
.
AA
k t
A
Ak t
kt
t
o
f
o
f
f
=
= [ ] = =
=
=
12 0 5 0 693
0 693
0
12
12
12
half life in time units
from above example
A= 1/2 Aowhen t = 50 time units
thus kf = 0.693/50 = 0.0.01386 time-1
should run experiment through 2 - 3 half lives
Ascorbic acid half life data
system half-life days @ F
canned OJ 300 100
frozen veges 240 10
IMF no oxygen 70 82IMF in air 12 82
IMF in air 5 100
dry tomato 180 100
dry potato 24 100
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Topic 2 Or der Det er m in a t ion P age # 3 4
2. life for any fractional decrease
where f = decimal fraction (A/Ao) and n=1
ln
ln ln
ln
AA
k t
fA
A f k t
kf
t
t
o
f
of f
f
f
f
=
= [ ] =
=
=
0
fraction life in time units
(a) eg for decrease by 1 log cycle ie 1/10 of original
as in microbial death where f=0.1
ln
ln
.
ln .
ln . ..
log . .
.
.
.
A
Ak t
A
A k t
k D D D D
Dk
t D
o
f
o
f
f
f
=
= [ ] =
=
= = =
=
=
0 1
0 1
0 1 2 30262 3
0 1 2 3
2 3
00 1
0 1 value in time for 1 log cycle decrease
(b) drug stability time to 10% loss f=0.9
tk k
0 9
0 9 0 11.
ln . .=
=
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Topic 2 Or der Det er m in a t ion P age # 3 5
3. half life for second order where n=2
= [ ] = [ ]
[ ]
=
= =
=
dA
dt
k A k A
dA
Akdt
A A Akt
tkA
A
A
At
o o o
o
2 2
2
0 5
0
12
12
2
2 1 1
1
0
01
2.
4. half life for any order n except n=1
t k n A
n
o
n12
1
1
1
1
2 1
= [ ]
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Topic 2 Or der Det er m in a t ion P age # 3 6
D. Other methods for order determination
1. Initial rate for complex reactions
- assume pseudo first order with respect to A
- assume all others are constant (B etc.) ie small change
eg. A +B ----> C
dA
dt
k A B
dA
dtk A
k k B
B
B
= [ ][ ]
= [ ]
= [ ]
assume B >> A
log plot of [A] vs time ---> kB
kB [Bo] = k
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Topic 2 Or der Det er m in a t ion P age # 3 7
2. Powell Plot method (n 1)
1 11
1
1 1 1
1
1
1 1
1 1
1
1
1
1 1
A A
n kt
A A n kt
A
An A kt zt
z n A k
A
A
A
A
n
o
n
n
o
n
o
n
o
n
o
n
o
n
o
= [ ]
= [ ]
= + = +
=
=
( )
( )
nn
n
x p
o
o
zt
also y p yp y
n AA zt
A
A nz
nt
=
= = + + [ ]
= [ ]
=
[ ] +
[ ]
1
1 1 12
1
1
1
1
1
1
2
lnln
!
( ) ln ln
ln ln ln
for small change 3rd term small thus:
do linear regression of ln A/Ao vs ln time
slope = 1/1-n
get k from intercept value (1/1-n) ln z
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Topic 2 Or der Det er m in a t ion P age # 3 8
original paper has A/Ao vs log t lines for different n values
choose the one that is the closest
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Topic 2 Or der Det er m in a t ion P age # 3 9
3. Wilkinson plots ( Chem. & Ind. 9/2/61 pg. 1395)
1 11
1
1 1
1
1
1
1 1
1 1
1
1
1
A An kt
A A n kt
A
An A kt
set f A
A
A
Af
A
Af
n
o
n
n
o
n
o
n
o
n
c
o
o
c
o
n
c
= [ ]
= [ ]
= +
=
=
=
( )
(fraction consumed)
f = 1 when A = Ao
[[ ] = +
= = =
[ ] = +
[ ] = +
[ ]
1 1
2 3
1 2
1
1 1
0
1
1 11
2
1 2
3
1 1 1 12
1 1
n
o
n
c
x
cn
c c
c
n
n A t
for n f zt k
At
for n
y xyx x y x x x y
f n f n n f
f
( )
( )
!
( )( )
!
( ) ( )
series expansion
drop last term
++
= +
=
( )( )
( )n fn nf
n A t
f k t n
f
f
f k t
f k t n kt
f
f
tk
n kf
cc
o
n
c c
c
c
c c
cc
11
21 1
2
2
2
21
2
subtract 1s and divide by n -1
= A
if small (small extent) then
A (essentially zero order at start)
substitute in for one value of f
= AA
AA
o
n-1
o
n-1
o
n-1 o
n-1
o
n-1 o
n-1
tt
f k
nt= +
1
2Aon-1
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Topic 2 Or der Det er m in a t ion P age # 4 0
plot t/fcon vs time where fcon= fraction consumed =1-A/Ao
slope of line is n/2
derivation is not 1st order but paper says use anyway
at small t and fcon t/fcon-->
t /f
f
test of Wilkinson plot with sam e dat a
t /f con
0
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Topic 2 Or der Det er m in a t ion P age # 4 1
plot of all data
Note poor fit of line
from plot order = 2 * 0.21 = 0.42
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Topic 2 Or der Det er m in a t ion P age # 4 2
plot of data after time =10
from modified plot
order = 2 * 0.55 = ~ 1.2 so closer but not a good method
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Topic 2 Or der Det er m in a t ion P age # 4 3
E. Other important factors
1. Error of % basis for zero order
suppose 1 mg/day loss rate and:
A1=100 or A
2=200
then at t = 50 A=100-kt A=200-kt
A=100-1*50=50 A=200-1*50=150
% %
% / . % /
dayx
dayx
day day
=
=
= =
100 50
100
50100
200 150
200
50100
1 0 5
thus %/day depends on initial value of A for zero order
not a problem for 1st order
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Topic 2 Or der Det er m in a t ion P age # 4 4
2. fraction consumed - zero order
A A kt
A A
A
A
A f
A A
A
k
At f
fk
A
t
kA A
tf
A
A
t
t
kA
t
f kA
t
o
o
o orem
o
o o
con
rem
o
o s
s
cons
o s
o
s
con
o
= =
=
= =
= =
=
=
=
=
= =
amount lost
f
thus plot f vs time - - > straight line
or f vs time
if set A = 100% shelf life and A = 0% then
con
con
rem
o s
1 1
1
1
ttt
AA
fs o
rem= = 1 1
plot fraction consumed vs time -->straight line
slope = k/Ao-As = 1/ts
fraction consumed at any condition = time/total time to 100% done
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Topic 2 Or der Det er m in a t ion P age # 4 5
3. first order fraction consumed
ln
log( ) log( )
A
A
kt
A
Ae
f f e
f e
f e
o
o
kt
rem con
kt
con
kt
con
kt
=
=
= =
= =
1
1
1
plot log fraction remaining vs time as before
plot of log fraction consumed vs time not a straight line !!!
while plot of fraction remaining is ie ln A/Ao vs time
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Topic 2 Or der Det er m in a t ion P age # 4 6
4. Choice of best order
a. best fit with highest r2
? of meaning of linear regression
with 6-8 points need r ~ 0.95 (r2 = 0.9)
change order regresses something different
standard error (SE) also different units
b. # of data points
should have at least 6-8 points spread over 30-50% loss
c. degree of change
For example if:
at t =0 A=100
at t=50 A=50
thenkz= (100-50)/50 = 1 unit/day
kf=0.693/50=0.01386 (units time-1)
time zero first
0 100 100
10 90 87
20 80 75.830 70 65.98
40 60 57.4
50 50 50
thus makes little difference for first 50% loss because of
analytical errors
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Topic 2 Or der Det er m in a t ion P age # 4 7
d. maximum analytical error (From Benson) error propagationx F p q
F
p
F
q
F
p
F
q
k C C t
x p q pq
k C C
o
t
C C
o
o
=
=
+
+
=( )
+
( , , ....)
( )
( )
2
2
2
2
2
2
2
2
2
2
2
2
2
2
for zero order
C = C - kt k =C - C
to
o
22 2 2= + C Co
if at time = 1 Co = 1 and C = 0.9 with relative error (analytical precision) of 1% in
measurement of C and 1% error in measurement of time then:
( ) ( . ) . .
.
.
..
. .
C C C C
k
k
k
o ox
k
x
kk
= + = + ( ) =
= +
[ ] =
= =
2 2 2 2 2 4
2
2
4
2
2
2
0 01 0 009 1 81 10
1 81 10
0 1
0 01
10 0182
0 135 0 135
if at t= 10 C = 0.5 then
( ) ( . ) . .
.
.
..
. .
C C C C
k
kk
o ox
k
xx
kk
= + = + ( ) =
= +
[ ] =
= =
2 2 2 2 2 4
2
2
4
2
2
2
4
0 01 0 005 1 25 10
1 25 10
0 5
0 01
105 01 10
0 0024 0 0224
Thus inaccuracy depends on magnitude of Co-C and time as well as
inaccuracies in time and concentration measurement
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Topic 2 Or der Det er m in a t ion P age # 4 8
% error in k at % change in reactant monitored
Analytical
precision % 1% 5% 10% 20% 30% 40% 50%
0.1 14 2.8 1.4 0.7 0.5 0.4 0.3
0.5 70 14 7 3.5 2.5 2 1.5
1 >100 28 14 7 5 4 3
2 >100 56 28 14 10 8 6
5 >100 >100 70 35 25 20 15
10 >100 >100 >100 70 50 40 30
e. should do multiple zero time values for precision
f. problem of extraction vs analytical test
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Topic 2 Or der Det er m in a t ion P age # 4 9
IV. Statistical analysis of rate constants
A. Evaluation of variance
1. Gaussian distribution
(a) mean distribution is 2 for 95.43% confidence
is measure of variation of individuals in population
2is the variance
Range of confidence = x_ 2
Se=standard error = variation of sample means
(b) for a large amount of data, the 95% confidence limits are:
x
n 1 96.
where2
2
1=
[ ]
x xn
(c) coefficient of variation (CV) is:
cvx
=
100
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Topic 2 Or der Det er m in a t ion P age # 5 0
(d) variability at any %CL is:
x tn
t t
n
T
T
=
=
value at degrees of freedom
and desired probability
= n
# of data points
T 2
Statistical significance says that data are adequate to reject thenull hypothesis that two systems are the same. Practicality is
based on how big a difference is important which must be
answered on other than statistical reasons.
Type I () error: hypothesis A=B when data says A B
Type II()error: hypothesis AB when data says A=B
Table shows need for high # of points to lower t value (>8)
? of time, cost, and reliability
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Topic 2 Or der Det er m in a t ion P age # 5 1
(e) Student t Table = degrees of freedom = n-2
p=90% p=95% p=99%
1 6.31 12.7163.66
2 2.92 4.30 9.93
3 2.35 3.18 5.84
4 2.13 2.78 4.60
5 2.02 2.57 4.03
6 1.94 2.45 3.71
7 1.90 2.37 3.50
8 1.86 2.31 3.36
9 1.83 2.31 3.3610 1.81 2.23 3.17
11 1.80 2.20 3.11
12 1.78 2.18 3.06
13 1.77 2.16 3.01
14 1.76 2.15 2.98
15 1.75 2.13 2.95
16 1.75 2.12 2.92
17 1.74 2.11 2.90
20 1.73 2.09 2.85
25 1.71 2.06 2.79
30 1.70 2.04 2.75
40 1.68 2.02 2.70
50 1.68 2.02 2.70
60 1.67 2.00 2.66
80 1.66 1.99 2.64
100 1.66 1.98 2.63
above 6-8 points, doubling the points reduces the t by about
10% so question of error allowed vs cost of doing more points
Going from 10 to 20 points decreases error by only 5%
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Topic 2 Or der Det er m in a t ion P age # 5 2
B. Linear regression
1. definition : minimize sum of squares of y from fit of linearized
function.
ie least vertical deviation from straight line fit.
Major assumption (time is exact - ie not a variable!!!)
cannot really compare different type functions
same is true for standard error comparison
2. equations
Intercept = I =
I y x x xy
n x x=
[ ][ ] [ ][ ][ ] [ ]
2
22
Slope = k =
k n xy x y
n x x= [ ] [ ][ ]
[ ] [ ] 2
2
True intercept = =
= +[ ]
[ ] [ ]
I t sn
x
n x n xe
12
2 22
12
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Topic 2 Or der Det er m in a t ion P age # 5 3
True slope = =
=
[ ] [ ] k
t s
n x x
n
e
2
2
Standard error = se =
`
s y I y k xy
ne =
[ ] [ ] [ ]
21
2
2
Coefficient of determination r2 (r =correlation coefficient)
rn xy x y
n x x n y y
2
2
2 1 2
2
2 1 2
2
=[ ] [ ][ ]
[ ] [ ][ ] [ ] [ ][ ]
3. Predicted future value and 95% confidence limits:
within data limits first term of 1 is deleted
y I kx t sn
n x x
n x xo e0
02
22
12
11= + + [ ]
[ ] [ ]
( )
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Topic 2 Or der Det er m in a t ion P age # 5 4
4. Comparison of rate constants
If hypothesis is k1 = k2 then they are the same if
(a) the 95% confidence limits overlap
significant difference (actually > 95%) rigorous test
k1k2
(b) t test of significance - Two tailed
= 2(n-1)
tk k
S S
t t
e e
table
=+
>
2 1
2 1
2
95significant if @ %
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Topic 2 Or der Det er m in a t ion P age # 5 5
5. Dealing with the zero/zero time point in regression
force fit or error of measurement
6. Point by point method: (can use spread sheet)
treat each data point as one experiment and get k and 95%CL
from table
kA A
t
k
AA
t
kk
n
t
n
zo
f
avg
=
=
=
ln0
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Topic 2 Or der Det er m in a t ion P age # 5 6
7. Quality of the regression value
r (correlation coefficient) must be greater thanat given level of significance
n=# of data pairs =95% =99% =99.53 0.9974 0.950 0.990 0.9995 0.878 0.934 0.9596 0.811 0.882 0.9177 0.754 0.833 0.8758 0.707 0.789 0.8349 0.666 0.750 0.79810 0.632 0.715 0.76511 0.602 0.685 0.73512 0.576 0.658 0.70813 0.553 0.634 0.68414 0.532 0.612 0.66115 0.514 0.592 0.641
16 0.497 0.574 0.62317 0.482 0.558 0.60618 0.468 0.543 0.59019 0.456 0.529 0.57520 0.444 0.516 0.56121 0.433 0.503 0.54922 0.423 0.492 0.53727 0.381 0.445 0.48732 0.349 0.409 0.44937 0.325 0.381 0.41842 0.304 0.358 0.39347 0.288 0.338 0.37252 0.273 0.322 0.35462 0.250 0.295 0.325
72 0.232 0.274 0.30282 0.217 0.256 0.28392 0.205 0.242 0.267
from R.Fisher and Y. Yates Statistical Tables for Biological, Agricultural, and Medical Research Oliver &Boyd Ltd., Edinburg.
Note that with a good r2 of 0.95 you can be assured of a high quality (95%CL) at only 4 data points. For 6 data points if the r2 exceeds 0.66 you havehigh quality (p = 0.95)
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Topic 2 Or der Det er m in a t ion P age # 5 7
8 Other considerations
a. minimize residual sum of squares in estimate vs actual value
Y f x t e
RSS y f x t
i i i
i i
i
n
= +
= [ ]=
( , )
( , )1
2
where
yi = actual value
model = f(xi, t)= estimate at xi, ti
residual difference = ei = yi- ymodel
b. make residual plots (ei vs time)
homoscedastic = equal propagation
var(ei) = constant= 2
yi-f(x,t)
time
0
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Topic 2 Or der Det er m in a t ion P age # 5 8
heteroscedastic = convergence or min/maxvar(ei) = non-constant = 2/wi
yi-f(x,t)
time
0
convergeor diverge
yi-f(x,t)
time
0
min or max
ie weighted by some unknown or error factor
should always make residual plot (good test of quality of
model)
c. Linear models not necessarily straight line
rule : derivative with respect to parameter is independent
of parameter, thus for
y ax bx cx
y
ax
y
bx
y
cx
= + +
= = =
2 3
2 3
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Topic 2 Or der Det er m in a t ion P age # 5 9
d. transformation of non-linear model to linear form
y ae
y a bx
y xa b
y yx
bx=
= +
= += =
= =
ln ln
ln
* *
*
1
e. transformation weights wi where
RSS w y f x t
Transformation weight
yy
yy
y y
i i i
i
n
i
i
i
ii
i
i i i
= [ ]=
( , )
ln
1
2
4
2
2
2
2
1
f. non-linear regression
need to estimate initial values
use iterative solution of some search procedure
minimize RSS
gives approximate confidence limits
danger of local minimum
gives different value at t = 0
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Topic 2 Or der Det er m in a t ion P age # 6 0
9. Use of Mac Program -
a. review of steps
(1) enter data for A vs time
(2) choose possible orders
(3) calculate k and 95% CL for given temperature
note for n not = 1 y value is A(1-n)*(1/1-n)
(4) predict future values
(5) make plot with 95% CL for order
(6) repeat for each temperature
(7) choose temps and calculate EA and Q10
(8) make Arrhenius plot
(9) calculate k for any temperature
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Topic 2 Or der Det er m in a t ion P age # 6 1
b. initial screen
b. data inpu t : exam ple of class da ta
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Topic 2 Or der Det er m in a t ion P age # 6 2
d, order determina tion
e. zero order plot
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Topic 2 Or der Det er m in a t ion P age # 6 3
f. firs t ord er plot
g. second order plot
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Topic 2 Or der Det er m in a t ion P age # 6 4
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