cinetica teoria - FANTASTICO

8/6/2019 cinetica teoria - FANTASTICO

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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


3/64


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


4/64


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


5/64


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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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


7/64


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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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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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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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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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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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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4. Class example


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` 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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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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b. data for amount vs time for zero order plot

c. draw line on graph and calculate slope (next page)


18/64



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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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Figure 2-3 Example from nonenzymatic browning


21/64


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


22/64


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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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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Figure 2-4 Example of first order Ascorbic acid loss


26/64


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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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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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)


29/64


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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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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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


32/64


(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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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


34/64


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 . .=

=


35/64


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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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


37/64


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


38/64


original paper has A/Ao vs log t lines for different n values

choose the one that is the closest


39/64


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


40/64


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


41/64


plot of all data

Note poor fit of line

from plot order = 2 * 0.21 = 0.42


42/64


plot of data after time =10

from modified plot

order = 2 * 0.55 = ~ 1.2 so closer but not a good method


43/64


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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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


45/64


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


46/64


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


47/64


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


48/64


% 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


49/64


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


50/64


(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


51/64


(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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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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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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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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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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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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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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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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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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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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b. initial screen

b. data inpu t : exam ple of class da ta


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d, order determina tion

e. zero order plot


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f. firs t ord er plot

g. second order plot


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cinetica teoria - FANTASTICO

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