VI. Determination of Reaction Orders - Cornell...

VI. Determination of Reaction Orders:

A. Given:

-d[A]/dt = k[A]α[B]β

If we use a large excess of B to establish pseudo-first-order conditions...

[B0] >> [A0]

[B] ≈ [B0] ≈ constant

-d[A]/dt = kobsd[A]α

such that...

kobsd = k[B]β

How do we determine "α" and "β"?

k!A + B P


B. Order in Limiting Reagent (A):

1. Graphical Method: looking for linearity in plots

a. α = 0:

A plot of [A] vs. t is linear (zeroth-order plot)...

• Observed when the rate limiting step does not involve thelimiting reagent, A.

t (sec)

[A]


b. α = 1:

A plot of ln[A] vs. t (or ln{[A]/[A0]} vs. t) is linear (1st-orderplot; Section II.C.2).

c. α = 2:

A plot of 1/[A] vs. t (or 1/[A] - 1/[A0] vs. t) is linear (2nd-orderplot; Section III.C.3).

d. α = n:

A plot of 1/[A](n-1) vs. t (or 1/[A](n-1) - 1/[A0](n-1) vs. t is linear(nth order plot; Section IV.C).


e. Problems and Questions:

i. How linear is linear?

ii. What would curvature look like?

iii. How many half-lives?

iv. Can fractional orders be distinguished?

• The method is highly susceptible to biased interpretation.


2. Integral Form, Least Squares Analysis:

a. Given: (from Section IV.B)

(1)

Rearranging...(Appendix 1)

!

1"#1

1A[ ]" #1

#1

Ao[ ]" #1$

% & &

'

( ) ) = kobsdt

!

A[ ] = "#1( )kobsdt + Ao[ ]# " #1( ){ }#1/(" #1)


b. Least Squares Analysis:

Let...

f(x) = (ax + b)c

such that...

f(x) = [A]

x = t a = (α - 1)kobsd b = [A0](1 - α) c = -1/(α - 1)

• a, b and c are adjustable parameters, and f(x) and x are measured experimental values.

• This method is nicely user blind, but not well-tested. It is most likely tosucceed if applied to each rate constant determination with averaging.


3. Integral Form, Fractional Lives:

a. The Experiment:

Vary [A0] and measure time, t, required to attain fixed percent conversion.

For example...

Let [A] = 1/2[A0]

• Although the time required to attain a half-life (50% conversion)is commonly employed, any percent conversion will suffice.

Substitute for [A] into eq (1) and rearrange...

• “Noyes Equation”

!

t1/2 =1

kobsd "#1( )2" #1#1Ao[ ]" #1

$

% & &

'

( ) )


b. Graphics:

i. Non-linear:

Let...

such that...

f(x) = t1/2

x = [A0] a = α - 1 b = kobsd

• The expression is undefined at α = 1 and is also untested. Overall, it is not terribly useful looking to me. However...

!

f x( ) =1ab

2a "1xa

#

$ %

&

' (


ii. Linear:

Taking the natural log of the Noyes Equation

ln(t1/2) = {ln(2α-1 - 1) - lnk - ln(α - 1)} - (α - 1)ln[A0]

Let...

f(x) = a + bx

such that...

f(x) = ln(t1/2)

x = ln[A0]

a = {ln(2α-1 - 1) - lnk - ln(α - 1)} = constant

b = -(α - 1) = (1 - α)

• f(x) is undefined at α = 1.


“Noyes Plot”...

• For example, the slope = -1 for α = 2.

• A discontinuity (change in slope) shows a change in mechanism with changing [A0].

ln[A0]

slope(s) = 1- !

ln(t1/2)! < 1

! > 1

! = 1


d. Fast and Dirty:

i. Recall:

ln{[A]/[A0]} = -kt

If we let...

[A]/[A0] = 1/2

and...

t = t1/2

Then...

t1/2 = 0.69/k

The half-life, t1/2, will show no [A0] dependence.

In contrast...


For a 2nd-order reaction...

[A]/[A0] = 1/(1 + [A0]kt)

Therefore...

t1/2 = 1/([A0]k)

• For a given half-life, t1/2 is inversely proportional to [A0].

Thus…

[A]

t1/2t1/2 =

[A]

t1/2t1/2 <

1st order 2nd order


ii. The Experiment:

• Change [A0] 10-fold. If there is no change in k (determined using a 1st-order function), then the reaction is 1st order in A.

• Since you are usually confirming lst-order behavior, you must show that t1/2 is independent of [A0].


4. Differential Form, Instantaneous Rates:

a. Given:

-d[A]/dt = kobsd[A]α

ln{-d[A]/dt} = lnkobsd + αln[A]

b. In principle:

To get α you could plot...

-d[A]/dt vs. [A]

or...

ln{-d[A]/dt} vs. ln[A]

How do you express d[A]/dt numerically?


c. In practice:

t (sec)

[A]!"[A]/"t # -d[A]/dt

!

"# A[ ]#t

="( A[ ]n+1" A[ ]n )

tn+1" tn


Let...

[A] = [A]n or [A]n + 1 or [A]avg

(doesn’t really matter)

Then...

-Δ[A]/Δt = kobsd[A]αavg (f(x) = axb)

or...

ln{-Δ[A]/Δt} = lnkobsd + αln[A]avg (f(x) = a + bx)


d. Graphics:

i. In Theory:

ln[A]avg

ln{-!A/!t}slope = "

-!A/!t

[A]avg

" > 1 " = 1

" < 1


ii. In Practice:

• Incredible scatter due to wildly oscillating slopes still gives a validanswer, but prevents you from ever seeing this plot in a paper!

ln[A]avg

ln{-!A/!t}


5. Differential Form, Time-Zero Rates:

a. The Experiment:

Monitor rates at early percent conversion for different [A0]’s.

b. Rate Equation:

-d[A]/dt ≈ -Δ[A]/Δt = k[A]α ≈ k[A0]α

(f(x) = axb)

[A]

t (sec)

Approximately linear!A/!t as t approaches 0.


c. Graphics:

! > 1 ! = 1

! < 1

-"A/"t(initial rates)

[A0]


6. Complementarity of Methods:

a. “Reaction Order Over Time”: Monitoring the disappearance of A over the course of a reaction (sections VI.B.1-4) affords the “ReactionOrder Over Time”. These methods provide spurious (unusual) results ifautoinhibition or autocatalysis occurs.

b. “True Reaction Order”: Monitoring the disappearance of A at low conversion (Section VI.B.5) affords the “True Reaction Order”. Thismethod excludes the effects of autoinhibition and autocatalysis.

c. Taken together, the two fundamentally different strategies afford evidence that the products from the reaction are (or are not) influencingthe rate.


7. Summary:

a. Integral Form, Graphical Method:

i. Most common

ii. Problems with bias.

b. Integral Form, Least Squares:

i. Experimentalist blind

ii. Order is “determined” not “confirmed”

iii. Could be temperamental at α ≅ 1 (undefined at α = 1)

iv. Relatively untested

v. Potentially powerful


c. Integral Form, Fractional Lives:

i. Crude

ii. Good cross-check to confirm 1st order in conjunctionwith graphical method; k will be independent of [A0]

d. Differential Form, Instantaneous Rates:

i. Clumsy; needs lots of good data.

e. Differential Form, Time-Zero Rates:

i. Good for detecting auto-feedback mechanisms

ii. Shortcut: double injections.


C. Reaction Order in B: (where B is in large excess; [B] ≈ [B0])

1. Given:

-d[A]/dt = k[A]α[B]β = kobsd[A]α

such that...

kobsd = k[B]β

2. Experiment: monitor the pseudo-first-order rate constants,kobsd, as a function of different [B0]’s.


3. Graphics:

Let...

f(x) = axb

such that... f(x) = kobsd x = [B0] a = k b = β

Alternatively, we could use the linear variant...

f(x) = a + bx

such that... f(x) = lnkobsd x = ln[B0] a = lnk b = β

! > 1 ! = 1

! < 1

kobsd

[B0]


D. Kinetics Under Non-Pseudo-1st-Order Conditions:

Although the focus of this presentation is on carrying out kinetics underpseudo-1st-order conditions, the methods for determining reaction orders atearly conversion allow rate laws to be determined under more standardconditions as well....

1. Given:

-d[A]/dt = k[A]α[B]β

such that neither A nor B is in large excess. Let us also assume that our analytical methods allow us to follow the loss of A over time.

!A + B Pk


2. Reaction Order in A:

The order in A is determined by plots of ΔA/Δt vs. [A0] as follows:

[A]

t (sec)

!A/!tat t=0

! > 1 ! = 1

! < 1

-"A/"t(initial rates)

[A0]


3. Reaction Order in B:

Analogously, the order in B (or any number of additional reagents)is determined from plots ΔA/Δt vs. [B0] as follows:

Although the zeroth-order plots for loss of [A] appear different when [A0] and [B0] are varied, the dependencies of the initial slopes(ΔA/Δt vs. [A0] or [B0]) are treated identically.

-!A/!tat t = 0

time

[A]! > 1

! = 1

! < 1-"A/"t

(initial rates)

[B0]

VI. Determination of Reaction Orders - Cornell...

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