cre lecture 15

8/16/2019 CRE Lecture 15

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5 14 2016

Lecture 15

Example – Finding theRate Law

2

t(min) 0 1 2 3

C A(mol/L) 1 0.7 0.5 0.35

0.3 0.2 0.15t

C A

t

C A

.1

.2

.3

t1 2 3

Areas equal for both

sides of the histogram



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Find f(t) of using equal area differentiationt

C A

C A 1 0.7 0.5 0.35

-dC A/dt 0.35 0.25 0.175 0.12

Plot (–dC A/dt) as a function of C A

ln

C A

Slope = α

dC A/dt

ln

3


Choose a point, p, and find the concentration and

derivative at that point to determine k.

4


Ap

p

A

C

dt

dC

k

ln

C A

Slope =α

dC A/dt

ln

p

A

dt

dC

p AC



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3

Linear Least-Square Analysis

5

In the case where the rate law is dependent onthe concentration of more than one reactants andmethod of excess is not applicable.

B A A

A C kC r dt

dC

0000)(

B A A

A C kC r dt

dC

000lnlnln)(ln

B A

A C C k dt

dC

Linear Least-Square Analysis

6

1002010

,ln,ln,ln,)/ln( ak aC X C X dt dC Y B A A

22110 X a X aaY

000 lnlnln)(ln

B A

A C C k dt

dC

N

j j

N

j j j

N

j j

N

j j j

N

j j j

N

j j

N

j j

N

j j j

N

j j

N

j j

N

j j

X a X X a X aY X

X X a X a X aY X

X a X a NaY

1

2

221

2111

201

2

1 212

1

2

111

101

1

1 22

1 110

1



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Non-Linear Least-Square Analysis

7

We want to find the parameter values (α, k, E) forwhich the sum of the squares of the differences, the

measured rate (Cm), and the calculated rate (Cc) is

a minimum.

That is, we want to be a minimum. 2

n

i

icim

K N

S

K N

C C

1

22

2

8

For concentration-time data, we can combine the

mole balance equation for to obtain:

Rearranging to obtain the calculated concentration

as a function of time, we obtain:

r A kC A

00

A A

A A

dC kC

dt

t C C

1 1

0 (1 )

A AC C kt

1 1/(1 )

0[ (1 ) ] Ac A A

C C C kt



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9

Now we could use Polymath or MATLAB to find the values of α and kthat would minimize the sum of squares of differences between themeasured (C Am) and calculated (C Ac) concentrations.

That is, for N data points,

Similarly one can calculate the time at a specified concentration, tc

and compare it with the measured time, tm, at that same concentration.

That is, we find the values of k and α that minimize:

Example

10

Find reaction order and specific reaction rate by nonlinear least square analysis .



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1 1

M ethod of Init ial Rates

1 2

A series of experiments are carried out at different

initial concentrations, C A0 and initial reaction rate

–r A0 is determined for each run.

The initial rate can be found by differentiating the

data and extrapolating to zero time.



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Example

1 3

Solution

1 4



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1 5

M e thod of Ha lf-Lives

1 6

The half-live of a reaction is the time it takes for theconcentrationofthereactanttofall tohalf of itsinitial value.

By determining the half-life as a function of the initialconcentration, the reaction order and constant can bedetermined.

If two reactants are involved, use the method of excess inconjunctionwiththemethodofhalf-livestoarrangetherateoftheform:



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1 7

1 8



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1 9

Differential Reactors

2 0



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2 1

Using very little catalyst and large volumetric flow rates:

CA0-CAecan be made quite small

Use method of initial rates to find reaction order and specificreaction rate.

Example

2 2

T=500 F



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2 3

Solution

2 4



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2 5

2 6



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cre lecture 15

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