P5-5

The reaction

A ---> B + C

was carried out in a constant-volume batch reactor where the following concentration measurements were recorded as a function of time.

t (min) 0 5 9 15 22 30 40 60

CA (mol/dm3) 2 1.6 1.35 1.1 0.87 0.70 0.53 0.35

(a) Use nonlinear least squares (i.e., regression) and one other method to determine the reaction order and the specific reaction rate.(b) If you were to take more data, where would you place the points? Why?

SOLUTION

This problem is more easily solved in Excel than MathCAD since the linear regression package is easier to utilize. But we'll solve it using MathCAD. First enter the data

t

0

5

9

15

22

30

40

60

min⋅:= CA

2

1.6

1.35

1.1

0.87

0.70

0.53

0.35

mole

liter⋅:=

Since we don't know the reaction order let's do a differential technique to get an estimate of the order. To do this start with the rate expression for a constant volume batch reactor

dCA/dt = -kCAa

and take the log of both sides to get

ln(-dCA/dt) = ln(k) + a ln(CA)

Thus a plot of ln(-dCA/dt) versus ln(CA) should be a straight line with a slope = reaction order and an intercept = ln (k). Let's use a central finite difference to approximate the differential. Thus we get

ii 1 6..:=

dCAii

CAii 1+

CAii 1−

−( )tii 1+ tii 1−−( )

:=

dCA

0

0.072−

0.05−

0.037−

0.027−

0.019−

0.012−

mol

liter min⋅=

Now take the logs of this data and the corresponding CA terms and put them into vy and vx vectors, respectively.

vxii 1− ln CAii

liter

mole⋅

:=

vyii 1− ln dCA−( )ii

min liter⋅

mole⋅

:=

Doing the regression gives

n slope vx vy,( ):=n 1.597=

lnk intercept vx vy,( ):=lnk 3.424−=

corr vx vy,( ) 0.998=

ij 0 5..:=

Yij lnk n vxij⋅+:=

0.8 0.6 0.4 0.2 0 0.2 0.4 0.64.5

4

3.5

3

2.5Differential Analysis

ln(CA)

ln(-

dCA

/dt)

let's say that the reaction is 1.6 order. The reaction rate expression then is given by

dCA/dt = -kCA1.6

Separating variables and integrating gives

CA-0.6 - CAo

-0.6 = 0.6kt

Thus a plot of CA-0.6 versus time should give a straight line with a slope of 0.6k and an intercept of CAo

-0.6. The analysis and the plot are shown below

i 0 7..:=

vyyi CAi( ) 0.6−:=

kp slope t vyy,( ):=

kkp

0.6:=

k 0.034liter0.6

mol0.6 min⋅=

CAo intercept t vyy,( )

10.6−:=

CAo 2.06mole

liter=

Ycalci

intercept t vyy,( ) 0.6 k⋅ ti⋅+:=

vxxi CAi( ) 0.6−:=

0 10 20 30 40 50 600.01

0.015

0.02

0.025

0.03Integral Analysis

Time (min)

CA

^-0.

6

For a 1.6 order reaction the rate constant would be 0.034 liter0.6/mole0.6-min.

(b) The data is reasonable well distributed. The intercept lies close to the zero time concentration (2.06 mole/liter versus 2.0 mole/liter). The best place for additional data would be a longer times.

P5-7

The following data were reported [C. N. Hinshelwood and P. J. Ackey, Proc. R. Soc. (Lond)., A115, 215 (1927)] for a gas-phase constant-volume decomposition of dimethyl ether at 504oC in a batch reactor. Initially, only (CH3)2O was present.

Time(s) 390 777 1195 3155 infiniteTotal Pressure (mmHg) 408 488 562 799 931

(a) Why do you think the total pressure measurement at t = 0 is missing? Can you estimate it?(b) Assuming that the reaction

(CH3)2O ----> CH4 + H2 + CO

is irreversible and goes to completion, determine the reaction order and specific reaction rate k. (c) What experimental conditions would you suggest if the were to obtain more data?(d) How would the data and your answers change if the reaction were run at a higher or lower temperature?

SOLUTION

The reaction rate law can be written as

dCA/dt = -kCAa

where A = (CH3)2O. However, the data is given in terms of total pressure. Thus we need to relate CA to total pressure (P0 to solve this problem. We know that in a constant volume system the pressure and total number of moles present are related by

PT = nT (RT/V) = CT(RT)

We know that

nT = nA + nB + nC + nD

where B = CH4, C = H2 and D = CO. We also know that

nB = nC = nD = nAo - nA

This gives

nT = 3nAo - 2nA

or

nA = 0.5( 3nAo - nT)

Dividing by V gives

CA = 0.5(3CAo - CT)

Substitute this into the rate expression to get

- 0.5(dCT/dt) = - k (0.5)a(3CAo - CT)a

Then replace CT and CAo by using the relationship between Ci and Pi (Ci = Pi/RT) to get

(- 0.5/RT) (dPT/dt) = - k (0.5/RT)a (3 PAo - PT)a

But since the initial charge was only A, PAo = PTo so we finally get

dPT/dt = (0.5/RT)a-1 k (3 PTo - PT)a

(a) In order to use this expression we are going to need PTo. A likely reason why this was not in the original data is that the experiment was probably conducted by first filling the reactor with the dimethyl ether, then heating it to the reaction temperature. Since the temperature increase was not instantaneous an accurate value for PTo would be difficult to obtain. However, since the reaction runs to completion we know that when t = ∞

CA = 0 = 0.5(3CAo - CT)

or

CT = 3 CAo

so

PT = 3 PAo = 3 PTo

Since PT = 931 mmHg at t = ∞, PTo = 931/3 = 310.3 mmHg.

(b) To find the order and rate constant let's start by doing a differential analysis. Taking the log of both sides of the rate expression (given in total pressure) gives

ln (dPT/dt) = ln [(0.5/RT)a-1k] + a ln(3PTo - PT)

Now enter the data, take a forward finite difference approximation for the differential and perform the regression (as in P5-5).

t

390

777

1195

3155

s⋅:=

PT

408

488

562

799

torr⋅:=

ii 0 2..:=

dPTii

PTii 1+

PTii

−

tii 1+ tii−:=

dPT

0.207

0.177

0.121

torr

s=

vx2ii ln dPTii

s

torr⋅

:=

vy2ii ln 931( ) PTii

1

torr⋅−

:=

a slope vx2 vy2,( ):=

a 0.618=

Y2ii intercept vx2 vy2,( ) slope vx2 vy2,( ) vx2ii⋅+:=

corr vx2 vy2,( ) 0.978=

2.2 2.1 2 1.9 1.8 1.7 1.6 1.55.8

5.9

6

6.1

6.2

6.3Differential Analysis

ln(3PTo - PT)

ln (d

PT/d

t)

The order is close to 0.6 so let's perform an integral analysis assuming a reaction order of 0.6. This would give a rate expression of

dPT/dt = (0.5/RT)-0.4 k (3 PTo - PT)0.6

that can be rearranged to give

dPT/(3PTo - PT)0.6 = (0.5/RT)-0.4 k dt

This can be integrated to give

2.5[(2PTo)0.4 - (3PTo - PT)0.4] = (0.5/RT)-0.4 kt

So a plot of (3PTo - PT)0.4 versus time should give a straight line with a slope = - 0.4 (0.5/RT)-0.4 k and an intercept of (2PTo)0.4. The integral analysis is performed below.

ii 0 3..:=

vy3ii 931 torr⋅ PTii

−( )0.4:=

vx3ii tii:=

m slope vx3 vy3,( ):= m 0.013−kg0

m0 s1=

inter intercept vx3 vy3,( ):=

inter 12.907 torr0.4=

corr vx3 vy3,( ) 1−=

Y3ii inter m tii⋅+:=

0 500 1000 1500 2000 2500 3000 35007

8

9

10

11

12

13Integral Analysis

Time (s)

(3PT

o - P

T)^

0.4

(tor

r^0.

4)

The straight line fit looks very good (r2 = 1.0) so a 0.6 order rate expression is reasonable. The rate constant is given by

km

0.4−0.5

1.987cal

mole K⋅⋅ 777⋅ K⋅

0.4−

⋅

:=

k 4.711 10 5−×mole0.4

liter0.4 s⋅=

Calculating PTo from the intercept gives

PTointer

10.4

2:=

PTo 299.271 torr=

This is very close to the 310.3 torr we calculated from the t = ∞ data.

(c) Additional data in the time range between 1000 and 300 seconds would be helpful as would data at shorter times (t < 400 s).

(d) If the reaction were run at higher or lower temperatures the order should not change. The rate constant would go up at higher temperatures and down at lower temperatures. Running at lower temperatures would require more time. Running at higher temperatures may make data collection difficult.