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 ( t (min)
0
5
9
15
22
30
40
60
3 C A (mol/dm )
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 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
0 5 9 15 t := ⋅ min 22 30 40 60
2 1.6 1.35 1.1 mole C A := ⋅ 0.87 liter 0.70 0.53 0.35
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 dC A/dt = -kC Aa and take the log of both sides to get ln(-dC A/dt) = ln(k) + a ln(C A) Thus a plot of ln(-dC A/dt) versus ln(C A) 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 dCA
ii
:=
( CA +
ii 1
− CA
)
ii− 1
( tii+1 − tii−1)
0 −0.072 −0.05 mol dCA = −0.037 liter⋅ min 0.027 − −0.019 −0.012 Now take the logs of this data and the corresponding C A terms and put them into vy and vx vectors, respectively.
⋅ liter mole min⋅ liter vyii−1 := ln ( −dCA) ⋅ ii mole vxii−1 := ln CA
ii
Doing the regression gives n := slope( vx , vy)
n
lnk lnk := inte interc rcep eptt( vx, vy) corr( vx, vy)
= 1.597
lnk
= 0.998
= −3.424
ij := 0 .. 5 Yij := lnk
+ n⋅ vxij
Differential Analysis 2.5
3
) t d / A C d ( n l
3.5
4
4.5
0.8
0.6
0.4
0.2 ln(CA)
0
0.2
0.4
0 .6
let's say that the reaction is 1.6 order. The reaction rate expression then then is given by dC A/dt = -kC A1.6 Separating variables and integrating gives C A-0.6 - C Ao-0.6 = 0.6kt Thus a plot of C A-0.6 versus time should give a straight line with a slope of 0.6k and an intercept of C Ao-0.6. The analysis and the plot are shown below below i := 0 .. 7 vyyi := CA
( )
− 0.6
i
kp := slop slopee( t , vyy) k :=
kp 0.6 0.6
k = 0.034
liter 0.6
mol
⋅ min 1
C Ao := intercept intercept( t , vyy vyy) C Ao = 2.06
Ycalc
i
:=
− 0.6
mole liter
intercept( t , vyy)
vxxi := CA
( ) i
− 0.6
+ 0.6⋅ k⋅ t i
Integral Analysis 0.03
0.025
6 . 0 ^ A C
0.02
0.015
0.01
0
10
20
30
40
50
60
Time (min)
For a 1.6 order reaction the rate constant would be 0.034 liter 0.6/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). mole/liter). The best place for additional data 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 Total Pressure (mmHg) 408
777 488
1195 562
3155 799
infinite 931
(a) Why do you think the total pressure measurement at t = 0 is 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 the reaction were run at a higher or lower temperature? SOLUTION The reaction rate law can be written as dC A/dt = -kC Aa 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 = n A + nB + nC + nD where B = CH4, C = H2 and D = CO. We also know that nB = nC = nD = n Ao - n A This gives nT = 3n Ao - 2n A or n A = 0.5( 3n Ao - nT)
Dividing by V gives C A = 0.5(3C Ao - CT) Substitute this into the rate expression to get - 0.5(dCT/dt) = - k (0.5)a(3C Ao - CT)a Then replace CT and C Ao by using the relationship between Ci and Pi (Ci = P i/RT) to get (- 0.5/RT) (dP T/dt) = - k (0.5/RT)a (3 P Ao - PT)a But since the initial charge was only A, P Ao = PTo so we finally get dPT/dt = (0.5/RT)a-1 k (3 P To - PT)a (a) In order to use this expression 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 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 = ∞ C A = 0 = 0.5(3C Ao - CT) or CT = 3 C Ao so PT = 3 P Ao = 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).
390 777 t := ⋅s 1195 3155 408 488 PT := ⋅ torr 562 799 ii := 0 .. 2 dPT
ii
:=
PT
ii+ 1
tii+1
− PT
ii
− tii
0.207 torr dPT = 0.177 s 0.121 s vx2ii := ln dPT ⋅ torr 931) − PT ⋅ 1 vy2ii := ln ( 931 torr ii
ii
a := slop slopee( vx2 vx2 , vy2) a
= 0.618 Y2ii := intercept intercept( vx2, vy2)
corr( vx2, vy2)
= 0.978
+
slope( vx2, vy2) ⋅ vx2ii
Differential Analysis 6.3
6.2
) t d / T P d ( n l
6.1
6
5.9
5.8
2.2
2.1
2
1.9
1.8
1.7
1.6
1 .5
ln(3PTo - PT)
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 (2P To)0.4. The integral analysis is performed performed below. ii := 0 .. 3 vy3ii := 931⋅ torr − PT
(
)
0.4
ii
vx3ii := tii 0
m := slop slopee( vx3 vx3, vy3)
m
= −0.013
kg
0 1
m s inter inter := inte interce rcept pt( vx3 vx3, vy3 vy3)
0.4
inter inter = 12.90 12.907 7 torr torr
corr( vx3, vy3) Y3ii := inte nter
= −1
+ m⋅ tii Integral Analysis
13
12
) 4 . 0 ^ r r o t ( 4 . 0 ^ ) T P o T P 3 (
11
10
9
8
7
0
500
1000
1500
2000
2500
3000
3500
Time (s)
The straight line fit looks very good (r 2 = 1.0) so a 0.6 order rate expression is reasonable. The rate constant is given by
k :=
m
−0.4⋅
1.987⋅ cal ⋅ 777⋅ K mole⋅ K 0.5
− 0.4
0.4
k = 4.711 ×
− 5 mole 10 0.4 liter ⋅ s
Calculating PTo from the intercept gives 1 0.4
PTo :=
inter 2
PTo = 299.271 299.271 torr 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 300 300 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 down at lower temperatures. Running at lower temperatures would require more time. Running at higher temperatures may make data collection collection difficult.
