Chapter 5 Collection and Analysis of Rate Data Two common types of reactors for obtaining kinetics data: Batch & Differential reactors Seven methods for data analyses: (1) Differential method; (2) Integration method; (3) Half-life method; (4) Initial-rate method; (5) Linear regression method; (6) Non-linear regression method; (7) Excess method.
§ Batch Reactors
Differential method (A) For constant V
dC A dC A α − r = k C = − ⇒ ln − = ln k A + α ln C A A A A A → P, dt dt ln(−
dC A dt
ln(−
S =α
ln(CA)
dC A ) dt
−
dC A dt
dC A dt P kA = (C AP )α −
P
CAP
ln(CA)
dC A ln C = 0 ⇒ − = − rA = k A (intercept) A Remark: Let dt To obtain − (a)
dC A dt
graphical, numerical, or polynomial fit:
Graphical method CA
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t
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∆CAC /∆tA
The area under ∆CA/∆t is the same as that under the curve.
dC A from the curve Read estimates of dt (b)
Numerical method CA t
t
CA0 CA1 CA2 t0
t1
t2
CA3 CA4 … t3
t4 …
By three-point differentiation formulas Initial：
dC A dt
= t0
1 ( −3C A0 + 4C A1 − C A 2 ) 2 ∆t
dC A dt
Interior：
dC A dt
Final： (c)
= ti
= tf
1 (C Ai +1 − C Ai −1 ) 2 ∆t 1 (C Af − 2 − 4C Af −1 + 3C Af ) 2∆t
Polynomial fit
Ca = a0 + a1t + a2t 2 + LL + an t n ⇒ Least-square fitting ⇒ a0, a1, a2, …, an can be obtained!
dC A ∴ = a1 + 2a2t + 3a3t 2 LL + nan t n −1 dt Remarks: (i) The order of polynomial have to be carefully chosen. (See Levenspiel, p.65, ex3-2 & p.259-260) (ii) Excess + Differential method.
−
dC A α β = −rA = kC A C B dt
(a ) C B 0 >> C A0 ⇒ C B ≈ C B 0 = constant α
β
β
α
⇒ − rA = kC A C B = kC B 0 C A = k 'C A
α
(b) C A0 >> C B 0 ⇒ C A ≈ C A0 = constant α
β
α
β
⇒ − rA = kC A C B = kC A0 C B = k ''C B From (a) & (b) ⇒ α, β can be obtained ⇒ obtain k! (B) For total pressure-time data
For gas-phase reaction
P T V = (1 + ε X A )( 0 )( ) V0 P T0
At V = constant, and isothermal operation ⇒ P = P0 (1 + ε X A ) ⇒ X A = Since − rA = − = ∴ − rA =
P − P0 εP0
dC A dX A PA0 d P − P0 = C A0 = ( ) dt dt RT dt εP0
PA0 dP PA0 dP N = , (ε = δ A0 = δy A0 ) RTεP0 dt RTδ y A0 P0 dt NT 0
1 dP α = kC A LLL ( A) δ RT dt
β
又 Q C A = C A0 (1 − X A ) = 代入式( A), ⇒
PA0 P − P0 P − [( P − P0 ) / δ ] (1 − ) = A0 RT RT εP0
dP = f ( P) ( See Ex. 5.1, p.260) dt
(C) For constant pressure operation
QV ≠ Constant , V = V0 (1 + ε X A )(
P0 T )( ) P T0
At constant P & isothermal ⇒ V = V0 (1 + ε X A ) ⇒ CA =
N A N A0 (1 − X A ) (1 − X A ) = = C A0 V V0 (1 + ε X A ) (1 + ε X A )
∴ −rA =
− 1 dN A −1 N A0 d (1 − X A ) = V dt V0 (1 + ε X A ) dt
=
C A0 dX A (1 + ε X A ) dt α
α
Since − rA = kC A
C A0 dX A α 1− X A = = kC A0 (1 + ε X A ) dt 1+ ε X A α
dX A α −1 1 − X A ⇒ = kC A0 (1 + ε X A ) dt 1+ ε X A 1
1− X A dX A 1 = α ln + (α − 1) ln C A0 + ln k ⇒ ln (1 + ε X A ) dt 1+ ε X A
XA
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t
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XA Plot XA vs. t
dX A
Slope = dt
t dX A dt
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1 dX A 1 + εX A dt
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XA
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1− X A 1 + εX A
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1 dX A ln( ) 1 + εX A dt
Slope = α
1− X A ln( ) 1 + εX A
Find slope = α, From intercept
(α-1) ln(CA0)+ln(k)
k obtained!!
Integral Method Key concepts: (1) The integral method is used most often when the reaction order is known. (2) It is desired to evaluate the specific reaction rate constants at different temperatures to determine Ea. (3) We are looking for the appropriate function of concentrations corresponding to a particular rate law that is linear with time.
(A) Constant-V Batch Reactors
− 1 dN A dC = − A = k f (C A ) V dt dt C A dC dC A A ∴− = kdt ⇒ − ∫ = kt C A 0 f (C A ) f (C A ) − rA =
−∫
CA
C A0
dC A f (C A )
Slope= k
Remark: The integral method uses a trial-and-error procedure to find the reaction order!
t
k A → P For
(i) zero-order reactions:
CA
Slope = -k
− rA = k ∴
dC A = −k ⇒ C A = C A0 − kt dt
t
(ii) first-order reactions:
− rA = kC A ∴−
dC A dC = kC A ⇒ − A = kdt dt CA
⇒ ln
ln
C A0 CA
C A0 = kt CA
N A N A0 (1 − X A ) = C A0 (1 − X A ) = V V dC dX A ⇒ − A = C A0 = kC A = kC A0 (1 − X A ) dt dt XA dX A dX A ∴ = kdt ⇒ ∫ = kt 0 (1 − X A ) (1 − X A ) Since C A =
⇒ ln
Slope = k
1 = kt ( see P.132, ex 4 − 1) 1− X A
t
ln
1 1− X A
S=k
t
(iii) second-order reactions:
− rA = kC A
1 CA
2
C A dC dC A 2 A = kC A ⇒ ∫ = − kdt ∴− 2 C A0 C dt A
⇒
1 1 − = kt C A C A0
1 1 ⇔ − = kt C A0 (1 − X A ) C A0
S=k 1 C A0
XA 1− X A
1 XA ⇒ C A0 kt = −1 = 1− X A 1− X A
Slope= CA0 k t
For A + B → P, -rA = kCACB (iv) QV = constant ∴ CB = C A0 (θ B −
t
b X A) a
CB 0 = C A0 (θ B − X A ), θ B = C A0
dC A dX A 2 = C A0 = kC A0 (1 − X A )(θ B − X A ) ∴ − rA = − dt dt XA 1A dX ⇒∫ = C A0 kt 0 (1 − X A )(θ B − X A ) Let
( Aθ B + B ) − X A ( A + B ) 1 A B = + = (1 − X A )(θ B − X A ) (1 − X A ) (θ B − X A ) (1 − X A )(θ B − X A )
A = −B A+ B = 0 ⇒ B = 1 , where θ B ≠ 1; if θ B = 1, belong − to < iii > ⇒ Aθ B + B = 1 1−θB 1 1 X A θ −1 X A 1−θ B B ⇒∫ dX A + ∫ dX A = C A0 kt 0 1− X 0 θ − X A B A
∴
1 θ − XA 1 ln(1 − X A ) + ln( B ) = C A0 kt 1−θB θB −1 θB
θB − X A ⇒ ln = (θ B − 1)C A0 kt θ B (1 − X A )
(v) Reactions of shifting order dC A k1C A − = − = r A A→R, dt 1+ k 2C A
At high CA → − rA = k1 / k 2 At low CA→ − rA = k1C A
(1) (zero order) (1st order)
C A, 0 ln + k 2 (C A, 0 − C A ) = k1t Integrating equation (1), CA ln C A, 0 / C A
ln
C A, 0 − C A
ln C A, 0 / C A k1t ∴ = −k2 + (C A, 0 − C A ) C A, 0 − C A
S=k1
-k2
t C A, 0 − C A
☆ The other reaction mechanisms for integral methods, see P.41~P.63, Levenspiel. (B) For total pressure-time data (V = V0, T = T0)
Q P = P0 (1 + εX A ) ⇒ X A = and − rA =
P − P0 εP0
P − P0 1 dP 1 (ε + 1) P0 − P , CA = ( PA0 − )= RT δRT dt δ δ RT
(Concept : − rA = kf (C A ), ⇒ f (C A ) = f 2 ( P )) C A = C A0 (1 − X A ) =
PA0 P − P0 P (ε + 1) P0 − P PA0 [(ε + 1) P0 − P] (ε + 1) P0 − P = (1 − ) = A0 = = CA RT RT RT y P RT εP0 εP0 δ δ A0 0
Similarly , b P − P0 b b PB 0 − ( ) (θ Bε + ) P0 − P b a a a δ C B = C A0 (θ B − X A ) = = a RT δ RT c P − P0 c c PC 0 + ( ) (θ C ε − ) P0 + P c a a a δ CC = C A0 (θ C + X A ) = = a RT δ RT C D = C A0 (θ D +
d X A) = a
(See ex. 5.2, P269)
PD 0 +
d P − P0 c d ( ) (θ Dε − ) P0 + P a a δ a = RT δ RT
Steps : (i ) − rA = (ii ) ∫
P
P0
1 dP = kf 2 ( P ) δ RT dt
t dP = kδ RT ∫ dt = kδ RTt 0 f 2 ( P)
(iii ) Plot ∫
P
P0
∫
P
P0
dP f 2 ( P) Slope= kδRT
dP vs. t f 2 ( P)
t
(C) Constant P, T, V ≠ Constant, Batch Reactors
V = V0 (1 + εX A ); C A = C A0
1− X A 1 + εX A
C A0 dX A − rA = 1 + εX A dt
C A0
ε
ln(1 + εX A )
< i > zero − order reactions : − rA = k C A0 dX A =k 1 + εX A dt ⇒ C A0 ∫
XA
0
dX A C = A0 ln(1 + εX A ) = kt 1 + εX A ε
Slope= k
t
< ii > first − order reactions : − rA = kC A 1− X A C A0 dX A = kC A0 1 + εX A dt 1 + εX A
⇒ ln(1 − X A ) = −kt
< iii > second − order reactions : − rA = kC A
2
2
X A 1 + εX C A0 dX A 2 1− X A A ⇒ ∫ = kC A0 dX A = kC A0t 2 0 1 + εX A dt (1 − X A ) 1 + εX A
⇒ (1 + ε )
XA + ε ln(1 − X A ) = kC A0t 1− X A
Method of Initial Rates
if
n
− rA = kC A ⇒ inital − rA0 = kC A0
n
⇒ ln(−rA0 ) = ln k + n ln C A0 Steps: (i) A series of experiments is carried out at different CA0.
必須進行多次實驗
(ii) Plot CA vs. t (iii) Determine –rA0 by differentiating the data and extrapolating to t = 0. (iv) Plot ln(–rA0) vs. ln(CA0).
C A0 C A04 C A03 C A02 C A01
dC A slope = dt
= rA0
ln(−rA0 )
CA0
S=n
ln k t
ln C A0
Remark: When a significant reverse reaction is present, the use of the differential method for data analysis to determine reaction orders and specific reaction rates is unsuitable.
⇔ Initial rate method!
See Ex. 5-4, p. 2778.
Method of Half-Lives Definition t1/2 : half-life of a reaction is defined as the time it takes from the concentration of the reactant to fall to half of its initial value. (A) Constant V
− rA = −
dC A α = kC A , (Constant V ) dt
1− α
⇒ CA
1 −α
− C A0
= (α − 1) kt ,
C α −1 1 1 1 1 A0 α −1 − = − 1 ⇒ t= 1 1 α − α − k (α − 1) C A C A0 kC A0 (α − 1) C A 1 2α −1 − 1 1 at t1 / 2 , C A = C A0 ⇒ t1 / 2 = k (α − 1) C A0α −1 2 Similarly , t1 / n
nα −1 − 1 1 2α −1 − 1 = ,∴ ln t1 / 2 = ln + (1 − α )lnC C A0A0 k (α − 1) C A0α −1 k (α − 1)
Steps: (i) Conduct CA vs. t, (different CA0)
必須進行多次實驗
(ii) Get table. C A0
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t1/ 2
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(iii) Plot ln t1/ 2 vs. ln C A0 .
ln t1/2 S = 1-α ln
2α −1 − 1 k (α − 1)
ln C A0
2α −1 − 1 α = 1 - S; and from intercept, I = ln k (α − 1) ,
(iv) from slope, S = 1 - α,
k.
Remark: If two reactants are involved in the chemical reaction, use the method of excess in conjunction with the method of half lives. (B) For total pressure-time data (ε + 1) P0 − P (ε + 1) P0 − P 1 dP α Q CA = ⇒ − rA = = kC A = k δ RT δ RT dt δ RT P t dP k kt dt ⇒∫ = = P0 [(ε + 1) P − P ]α (δ RT )α −1 ∫0 (δ RT )α −1 0 1 1 α − 1 (ε + 1) P0 − P
α −1 P
1 ⇒ ( ε 1 ) P P + − 0
α −1
εP0 ⇒ (ε + 1) P0 − P
α −1
At P =
= P0
kt (δ RT )α −1
1 − εP0
α −1
= (α − 1)
kt (δ RT )α −1
α −1
εP − 1 = 0 δ RT
(α − 1)kt
1 P0 ⇔ t = t 1' / 2 ; or P = 2 P0 ⇔ t = t 1''/ 2 2
α
α −1
α −1
2ε ' εP0 ⇒ − 1 = (α − 1)kt 1 / 2 2ε + 1 δ RT 2ε α −1 α −1 − 1 (δ RT ) 2ε + 1 1−α ' P0 ⇒ t 1/ 2 = α −1 (α − 1)kε
⇒ ln t 1' / 2
2ε α −1 α −1 − 1 (δ RT ) 2ε + 1 = (1 − α ) ln P0 + ln (α − 1)kε α −1
Similarly , ln t 1''/ 2
ε α −1 α −1 − 1 (δ RT ) ε + 1 = (1 − α ) ln P0 + ln (α − 1)kε α −1
Steps: (i) P
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P0
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T
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t 1' / 2
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(ii) Plot ln(t1/2’) vs. ln(P0),
k, α obtained!
P0
ln t 1' / 2
P01 P02 P03 P04
Slope= 1-α
k
t '
t1/ 2
Method of Excess −
A+B→P
dCA = − rA = kC Aα CBβ dt
(i) CB,0 >> CA,0
ln P0
(Constant V)
=> CB ≈ CB,0 = constant
α β β α α β β ∴ − rA = kC A CB = kCB ,0C A = k ′C A , where k ′ = kCB = kCB,0
Applying “differential, integral, initial or half-lines methods” to obtain α (ii) CA,0 >> CB,0
=> CA ≈ CA,0 = constant
∴ −rA = kC αA CBβ = kC αA,0CBβ = k ′′CBβ , where k ′′ = kC αA ≈ kCαA,0 β can be determined !
§ Differential Reactors (initial rate) A differential reactor consists of a tube containing a very small amount of catalyst usually arranged in the form of a thin wafer or disk. Properties: (i) Conversion is extremely small (ii) Reactant concentration through the reactor is essentially constant (iii) No concentration gradient (iv) Heat release is small => isothermal (v) No by pass or channeling (vi) No catalyst decay !
FA,e
FA, ∆L
catalyst
′
Mass balance FA, 0 − FA, e + γ A ⋅ w = 0
V0C A, 0 − VC A, e FA, 0 X A ∴ − rA ' = = w w If
FA, 0 X A Fp ⇒ = A → P (stoichiometric ratio = 1), w w
(Fp: product flow rate)
V0 (C A, 0 − C A,e ) V0C p = For constant volumetric flow, V0 = Ve = V, − rA = w w ′
As w → 0, ⇒ C A.b =
V0 → ∞ ⇒ CA,0 - CA,e → 0
C A, 0 + C A, e or CA,b ≈ CA,0 , where CA,b: bed concentration (see Ex 5.5, p.284) 2
§ Least-Square Analysis dC A α β − = − γ = kC A A CB For a constant-volume batch reactor, dt dC ⇒ − A t = 0 kC αA, 0CBβ, 0 dt dC A ⇒ ln − = ln k + α ln C A, 0 + β ln CB , 0 dt t =0 ⇒ Y = a0 + a1 X 1 + a2 X 2 =>Using multiple linear regression
(利用統計)
(see Ex.5-5 P. 251) for the nonlinear least-squares analysis γc : rxn rate estimated from calculation γm : rxn rate measured
(
L = ∑ γ C i − γ mi
) ; 2
σ2 =
Minimize L/N-K
s2 L = N −K N −K
(see Ex. 5.6 P. 255) (先看 p. 253 Table 5-2) N: # of runs , K: # of parameters
Remarks:
(i)
We can compare the residual plots for model checking (是否缺少某一參數，比較不同 model 之適切性…)
(ii) The model with physical meanings is preferred! (iii) It is also possible to use the nonlinear regression to determine the rate law parameters from conc.-t data!
