reversible first-order reactions - mywebmyweb.scu.edu.tw/~kguo/pchem3/9502/l05_ch18.pdf ·...
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2006/03/17 51
Reversible First-Order Reactions
• If a reaction does not go to completion, the rate law is of the form υ=υf –υb, where is υf the rate of the forward reaction and υb is the rate of the backward reaction.
• A reversible process in kinetics means the backward reaction is significant.
• The effect of a product on the reaction rate may be determined by adding it initially.
• The products may be inhibitory. A reaction is said to be autocatalytic if a product of the reaction cause it to go faster.
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Reversible First-Order Reactions
• For a reversible reaction A ⇌ B the rate law is of the form
where kf is the rate constant of the forward reaction and kb is the rate constant of the backward reaction.
• If both A and B are present initially, [A]o = [A] and [B]o = [B] then [B] = [A]o +[B]o – [A] as a boundary condition:
• The plot of –ln (kf[A] -kb[B]) versus time is linear, and the slope is used to calculate (kf + kb):
[ ] [ ] [ ] [ ] [ ]( ) [ ] [ ]( ) ( ) [ ]A kkBAk- A-BAk - Akd
A d- bfo o bo o bf +++=+=t
( ) [ ] [ ] [ ]( )
++
+= o o bf
bbf BA
kkk-A kk
[ ] [ ][ ] [ ] ( )tbf
o bo f
bf kk- Bk-AkBk-Ak ln +=
[ ] [ ] [ ]Bk-AkdAd- bf==t
υ
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Reversible First-Order Reactions
• If initially only A is present, [A]o = [A] and [B]o = 0 then [B] = [A]o – [A] is a boundary condition,
• The expression for [A]eq is obtained as follow:
• The equilibrium constant K can be used to eliminate kb.
[ ] [ ] [ ] [ ]( ) [ ] ( ) [ ]A kkAk A-Ak A-kd
A dbfo bo bf +−=+=
t( ) [ ] [ ] ( ) [ ] [ ]( )eq b f o
b f
b b f AA kk- A
kkk-A kk- −+=
++=
[ ][ ]
[ ] [ ][ ] eq
eq o
eq
eq
b
f
AA-A
AB K
kk === [ ] [ ] o
b f
b eq A
kkk A+
=
2006/03/17 54
Reversible First-Order Reactions
• Since
• The reaction has an initial rate of kf[A] and slows down as B accumulated. When [B]eq/[A]eq = K. The reaction is at equilibrium and the rate is zero.
• The integration is obtained as follow:.
[ ] [ ] [ ] [ ] [ ][ ]
=+=K A
B-1 A-kBKkA-k
dA d
ff
ft
[ ] ( ) [ ] [ ]( )eq b f AAkk- dtA d −+=
[ ] [ ] ( )∫∫ +=t
0 bf[A]
[A]eq
dkk- A-A
[A] do
t
Kkk
b
f =
[ ] [ ][ ] [ ] ( )t kk-
A-AA-A
ln b f eq o
eq +=
[ ] [ ]( ) [ ] [ ]( ) ( )t kk-eq o eq b f e A-AA-A +=
2006/03/17 55
Reversible First-Order Reactions
• The concentrations of A and B will be halfway to their equilibrium values in a time of ln(2)/(kf + kb)
• The plot of –ln([A] - [A]eq) versus time is linear, and the slope is used to calculate (kf + kb).
• The integration of [A] and [B] are in exponential forms and the boundary condition are [A] + [B] = [A]o
[ ] [ ] ( )( )
[ ] o bf
kk-fbkk-
b
f
bf
o b Akk
ekkekk1
kkAk A bf
bf
++=
+
+=
++
tt
[ ] [ ] ( )[ ]( )
[ ] o bf
kk-ffkk-
bf
o f Akk
ek-ke-1kk
Ak Bbf
bf
+=
+=
++
tt
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Reversible First-Order Reactions
Fig. 18.5 Reversible first-order reaction starting with A at concentration [A]o. The values of the rate constants are k1 = 3 s-1 and k2 = 1 s-1 .
B A 1k
2k
→←
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Experimental Kinetics and Gas Reactions
6 The approach of concentrations to their equilibrium values for a reaction A ⇄ B that is first-order in each direction, and for which k = 2k ’.
[ ]( )
[ ] o
'-
A 'e ' Akk
kk tkk
++=
+
B A '
→←
k
k
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Consecutive First-Order Reactions
• For a consecutive irreversible reaction represented by:
the change of concentrations depend on time, rate law is these simultaneous differential equations:
• If initially only A is present, [A]o = [A], [B]o = 0 and [C]o = 0 at t=0, then [A]o = [A]+ [B]+[C] as a boundary condition:
[ ] [ ]A-kd
A d1=
t
CBA 21 kk → →
[ ] [ ] [ ]Bk-Akd
B d21=
t[ ] [ ]Bk
dC d
2+=t
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Consecutive First-Order Reactions
• First, the integration form for A is : • Substituted into
• Integrated to
• Now
[ ] [ ] t -ko 1eAA =
[ ] [ ] [ ] [ ] [ ]Bk-eA kB k-A kd
B d2
k-o 121 1t
t==
[ ] [ ] [ ]tt 21 k-k-
12
1o e-e
k-kk AB
=
[ ] [ ] [ ] [ ]ttt 211 k-k-
12
1o 2
k-o 1 e-e
k-kkA k-eA k
dtB d
=
[ ] [ ]tt k-2
k-1
12
1o
21 e k-e k k-k
kA-
=
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Consecutive First-Order Reactions
To solve an ODE form like
First substitute with y(t ) = [B], r(t ) = k1[A]o e-k1t , b = k2 ,
Now as
Use integration factor F = eh = exp( ∫ b dt ) = ebt
eh (y’ + b y) = eh r(t )
or y eh = ∫ eh r(t ) dty = e-h [ ∫ eh r(t ) dt ]
= e-bt [∫ ebt r(t ) dt ]= e-bt ∫ ebt r(t ) dt
[ ] [ ] [ ] [ ] [ ]Bk-eAkBk-Akd
B d2
tk-o 121 1==
t
( ) ( )tr e dey d h
h =
t
( ) ( ) )y( b -r d
)y(dy' ttttt ==
2006/03/17 61
Consecutive First-Order Reactions
Now y(t ) = e-bt ∫ ebt r(t ) dtsubstitute with b = k2, r(t ) = k1 [A]o e-k1t
B(t ) = e-k2t ∫ e k2t k1 [A]o e-k1t dt= e-k2t ∫ e (k2 - k1)t k1 [A]o dt
Since [B]o = B(0) = 0
[ ] ( ) e c eA k-k
k e k- k-k o
12
1 k- 2122 ttt +
=
[ ] e c eA k-k
k e 0 0 o
12
10 +
=
( ) [ ] ( ) [ ]
[ ] ( )tt
tttt
k- k-
12
1o
k- o
12
1 k-k o
12
1 k-
21
2122
e-e k-k
kA
eA k-k
k- eA k-k
k e B
=
=
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Consecutive First-Order Reactions
The concentration of [C] is:
[ ] [ ] [ ] [ ] [ ] [ ] [ ] ( )ttt k- k-
12
1o
k-o o o 211 e-e
k-kkA-eA-A B-A-A C
==
[ ] ( )
[ ] [ ] [ ] [ ]tt
tt
k-
12
1o
k-
12
2o
k-1
k-2
12o
21
21
e-1 k-k
kA - e-1 k-k
kA
ek-ekk-k
1-1A
=
=
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Consecutive First-Order Reactions
Fig. 18.6 Plots of concentrations of reactants in A1 → A2 → A3 when k1 = 1 s-1 and k2 is (a) 1, (b) 5, and (c) 25 s-1. These plots could be calculated using equations but they were actually calculated by solving the three simultaneous differential equations that describe the system.
2006/03/17 64
Consecutive First-Order Reactions
• The depletion of A does not depend upon B and C, it can be solved exactly:
• Accumulation of B approached its maximum in the induction period.
• If the production of C is faster than the rate of accumulation of the intermediate B, i.e., k2 » k1. The concentration of the intermediate does not have sufficient time to build up. Consequently, its concentration remains small throughout the reaction.
[ ] [ ] t-ko 1eAA =
2006/03/17 65
Consecutive First-Order Reactions
• In the steady state, the rate of change of [B] is zero at t = tm, so
• The steady state of [B] is: . Also
• So, the time for [B]ss is
[ ] [ ] [ ] [ ] [ ] 0B k-eA kB k-A kd
B d2
k-o 121 1 === t
t[ ] [ ] t k-
o 2
1SS 1eA
kkB
=
[ ] [ ] [ ] [ ]
[ ] [ ]tt
tttt
k-2
k-1
12
1o
k- k-
12
1o 2
k-o 1
21
211
e k-e k k-k kA -
e-e k-k k Ak-eA k 0
dB d
=
==
m -k
2 -k
1 at e ke k 21 tttt == m22m11 k-k lnk-kln tt =
=
2
1
21m k
k lnk-k
1t
2006/03/17 66
Consecutive First-Order Reactions
• At the steady state, the concentration of [A] at tm is:
• And further more, we have the
[ ] [ ] [ ]
==
2
1
21
1o m1o SS k
k lnk-k k-A lnk-A lnA ln t
[ ] [ ] [ ] kkA A or
kkA ln
12
1
12
1k-k
k
2
1o SS
k-kk
2
1o
=
=
[ ] [ ]
=
+=
12
1k-k
k
2
1o
2
1
12
1o k
kA lnkk ln
k-kk A ln
[ ] [ ] m1 -ko SS eAA t=
=
2
1
21m k
klnk-k
1t
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Consecutive First-Order Reactions
• At the steady state, the concentrations of [B] at tm is
• And further more, we have the
[ ] [ ] ( )m2m1 k- k-
12
1o SS e-e
k-kk AB tt
=
[ ] m1 k-
2
1o e
kkA t
=[ ]
= m1m1 k-
2
1 k-
12
1o e
kk-e
k-kkA tt
[ ] [ ] [ ]
+=
=
2
1
21
1
2
1o m1
2
1o SS k
k ln k-k
k-kk ln A ln k-
kkA lnB ln t
[ ] [ ] [ ] kkA B or
kkA ln
12
2
12
2k-k
k
2
1o SS
k-kk
2
1o
=
=
[ ] [ ]
=
=
21
2k-k
-k
2
1o
2
1
21
2o k
kA lnkk ln
k-kk- A ln
2006/03/17 68
Consecutive First-Order Reactions
• At the steady state, the concentration of [C] at tm is:
• And further more, we have the
[ ] [ ] [ ] [ ] B-A-AC o SS =
[ ] [ ]
−
−=
21
2
21
1k-k k
2
1k-k k
2
1o SS k
kkk A C 1
[ ] ( )
+= m1k-
2
12o e
kkk-1 A t
[ ] ( )
= m1m1 k-
2
1k-o e
kk-e-1 A tt
2006/03/17 69
Experimental Kinetics and Gas Reactions
12 The concentrations of A, I, and P in the consecutive reaction scheme A → I → P. The curves are plots with k1 = 10 k2. If the intermediate I is in fact the desired product, it is important to be able to predict when its concentration is greatest.
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Consecutive Reversible First-Order Reactions
Fig. 18.7 Plots of concentrations of reactants in a consecutive reactions that do not go to completion:
when k1 = k2 = k4 = 1 s-1 and k3 is (a) 1, (b) 3, and (c) 9 s-1. These plots were calculated by solving the three simultaneous differential equations that describe the system.
321 A A A 3k
4k
1k
2k
→ ←
→ ←
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Experimental Kinetics and Gas Reactions
Example 18.3 Parallel ReactionsDerive equations for the concentrations of B and C as functions of time as they are produced in the parallel reactions
k1
A −−→ Bk2 ↘
CThe rate equation for A is
–d [A]/dt = k1 [A] + k2 [A] = (k1+ k2) [A]Thus, the disappearance of A will be first order, and on the basis of the earlier discussion of first-order reactions we can write
[A] = [A]o e-(k1+k2)t
The rate equation for B is –d [B]/dt = k1 [A] = k1 [A]o e-(k1+k2) t
2006/03/17 72
Experimental Kinetics and Gas Reactions
Example 18.3 Parallel ReactionsIntegration yield
If [B]=0 at t = 0, the constant is
Now
Thus the fraction of A that is converted to B at infinite time is k1/ (k1+ k2). At any time the sum of [A], [B], and [C] must be equal to the initial concentration of A, [A]o. Consequently, if [C]o = 0, then
[ ] [ ] ( ) [ ] kk
kAe kk
kA-B21
1o
kk-
21
1o 21
+
+
+
= + t
[ ] ( )( )t21 kk-
21
1o e-1
kkk A +
+
=
[ ] [ ] [ ] [ ] [ ] ( )( )t21 kk-
21
2o o e-1
kkk A B-A-AC +
+
==
[ ] kk
kA21
1o
+
[ ] [ ] ( ) constante kk
kA-B 21 kk-
21
1o +
+
= + t
2006/03/17 73
Microscopic Reversibility and Detailed Balance
Fig. 18.8 (a) A particle with coordinates r(0) and velocity vector υ(0) moves under the action of a force to position r(t1), where it has the velocity vector v(t1). (b) We imagine that the direction of motion is reversed and the clock is set to read -t1. At t = 0, the particle will have returned to its original position, but its velocity components are reversed.
2006/03/17 74
微觀可逆性原理
• 如果一個反應的正向反應為一個基元反應,則其逆向反應也必為一個基元反應,而且其正逆向的反應進行時必定通過同一個過渡態。
• 例如: 2NH3 = N2 + 3H2
• 逆向反應為典型的複雜反應,因一個基元反應中沒有四分子的反應,因此可直接判定正向反應亦為一個複雜反應.
2006/03/17 75
Collision Model
•Key Idea: Molecules must collide to react.
•However, only a small fraction of collisions produces a reaction. Why?
•Arrhenius: An activation energy must be overcome.-Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy).
-Orientation of reactants must allow formation of new bonds.
2006/03/17 76
Arrhenius Equation
• The dependence of rate constants on temperature over a limited range can usually be represented by an empirical equation proposed by Arrhenius in 1889:
• k = rate constant• A = pre-exponential factor (same units as the k) • Ea = activation energy• T = temperature• R = gas constant
RΤΕaΑek /=
2006/03/17 77
Effect of Temperature
• In logarithmic form:
• In the Arrhenius plot, a straight line with a slope of –Ea/R is obtained when the logarithm of the rate constant is plotted against the reciprocal of the absolute temperature T.
• If differentiating equation with respect to the temperature yields:
• The definition of the activation energy Ea may be regarded as
• The rate constants between two temperature are
TEAkR
- ln ln a=
2a
R
dd
TE
Tkln =
=
TklnTE
d dR 2
a
=
21
12a
1
2
-
R ln
TTTTE
kk
2006/03/17 78
Experimental Kinetics and Gas Reactions
10 The Arrhenius plot of ln kagainst 1/T for the decomposition of CH3CHO, and the best straight line. The slope gives −Ea/R and the intercept at 1/T = 0 gives ln A.
2006/03/17 79
Experimental Kinetics and Gas Reactions
Fig. 18.9 Plot of ln k versus 1/T for the decomposition of N2O5 from which the Arrhenius activation energy EA may be calculated.
2006/03/17 80
Determination of the activation energy
Example 18.4 Determination of the activation energyThe rate constants for the first-order gas reaction are as follows:
T/K 273 298 308 318 328 338k/10-5s-1 0.0787 3.46 13.5 49.8 150 487
What are the values of the activation energy and the pre-exponential factor?
Ans: The plot of ln k versus 1/T is given in Fig. 18.9. The plot of points was fit by the least squares methods, which yielded a slope of 12.375 K-1 and an intercept of 31.27. Therefore, the activation energy Ea = (12.375 K-1) (8.3145 J K-1 mol-1) = 103x103 J mol-1. The pre-exponential factor is given by exp(31.27) = 3.96x1013 s-1. Thus equation becomes ( ) ( )
=
Tmol K J 8.3145103x10-exp s 3.96x10k 1-1-
3 1-13
2252 O212NO ON +=
2006/03/17 81
• For a bimolecular reaction: A+A = P, since the collision density ZAA is
• The number of effective collisions based on Maxwell Boltzmann distribution is ZAA exp(-Eo/RT), and the rate of disappear of A in unit volume is:
• Compare with Arrhenius equation:
/RT -E A
oeMRTdNk π222 =
[ ] 2 2 2A
2 2
A2 2AA A
MRT d2N
M 8RTd
VnN
21d
21Z π
ππυπρ =
==
[ ] /RTE- 2 2 2A
/RTE- AA
oo eAMRT d-2NeZ -
dd[A]
21- π==
t
2006/03/17 82
Experimental Kinetics and Gas Reactions
Example 18.5 Curved Arrhenius plotsFor a reaction that follows equation 18.70, what are the Arrhenius parameters at temperature T?
Ans: The arrhenius activation energy may be obtained as follows:
Thus Eo is a hypothetical activation energy at absolute zero. Substitution of this relations yields:k = a Tm em e-Ea/RT
Thus, for pre-exponential factor A at temperature T’ is equal to a (T’) m em. When the Arrhenius plot is curved and Ea is calculated at two temperatures, Eo and m can be calculated. Then the factor acan be calculated from the rate constant at either temperature.
TETakR
-ln m ln ln 0+=
mRTERT
mRd
dR o2o22
a +=
+=
=
TET
TklnTE
2006/03/17 83
Arrhenius: RTEa
Aek −=
碰撞: RTE
AARTE
A
CC
eZPeπµRTNdPk
−−== 822π
過渡態: RTH∆
RS∆
Brr
eehTkk
00≠≠ −
=
活化能
RTEE Ca 21+= nRTHE mra +∆= ≠
0
0mrCa H∆EE ≠≈≈常溫下:
P:Steric factor (空間因數)