Reversible First-Order Reactions - mywebmyweb.scu.edu.tw/~kguo/pchem3/9502/L05_Ch18.pdf ·...

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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• 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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• 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+

=

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• 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 +=

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• 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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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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• 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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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 ==

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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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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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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.

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• 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 =

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• 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

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• 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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• At the steady state, the concentrations of [B] at tm is


[ ] [ ] ( )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

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• At the steady state, the concentration of [C] at tm is:


[ ] [ ] [ ] [ ] 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

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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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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

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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

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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.

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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.

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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.

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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 /=

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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

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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.

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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.

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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 +=

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• 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


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

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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)

Reversible First-Order Reactions - mywebmyweb.scu.edu.tw/~kguo/pchem3/9502/L05_Ch18.pdf ·...

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