Chapter Five Rates of Chemical Reaction

Chemical

Reaction

Weather can occur ?

---chemical thermodynamics

How fast (reaction rate)?

---chemical kinetics

反应机理：化学反应所经历的途径或具体步骤， 又称为反应历程。

化学反应速度的快慢 ——主要决定于反应的内在机理。

化学动力学的基本任务就是研究反应的机理。

某一反应究竟是经过哪些步骤完成的， 了解各个步骤的特征和相互联系， 揭示化学反应速度的本质，使人们能够

自觉控制反应速度。

反应机理可以告诉我们：

问题

1. 为什么有些口服药物的服用方法是 2 次 / 天？

3 次 / 天 ?

2. 为什么静脉滴注某些青霉素类药物要加快滴注的速度？

二、酶作为生物催化剂的特性：

2.

)

5-1 Rates and Mechanisms of Chemical

Reactions

5-2 Theories of Reaction Rate

5-3 Reaction Rates and Concentrations

5-4 Effect of Temperature on Reaction

Rates

5-5 Effect of Catalyst on Reaction Rates

Chapter Five Rates of Chemical Reaction

Reaction rate: changes in a concentration of a product or a reactant per unit time.

t

cv

The rate is defined to be a positive number.

The unit of reaction rate is mol·L-1·S-1,

mol·L-1·min-1

or mol·L-1·h-1 et al

5-1 Rates and Mechanisms of Chemical Reactions

average reaction rate is obtained

by dividing the change in

concentration of a reactant or

product by the time interval over

which the change occurs

4)-(5 t

cv

Define reaction rate :

2N2O5 = 4NO2 + O21. Average reaction rate

SAMPLE EXERCISE : H2O2 = H2O + 1/2 O2

0 〃 : c1(H2O2) = 15.88×10-3 mol/L

5 〃 : c2(H2O2) = 12.8×10-3 mol/L

)(106.05

100.3)( 113

3

12

1222

sLmoltt

cc

t

cOHv

)(103.05

105.1)( 113

3

12

122

sLmoltt

cc

t

cOv

So:

Reaction Rates and Stoichiometry

H2O2 = H2O + 1/2 O2

t

O

t

OH

t

OHrate

][

2

1][][ 2222

用不同物质浓度变化所表示的反应速率之间存在着一定的关系是：它们之间的速率比正好等于反应式中各物质分子式的系数之比，

To generalize, for the reaction

aA + bB cC + dD

This equation can be used to establish the relationship between rate of change of one reactant or product to another reactant or product.

You have to be able to do this on the test, too!

t

c

t

2. Instantaneous reaction rateThe instantaneous rate of change at a point is the same as the slope of the tangent line. That is, it's the slope of a curve.

V = lim —————— = -————

- Δc

Δt0 Δt

dc

dt

c

v

dt

Odv

dt

OHdv

][

][ 222

Define reaction rate : 2N2O5 = 4NO2 + O2

5-1.2 The Mechanisms of Chemical Reactions

A reaction mechanism is a description of the path that a reaction takes.

elementary reaction ？overall reaction ？types of elementary reactions

5-1.2 The Mechanisms of Chemical Reactions

elementary reaction: is one that the reactants can convert directly into the products in a single step when they act each other. (one-step reaction)

基元反应：反应物间相互作用直接转化为生成物的反应。 （简称元反应）。

MClM2Cl

ClHClClH

HHClH Cl

M2ClMCl

2

2

2

2

例如：

overall reaction: Many reactions are actually made up of several elementary steps, which are combined to yield the overall reaction.

NO2 + CO → NO + CO2

NO2 + NO2 → NO3 + NO (slow)

NO3 + CO → NO2 + CO2 (fast)

This means that the rate of the overall reaction is dominated （控制） by the rate of the first reaction, this is the rate-determining step.

H2(g) + I2(g) 2 HI(g)

I2(g) → 2I (fast)

H2+ 2I → 2HI (slow)

——rate-determining step.

Types of Elementary Reactions

– unimolecular reaction: an elementary reaction in which th

e rearrangement of a single molecule produces one or more

molecules of product.

I2(g) ─→ 2I(g)

– bimolecular reaction: the collision and combination

of two reactants to give an activated complex in an elementar

y reaction.

NO(g) + O3(g) ─→ NO2(g) + O2(g)

molecularity --for Elementary Reactions:

_ termolecular reaction: an elementary reaction in

volving the simultaneous collision of any combi

nation of three molecules, ions, or atoms.

2 NO + H2 ─→ N2 + H2O2

5-2 Theories of Reaction Rate One is the collision theory : The collision theory is based on the kinetic theor

y and assumes a collision between reactants before a reaction can take place.

1918 年 Lewis 以气体分子运动论为基础提出

Another is the transition state theory: The transition state theory suggests that as reac

tant molecules approach each other closely they are momentarily in a less stable state than either the reactants or the products.

● Contents of Collision Theory:

⑴ reacting molecules must come so close that they

collide.

⑵ " effective" collisions : not every collision

between molecules creates products,

only few collisions between reactant

molecules will react.

5-2.1 Collision Theory and Activation Energy

● Contents of Collision Theory:

According to this theory,

product formation can only take place

when there are "effective" collisions

between reactant molecules involved in

the rate determining step of the process.

Br(g) + HI(g) ─→ HBr(g) + I(g)

Straight on collision, hydrogen facing incoming bromine. Reaction occurs.

Straight on collision, hydrogen facing away from incoming bromine. Reaction does not occur.

Straight on collision, bromine at 90 degrees. Reaction does not occur.

What constitutes an effective collision?

发生有效碰撞的两个基本前提 :

● enough energy

对 HCl 和 NH3 的气相反应

● proper orientation

• Activation molecule:

the average molecules must absorb some energy to become activation molecules

is the molecule have enough energy and can produce effective collision

活化分子一般只占极少数，它具有的最低能量为 Ec 。通常把活化分子具有的平均能量与反应物分子的平均能量之差称为反应的活化能，用符号 Ea 表示。

kca EEE

在一定温度下，反应的活化能越大，活化分子的分子分数越小，活化分子越少，有效碰撞次数就越少，因此化学反应速率越慢；

反应的活化能越小，活化分子的分子分数越大，活化分子越多，有效碰撞次数就越多，化学反应速率越快。

Figure: As the activation energy of a reaction decreases, the number of molecules with at least this much energy increases, as shown by the yellow shaded areas.

▲ 离子反应和沉淀反应的 Ea 都很小

一般认为 Ea 小于 63 kJ·mol-1 的为快速反应

小于 40 kJ·mol-1 和大于 400 kJ·mol-1 的都很难测定出

▲ 一些反应的 Ea

▲ 每一反应的活化能数值各异 , 可以通过实验和计算得 到 。活化能越小 , 反应速率越快。 Ea 是动力学参数。

N2(g) + 3H2(g) = 2NH3(g), Ea=175.5 kJ·mol-1

HCl + NaOH → NaCl+H2O, Ea≈20 kJ·mol-1

2SO2(g) + O2(g) = 2SO3(g), Ea=251 kJ·mol-1

活化能的特征Character of activation energy

活化能的特征Character of activation energy

1-r223 molkJ6.199- (g)O(g)NONO(g)(g)O •=+=+ H

★ 反应物的能量必 须爬过一个能垒 才能转化为产物

★ 即使是放热反应 (△rH 为负值 ) ， 外界仍必须提供 最低限度的能量， 这个能量就是 反应的活化能

Ea 与 △ rH 的关系energy

Ea 与 △ rH 的关系energy

5-2.2 The Transition State Theory

Transition state theory (TST) is also called

activated complex theory.

reactants pass through high-energy transition

states before forming products, they are associated in

an unstable entity called an activated complex,

then change into products.

要点： 化学键重排、 活化络合物形成

Example : NO2(g) + CO(g) ─→ NO(g) + CO2(g)

ΔH = Ea,f - Ea,r

=358 kJ·mol-1 -132 kJ·mol-1

ΔH = –226kJ·mol-1.

酶作用的机制——中间产物学说 酶与底物形成酶－底物中间复合物， 中间复合物再分解成产物和酶。 E + S = E-S P + E• 许多实验事实证明了 E － S 复合物的存在。

理论受限 活化络合物的结构无法在实验中加以确定 计算过于复杂

Factors That Affect Reaction Rates

Concentration of Reactants

Temperature

Catalysts ___Speed by changing mechanism

5-3 Reaction Rates and

Concentrations

Chemical reactions are faster when the concent

rations of the reactants are increased .

Because more molecules will exist in a given v

olume. More collisions will occur and the rate o

fa reaction will increase.

Concentration and Rate

Each reaction has its own equation that gives its rate as a function of reactant concentrations.

this is called its Rate Law

说明在一定温度下，反应速度与反应物浓度之间的定量关系 —— 质量作用定律

A rate law shows the relationship between the reaction rate and the concentrations of reactants.

5-3.1 The Rate Law

v [A]∝ m[B]n

aA + bB → cC + dD

v = k[A]m[B]n 反应速率方程或称质量作用定律

The rate of a reaction is proportional to the product of the concentrations of the reactants

raised to some power.

v = k[A]m[B]n

•k is a rate constant that has a specific value for each reaction.The value of k is determined experimentally.

“Constant” is relative here - k changes with T ，

　　 the unit( 量纲 ) depend on m + n ① when [A]=[B]=1mol·L-1, v =k ② the greater the k , the faster the rate

参见中文 P101

质量作用定律（ law of mass action ）

对于基元反应，反应速率与反应物浓度的幂乘积成正比。幂指数就是基元反应方程中各反应物的系数，这就是质量作用定律。

]M[[Cl] MClM2Cl

][H][Cl ClHClClH

]Cl][H[ HHClH Cl

][M][Cl M2ClMCl

22

22

22

22

kv

kv

k v

k v

对于复杂反应， 它适用于每一步的基元反应，它的反应速度取决于定速步骤。

2N2O5 （ g ）→ 4NO2 （ g ）＋ O2 (g)

实验证明 N2O5 → NO3 ＋ NO2 （慢） NO2 ＋ NO3 → NO2 ＋ O2 ＋ NO （快

） NO ＋ NO3 → 2NO2 （快

）][ 52ONkv

质量作用定律仅适用于基元反应。

速率方程应为

in general, m and n are not equal to the stoichiometric coefficients a and b

[A], [B] are the concentration of A and B; m and n are themselves constants for a given reaction, it must be determined experimentally

v = k[A]m[B]n aA + bB → cC + dD

The order of a reaction with respect to one of the reactants is equal to the power to which the concentration of that reactant is raised in the rate equation.

5-3.2 Order of A Reaction

The sum of the powers to which all reactant concentrations appearing in the rate law are raised is called the overall reaction order.

m is the order of the reaction with respect to A, n is the order of the reaction with respect to B.

For rate equation v = k[A]m[B]n

the exponents m and n are not necessarily relat

ed to the stoichiometric coefficients in the balan

ced equation, that is, in general it is not true tha

t for a A + b B → c C + d D,

m ≠ a and n ≠ b

m+n is overall order of the reaction

The rate law for the thermal decomposition

（热分解） of acetaldehyde (CH3CHO)

CH3CHO(g) →CH4(g) + CO(g)

has been determined experimentally to be

v =k[CH3CHO]3/2

and not rate = k[CH3CHO] ×

Example 5-1: Given the following data, what is the rate expression for the reaction between hydroxide ion and chlorine dioxide?

2ClO2(aq) + 2OH-(aq) →ClO3-(aq) + ClO2

-(aq) + H2O

[ClO2] (mol.L-1) [OH-] (mol.L-1) Rate (mol.L-1 s-1)

0.010 0.030 6.00×10-4 v1

0.010 0.075 1.50×10-3 v2

0.055 0.030 1.82×10-2 v3

Solution:

v = k[ClO2]m[OH-]n

m=?

32-

23-

1 4-

112

101.82 0.030 055.0

101.50 0.075 010.0

106.00 0.030 010.0

)( ][ ][

v

v

v

sLmolrateOHClO

m

m

mm

m

ClO

ClO

ClO

ClO

v

v

)5.5(3.30

)010.0

055.0(

1000.6

1082.1

)][

][(

][

][

4

2

12

32

12

32

1

3

By inspection, m = 2. The reaction is 2nd order in ClO2

By inspection, n = 1

32-

23-

1 4-

112

101.82 0.030 055.0

101.50 0.075 010.0

106.00 0.030 010.0

)( ][ ][

v

v

v

sLmolrateOHClO

n

n

n

OH

OH

v

v

)5.2(5.2

)030.0

075.0(

1000.6

1050.1

)][

][(

4

3

1

2

1

2

The overall rate expression is therefore

v = k[ClO2]2[OH-]

v = k[ClO2]m[OH-]n

n=?

想一想：

反应速度与速率方程？反应分子数和反应级数？

•First-order reactions

A first-order reaction is a reaction whose rate depends on the reactant concentration raised to the first power.

A →B

the rate equation is

ckAkv 1][ Differential form:

ckdt

dc

dt

Adv 1

][

Integrate the left side from c = c0 to c and the right from t = 0 to t.

ckdt

dc1

dtkc

dc1

Thus

tc

cdtk

c

dc0 1

0

速度方程的积分表达式

Can be rearranged to give:

tkc

ct1

0

ln

- { lnc – lnc0} = k1( t - 0 )

c0 is the initial concentration of c (t =0).

c is the concentration of c at some time t.

8)-(5 log303.2 0

1 c

c

tk

9)-(5 log303.2

log 01 ct

kc or

The characteristics of first-order reactions ：

1. A plot of lgc versus t (time) gives a straight line with a slope of -k1/2.30

3.

2. The rate constant, k, has units of [time]-1.

3. half-life (t1/2)

is the time it takes

for the concentration

of a reactant A to fall

to one half of its

original value.

8)-(5 log303.2 0

1 c

c

tk

By definition, when t = t1/2, , so

c

c

kt 0

1

log303.2

lg2 2.303

2

log303.2

10

0

12

1 kcc

kt

20c

c

恒温下，一级反应的半衰期是一个常数。 半衰期越长， k1 就越小，反应速率就越慢。

10)-(5 693.0

12/1 k

t

可以用半衰期衡量反应速度， t1/2 越大，反应越慢。

—— 一级反应的半衰期与速率常数 k1 成反比，与反应物的起始浓度无关。

常见药物的半衰期：

　 药物名称 t1/2 （ h ） 青霉素 0.5

链霉素 2 ～ 3

安定 20 ～ 40

放射性同位素 14C 半衰期为 5580 年

一级反应的特点是：

（ 1 ） 1gc 对时间 t 作图可得一条直线，直线的斜率为 － k1/2.303 。因此可由斜率求得反应速率常数 k1 。

（ 2 ） k1 的单位为 [ 时间 ] － 1 。 这说明 k1 的数值与时间单位有关， 而与浓度无关。

（ 3 ）恒温下，一级反应的半衰期是与速率常数 k1 成反比，与反应物的起始浓度无关的一个常数。

log303.2 0

1 c

c

tk =

12/1

693.0

kt

Example 5-2(a) What is the rate constant k for the first-order

decomposition of N2O5(g) at 25 if the ℃ half-life of

N2O5(g) at that temperature is 4.03×104 seconds?

Solution: (a) s

kt 4

2/1 1003.4693.0

)(1072.11003.4

693.0693.0 154

2/1

st

k ×=×

== -

Solution: (b)

Putting in the value for k and substituting t = 8.64×104 seconds (one day has 86,400 seconds) gives

303.2log

0

kt

c

c

645.0303.2

1064.81072.1log

45

0

c

c

(b) Under these conditions, what percent of the N2O5

molecules will not have reacted after one day?

Hence

Therefore, 22.6% of the N2O5 molecules will not

have reacted after one day at 25 .℃

226.00

c

c

226.0][

][

052

52 =ON

ON

Example 5-3SO2Cl2 （磺酰氯 ） decomposes to sulfur dioxide and chl

orine gas. The reaction is first order. If it takes 13.7 hour

s for a 0.250mol/L solution of SO2Cl2 to decompose into

a 0.117mol/L solution, what is the rate constant for the r

eaction and what is the half-life of SO2Cl2 decomposition?

Solution: 303.2

loglog 0

ktcc

303.2

7.13117.0log250.0log k

K = 0.0554 h-1

)(5.120554.0

693.0693.02/1 h

kt

一级反应很多，许多热分解反应、分子重排反应、放射性元素的蜕变等都是一级反应，许多化合物的水解反应在低浓度的水溶液中进行时也表现为一级反应，许多药物在生物体内的吸收、分布、代谢和排泄过程，也常近似地看作是一级反应。

[ 例 5-2] 已知四环素在人体内的代谢服从一级反应规律。设给人体注射 0.5g 四环素，然后在不同时间测定血液中四环素的含量，得如下数据：

服药后时间 t/h 4 6 8 10 12 14 16

血中四环素含量 c/mg·L-1 4.6 3.9 3.2 2.8 2.5 2.0 1.6

试求： ①四环素代谢的半衰期；

②若血液中四环素的最低有效量相当于 3. 7mg·L-1 ，

则需几小时后注射第二次？

解： ① 先求速率常数 k1 ，一级反应以 log c 对 t 作图，得直线， 见图 5-6 。由前后两点或由直线回归得 :

038.0416

6.4lg6.1lg

斜率

）（ -1h088.0)038.0(303.2 k

∴

(h) 9.7088.0

693.0

693.0

12/1

k

t

计算表明：要使血液中四环素含量不低 于 3.7mg·L-1 ，应于第一次 注射后 6.3h 之前注射第二次，临床上一般控制 在 6h后注射第二次， 即 4次 /天。

②由图 5-6 知， t＝ 0 时， lgc0＝ 0.81 ，最低有效量 c＝ 3.7

mg·L-1 lgc＝ 0.57, 将（ 5-8 ）重排得：

)(3.6088.0

)57.081.0(303.2

)lg(lg303.2

1

0

h

k

cct

8)-(5 log303.2 0

1 c

c

tk

另外，衡量药物分解的速率时，常用分解 10％所需的时间，称为十分之一衰期，用 t0.9 表示，恒温下 t0.

9 也是与浓度无关的常数。

119.0

1054.0

90

100lg

303.2

kkt

了解药物的半衰期，对于合理用药有着重要意义。 常见药物的半衰期，如 磺胺甲恶唑（新诺明），半衰期 12 小时左右； 头孢氨苄，半衰期为 1 ～ 2 小时； 地西泮为 20 ～ 40h ， 保泰松为 48 ～ 120h 。

• Second - order reactions

22ck

dt

dcv

v = k2[A]2 , v = k2 [A][B] [A]=[B]

dtkc

dc 22

二级反应是一类常见的反应，溶液中的许多有机反应像加成、取代及消去反应等都是二级反应。

11)-(5 11

20

tkcc

02

11

ctk

c )

11(

1

02 cct

k

The integrated rate law for a 2nd order reaction can be easily shown to be

tc

cdtk

c

dc0 22

0

The characteristics of second- order reactions ：1. A graph of 1/c against time is a straight line , the slope of which gives the rate constant for t

he reaction ；

2. The rate constant, k, has units of [c]-1[t]-1 ；

3. The half-life of 2th-order reactions

)11

(1

02 cct

k

(c = 2/c0 代入上式 )t1/2=1 / kc0.

Note that the half-life of a second-order

reaction is not independent of the initial

concentration, as in the case of a first-order

reaction. This is one way to distinguish a

first-order reaction from a second-order

reaction.

（ 1 ） 1/c 对时间 t 作图可得一条直线，

直线的斜率即为反应速率常数 k2 。

（ 2 ） k2 的单位为 [ 浓度 ]-1[ 时间 ]-1 。

k2 的数值与时间和浓度单位有关。

（ 3 ）二级反应的 。由此可见：二级反应

的半衰期与反应物的初始浓度成反比。反应物初始

浓度越大，半衰期越短。

二级反应有以下特点是：

Solution:

t = 3.6 (min)

tkcc 2

0

11

t 84.0500.0

1

200.0

1

Example 5-4Butadiene(C4H6丁二烯 ) dimerizes( 聚合 ) to form C8H12 （二聚物） . This reaction is 2nd order in butadiene. If the rate constant for the reaction is 0.84 Lmol-1min-1, how long will it take for a 0.500 mol/L sample of butadiene to dimerize until the butadiene concentration is 0.200 mol /L?

（ 5-11 ）

［例 5-3］乙酸乙酯在 25℃时的皂化反应为二级反应： CH ３COOC ２H ５十 NaOH→CH ３COONa十 C ２H ５OH

乙酸乙酯和氢氧化钠的起始浓度均为 0.0100mol·L －１，反应 20min后，氢氧化钠的浓度消耗了 0.00566mol·L －１，

求： ①反应速率常数； ②反应的半衰期。解：

)11

(1

02 cct

k

11 min52.6

0100.0

1

00566.00100.0

1

20

1

Lmol

022/1

1

ckt min3.15

0100.052.6

1

①

②

• Zero - order reactions

—— is one where the rate does not depend on the concentration of the species.

000ck

dt

dcv v = k0 [A]0

tkcc 00

integrated ： tc

cdtkdc

o 00

固体界面上的分解反应

c = - k0 t + c0

c = - k0 t + c0

1. A graph of c against t is a straight line with a

slope of -k0.

The characteristics of zero-order reactions ：

2. The rate constant, k,

has units of [c][ t ]-1 ；

3. The half-life of a zero-order reaction is

0

0

0

02/1

5.0

2 k

c

k

ct

（ 1 ） c 对 t 作图得一直线，斜率为－ k0

（ 2 ） k0 的单位为［浓度］［时间］－１

所以 k0与时间单位和浓度单位有关

（ 3 ）零级反应的半衰期：

所以零级反应的半衰期与最初浓度成正比。 反应物的初始浓度越大，半衰期越长。

0

0

0

02/1

5.0

2 k

c

k

ct

近年来发展的一些缓解长效药，其释药速率在相当长的时间范围内比较恒定，即属零级反应。如国际上应用较广的一种皮下植入剂，内含女性避孕药左旋 18 －甲基炔诺酮，每天约释药 3

0μg ，可一直维持 5 年左右。

零级反应的特点：

The decomposition of HI into hydrogen and iodine on a gold surface is 0th order in HI. The rate constant for the reaction is 0.050mol·L-1·s-1. If you begin with a 0.500mol/L concentration of HI, what is the concentration of HI after 5 seconds?

Solution:

[HI] = 0.500 － 0.050×5

= 0.250(mol·L-1)0cktc

c0= 0.500mol·L-1 k =0.050mol·L-1· s-1

t =5s

Example 5-5

5-4 Effect of Temperature on Reaction Rates

5-4 Effect of Temperature on Reaction Rates

Chemical reactions are faster when the temperature is increased. Why?

Figure: At a higher temperature,T2, more molecules have an energy

greater than E a , as shown by the yellow shaded area.

When we increase the temperature, kinetic energies and speeds will increase and so the average energy of a collision will also increase. As a result, the energy of any particular collision will be more likely to exceed the activation energy. Thus, chemical reactions are faster at higher temperatures than at lower temperatures.

The dependence of reaction rate on temperature has led to a common rule of thumb:

the rate of a chemical reaction will double for each 10 increase in the ℃temperature.

Rule of thumb ：

5-4.1 Rule of Thumb 经验法则 (Van’t Hoff Law)

The rate of a chemical reaction will double for

each 10 increase in the temperature. ℃

v (T+10) k (T+10)γ= ---------- = ----------- = 2 ～ 4 v T k T

a A + b B = c C + d D

T : v T = k T [A]a [B]b

(T + 10) : v (T+10) = k T+10 [A]a [B]b

5-4.2 The Arrhenius Equation

In 1889 Svante Arrhenius showed that the dependence of the constant of a reaction on temperature can be expressed by the following equation, now known as the Arrhenius equation

通过实验制作速率常数 k 随温度 T 升高而变大的曲线图，并分析曲线，可以得出速率常数随温度变化函数的经验方程 :

RTEaAek / Ea —is the activation energy of the reaction (in kJ/mol)

R — is the gas constant (8.314 JK-1mol-1) T — is the absolute temperature

e — is the base of the natural logarithm scale

A — represents the collision frequency, and is called

the frequency factor (指前因子 ).

The Arrhenius Equation:

阿伦尼乌斯方程式的重要假设： A 和 Ea 是不随温度改变的特征参数。大多数反应在一定温度区间内的这种近似是完全允许的，这种近似方法被称为线性化。

（ 1 ）对某一反应： Ea 基本不变，温度升高， 增大，

k 值也增大，反应速率加快，说明了温度对反应速率的影响。

（ 2 ）对不同反应：当温度一定时，活化能 Ea 值越小，

则 越大， k 值愈大，反应速率愈快。

反之， Ea愈大，反应速率愈慢。

)/(RTEae

)/(RTEae

阿伦尼乌斯方程式把速率常数 k、活化能 Ea和温度 T三者联系起来，由此关系式可以说明：

RTEaAek /

ATR

Ek a lg)

1(

303.2lg

Taking the natural logarithm of both sides, the equation becomes

or in terms of common logarithms :

RTEaAek /

Example 5-6The rate constants for the decomposition of acetaldehyde

CH3CHO(g) ─→ CH4(g) + CO(g)

were measured at five different temperatures. The data are s

how below. Plot lgk versus 1/T and determine the activatio

n energy (in KJ/mol) for the reaction.

────────────────────────────── T(K) 700 730 760 790 810

k [1/(mol/L)1/2s] 0.011 0.035 0.105 0.343 0.789

 

──────────────────────────────

Answer: We need to plot lg k versus 1/T . From the given data we obtain

1/T(1/K) 1.43×10-3 1.37×10-3 1.32×10-3 1.27×10-3 1.23×10-3

log k -1.96 -1.46 -0.979 -0.465 -0.103

The slope of the line is calculated from two pairs of coordinates:

31025.1 ;30.0

31041.1

;80.1

ATR

Ek a lg)

1(

303.2lg

lg k

1/T

Thus, a plot of lg k versus 1/T gives a straight line whose slope is equal to -E a / 2.303R and whose

intercept （截距） with the ordinate( 纵坐标 ) is log A.

Ea = 2.303×(8.314J/K·mol)× (9.38×103K)

= 1.80×105 J/mol

= 1.80×102 KJ/mol

5-4.3 Application of Arrhenius Equation

T1 → k1

According to this equation, we can calculate E a and k

T2 → k2

13)-(5 303.2

loglg1

1 RT

EAk a

14)-(5 303.2

loglg2

2 RT

EAk a

155 11

303.2lg

122

1

TTR

E

k

k a(5-14) - (5-13) ：

Example 5-7

The rate constant of a first-order reaction is

3.46×10-2 /s at 298 K. What is the rate constant at

350 K if the activation energy for the reaction is

50.2 KJ/mol?

Answer:

298

1

350

1

314.8303.2

102.501046.3log

3

2

2

k

155 11

303.2log

122

1

TTR

E

k

k a

31.11046.3

log2

2

k

)(71.0 12

sk

5-5 Effect of Catalyst on Reaction Rates

Catalyst: a substance that increase the rate of a chemical reaction , but is itself

neither consumed nor produced in the reaction

MnO2

• Catalysts increase the rate of a reaction by decreasing the activation energy of the

reaction.• Catalysts change the mechanism by which the process occurs.

Reaction O3 + O = O2

three types

homogeneous(均相 ) catalysis,

heterogeneous ( 非均相 ) catalysis

enzyme catalysis.

Catalysis can be classified into

Homogeneous catalysis

In homogeneous catalysis, the catalyst is

present in the same phase as the reactants,

as when a gas-phase catalyst speeds up a

gas-phase reaction, or a species dissolved

in solution speeds up a reaction in solution

but it can be catalyzed by Ag+ or Mn2+:

The rate is very Slow,

Heterogeneous catalysis

In heterogeneous catalysis, the catalyst is

present as a distinct phase. The most

important case is the catalytic action of

certain solid surfaces on gas-phase and

solution-phase reactions.

C2H4

+H2 C2H6

PE

RC

e.g., H2 + CH2=CH2 no reaction (k very small)25ºC

CH3–CH3 rapid (k larger)

25ºCPd

Ea (uncatalyzed)

Ea (catalyzed)

catalysis

Enzyme Catalysis

Enzymes are biological catalysts.

An average living cell may contain some 3000 different enzymes

E + S = E-S P + E

activated complex

酶催化作用的本质：降低反应活化能

E+S

P+ E

ES

能量水平

反应过程

G

E1

E2

酶（ E ）与底物（ S

）结合生成不稳定的中间物（ ES ），再分解成产物（ P ）并释放出酶，使反应沿一个低活化能的途径进行，降低反应所需活化能，所以能加快反应速度。

E…S

(1) 锁钥假说 (lock and key hypothesis)：• 认为整个酶分子的天然构象是具有刚性结构的，酶表面具有特定的形状。酶与底物的结合如同一把钥匙对一把锁一样 .

酶与底物结合形成中间络合物的方式（理论）

该学说认为酶表面并没有一种与底物互补的固定形状，而只是由于底物的诱导才形成了互补形。酶与底物接近→结构相互诱导→变形和相互适应 →结合。

(2) 诱导契合假说（ induced–fit hypothesis):

Zn

Zn

羧肽酶活性中心示意图

Tyr 248

Arg 145

Glu 270

底物

羧肽酶的诱导契合模式

与酶具有高效催化作用的因素

邻近效应与定向排列

亲电催化

多元催化 亲核催化

酸、碱催化

离子催化

表面效应

1. 邻近效应 (proximity effect)与 定向排列 (orientation arrange )：

邻近效应： 酶将诸底物结合到其活性中心，使它们的反应基团相互靠近，降低进入过渡态所需的活化能；局部底物浓度增加。

定向排列：酶使诸底物反应基团形成正确的几何定向关系，产生有效碰撞。

靠近

2. 多元催化 (multielement catalysis) 多个基元催化形式的协同作用。一般包括： 酸 -碱催化，共价催化（亲核催化 , 亲电子催化）等。

酶分子中广义的酸碱基团：氨基、羧基、巯基、酚羟基、咪唑基。

同种基团在不同的微环境下解离度亦不同。 这种多功能基团兼有酸、碱双重催化效能。

某些辅酶：焦磷酸硫胺素、 磷酸吡哆醛等也参与共价催化作用。

底物中典型的亲电中心：磷酰基、酰基、糖基

这类反应有 羰基的加成，酮基和烯醇的互变异构， 肽和酯的水解及磷酸和焦磷酸参与的反应

例 :胰凝乳蛋白酶催化肽键水解的多种类型催化机制

(1) “ 催化三元区 ”——活性中心的结构特性

稳定作用

Asp102 —— 起稳定 His57构象作用His57 —— 质子受体，促进 Ser195 羟基氧对底物肽酰基 亲核进攻作用Ser195 —— 羟基氧亲核进攻作用

(2) 亲核共价催化作用： ① 四面体结构中间体的形成： Ser195羟基氧对底物肽酰基亲核进攻时 , 形成一个共价四面体中间体

② His-57 广义酸催化过程 酰基一酶中间体生成 ：

a: His-57 给出质子，底物接受质子，伴随着 C-N 肽 键的断裂后 , R1NH2 （产物 1）从酶中离去；

b: 水亲核进攻羰基碳，另一个过渡态四面体产生；

产物 1

a b

酰基一酶中间体虽然 不稳定 , 但 它的存在已经被 X 射线衍

射实验所证实。

His57 作为碱 接受 H2O 给予的质子， 随后它作为酸将质子转移给 Ser195氧， 以此方式帮助酰基 - 酶中间体的解离。

产物 2

③His-57 作为广义碱 - 酸催化作用

酶活性中心的疏水环境排除了水分子对酶和底物功能基团的干扰、吸引或排斥，防止酶和底物之间形成水化膜，有利于酶和底物的密切接触。

——疏水的为微环境极大利于酶的催化作用

3. 表面效应 (surface effect) ：

-+

上学去