This presentation is the property of its rightful owner.
Sponsored Links
1 / 27

化学动力学 - 化学反应速度 PowerPoint PPT Presentation


  • 108 Views
  • Uploaded on
  • Presentation posted in: General

第三章. 化学动力学 - 化学反应速度. 学习要求:. 1. 了解化学反应速率的表示方法以及基元 反应、非基元反应等基本概念。. 2. 了解浓度对反应速率的影响;掌握质量作用 定律、速率方程及速率常数、反应级数等。. 3. 了解温度对反应速率的影响;掌握阿仑尼乌 斯公式的应用。. 4. 了解反应速率的碰撞理论及过渡态理论要点。. 【 内容提要 】 本章介绍了反应速率概念、反 应速率理论及反应机理,重点讨论了影响反应 速率的因素,阐明了浓度、温度和催化剂与反 应速率的关系。. 3.1 化学反应速率的概念

Download Presentation

化学动力学 - 化学反应速度

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


4055131

-


4055131

1.

2.

3.

4.


4055131

3.1

3.2

3.3

3.4


4055131

CO + NO = CO2 + H2O G=-343.8KJ<0


4055131

c/t

==dc/dt

3.1

1.

2.


4055131

aA + bB = gG + dD

moldm-3s-1


4055131

2N2O5=4NO2+O2


4055131

aA + bB = gG + hH


4055131

k

1mol L-1

a bAB

(a+b)


4055131

3-1 303K CH3CHO(g)=CH4(g)+CO(g)

c(CH3CHO)/ molL1 0.10 0.20 0.30 0.40

/ mol L1 s1 0.025 0.102 0.228 0.406

1

2k

3c(CH3CHO)=0.25 molL1


4055131

1 =kcn(CH3CHO)

14

0.025=k(0.10)n

0.406=k(0.40)n

n2 =kc2(CH3CHO)

23k

0.228 =k (0.30)2k = 2.53 mol1 L s1

3c(CH3CHO= 0.25 molL1

=kc2(CH3CHO)=2.530.25=0.158 mol L1 s1


4055131

3.2

(1918)

1.


4055131

2.

(E)

(E)

(Ea)

kJmol-1


4055131

Ea

E

E

3-3

E

EaEa

Ea


4055131

A BC

Ea

Ea

A +BC

H

AB+C

3-4

H = Ea Ea


4055131

3.3


4055131

vant Hoff

(Vant Hoff)

10K, 24

24kT , kT+10 ,

kT+n 10TK , (T+10)K (T+n10)K


4055131

Arrhenius

1889(Arrhenius)

Ea A

R T


4055131

T1T2

k1k2

ln k1= ln A EaRT1

ln k2= ln AEaRT2

EaAk1

k2


4055131

3-2 C2H4 (g) + H2 (g) = C2H6 (g)700K

k1=1.3108 mol1Ls1,730Kk2

Ea=180kJmol1

k2=4.6108 mol1Ls1


4055131

3.4


4055131

1


4055131

2

3

4

5


4055131

Ea

k

3-2

3-2


4055131


4055131

1986

JCPolan

197919863

4

Peter Debye 1989


4055131

1978


  • Login