1. Organic Chemical Process Industries
CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT
Organic Chemical Process Industries
Section 1
TA. Jasmine Amr
Eng. Hassan El- Shimi
Wednesday, September 22nd, 2010

2. Today’ Agenda General view on Organic Industries Course contents Marks
Review on Chemical Kinetics
Sheet 1

3. Course Contents Lectures Sections Laboratory
Organic Technology (Overview)
Nitration
Nitrobenzene
Dinitrobenzene
TNB
Aniline industry
TNT, and ….
Sulfonation
Intro to Petrochemical Industry
Review on Chemical Kinetics (2 sections)
Nitration Thermodynamics ( 2 sections)
Peroxide value
Saponification value
O & G Estimation
BOD5 & COD
Dying
Mwt. Determination
Prof. Dr. Samia
Prof. Dr. Hoda
Prof. Dr. Sahar
TA. Jasmine Amr
Eng. Hassan El- Shimi

4. Marks (acceptable to change)
Total marks = 150 mark
Final exam : 90 mark
Year work : 60 mark
1st midterm : 30 mark
2nd midterm : 10 marks
Oral exam : 10 marks
Lab : 10 mark

5. Chemical Reaction Engineering Chemical Reaction Kinetics
Optimization
Reactor Design
Control

6. Chemical Reaction Kinetics
For any reaction A+B → C (product)
Rate of reaction α no. of collisions between molecules A, B.
Rate of reaction α concentration of A, B
As rA α CA CB, hence rA = k CA CB
Where k: specific rate constant
Rate of reaction is rate of consumption or production of component (i)
rA= - dCA/dt = - dCB/dt = +dCC/dt = kCACB

7. If aA+bB → cC+dD (balanced equation)
Hence, rA = k CAa CBb (rate equation)
Where, a: order of reaction w.r.t. reactant (A) and b: order of reaction w.r.t. reactant (B)
Overall order of reaction = a+b
In case of reversible reaction:
aA+bB ↔ cC+dD
Rate eq.: rA= kf CAaCBb – kb CCcCDd
Order of reaction= (a+b) – (c+d)

8. Mechanism of reaction Elementary Reactions Non-elementary Reactions
Reactions occur in a single step and the rate expression is suggested from the stiochiometric equation.
2A+B → R+S
rA = k CA2CB
Reactions which we observe as a single reaction, but in the real, it’s the result of a sequence of elementary reactions and the rate equation can’t be detected from the stiochiometric equation.
2A+B → R+S
In fact: A+B → AB
AB+A → A2B +S
A2B → R
These 3 steps are called: the reaction mechanism
Rate eq. = rate eq. of determining step (slowest step) ≠ kCA2CB

9. Arrhenius Equation K= Ao e –E/RT K: specific rate constant
Ao: Arrhenius constant
E: activation energy
T: absolute temperature (Kelvin or oR)
R: universal gas constant= J/gmole.K or R= lit.atm/gmole.K or R= cal/gmole. K = Btu/Ib. R

10. Dependence of Rx Rate on Concentration:
r = k CAn CBm; n & m order of Rx
If B is in large excess; CB ≈ constant, i.e. –dCB/dt ≈ 0, hence
CBm = constant and rA = k’ CA n, where k’= k CBm, and it’s easy to get (n).
Molecularity:
It’s the number of reactant molecules in the stiochiometric equation (Elementary Rx) OR number of molecules in the rate determining step (Non-elementary Rx).
N.B.: if order of Rx = Molecularity, the reaction is called simple order reaction.

11. Sheet 1
Problem no. 1 A → B, r = kCAn; n=? 1) Assume zero order reaction (n=0) Hence, r= - dC/dt = k[C]n=0 = k Get CA-CAo = - k*t Hence, CA=CAo – k*t CA
Slope = k t

12. 2) Assume 1st order reaction ( n=1) r= -dCA/dt = k[CA] -dC/[C] = k dt Hence, Ln CA- Ln CAo = -k*t Finally, ln CA = ln CAo –k*t
Ln CA
Slope = k t

13. 3) Assume 2nd order reaction ( n=2) r= -dCA/dt = k[CA]2 -dC/[C]2 = k dt Hence, 1/CA- 1/CAo = k*t Finally, 1/CA = 1/CAo + k*t
1/CA
Slope = k
1/CAo t

14. Plot C vs t
Plot 1/C vs t
The order of reaction is 2nd order
Slope = k = 1.415E-06 lit/gmole. Sec

15. Problem no. 2 Given E= 75000 cal K at 650oC and 500oC
Hence, T1= 923ok & T2= 773ok
Since k= Ao e-E/RT
k1= Ao e-E/R(773) → 1
k2= Ao e-E/R(923) → 2
k1/k2 =

16. k1/k2 = Since E= 73000 cal & R= 1. 987 cal/gmol
k1/k2 = Since E= cal & R= cal/gmol .k Thus, k1/k2 = 2/3 and k1=(2/3) k2 i.e. k2 > k1 , thus the reaction at (T2= 650oC) is faster than the reaction at (T1= 500oC).

17. Chemical Third Year, CHE2012 Any Question

18. For any problem, please contact me:
Eng. Hassan El- Shimi
Mobile:

19. Thank You