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

CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

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Page 1: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

CAIRO UNIVERSITYCAIRO UNIVERSITYFACULTY OF ENGINEERINGFACULTY OF ENGINEERINGCHEMICAL ENGINEERING DEPARTMENTCHEMICAL ENGINEERING DEPARTMENT

Organic Chemical Process Industries

TA. Jasmine AmrEng. Hassan El- Shimi

Section 1

Wednesday, September 22nd, 2010

Page 2: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

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

Page 3: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Course ContentsCourse Contents

Lectures Sections

Organic Technology (Overview)

o Nitrationo Nitrobenzene

o Dinitrobenzene

o TNB

o Aniline industry

o 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

Laboratory

Prof. Dr. SamiaProf. Dr. HodaProf. Dr. Sahar

TA. Jasmine AmrEng. Hassan El-

Shimi

Page 4: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

MarksMarks (acceptable to change) (acceptable to change)

Total marks = 150 markFinal exam : 90 markYear work : 60 marko 1st midterm : 30 marko 2nd midterm : 10 markso Oral exam : 10 markso Lab : 10 mark

Page 5: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section
Page 6: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Chemical Reaction KineticsChemical Reaction KineticsFor any reaction A+B → C (product)Rate of reaction α no. of collisions

between molecules A, B.Rate of reaction α concentration of A, BAs rA α CA CB, hence rA = k CA CB

Where k: specific rate constantRate of reaction is rate of consumption or

production of component (i) rA= - dCA/dt = - dCB/dt = +dCC/dt = kCACB

Page 7: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

If aA+bB → cC+dD (balanced equation)Hence, rA = k CA

a 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+dDRate eq.: rA= kf CA

aCBb – kb CC

cCDd

Order of reaction= (a+b) – (c+d)

Page 8: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Mechanism of reactionMechanism 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

Page 9: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Arrhenius EquationArrhenius Equation

K= Ao e –E/RT

K: specific rate constantAo: Arrhenius constantE: activation energyT: absolute temperature (Kelvin or oR)R: universal gas constant= 8.314 J/gmole.K or

R= 0.082 lit.atm/gmole.K or R= 1.987 cal/gmole. K = 10.73 Btu/Ib. R

Page 10: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Dependence of Rx Rate on Concentration:

r = k CAn CB

m; 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:o 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).

o N.B.: if order of Rx = Molecularity, the reaction is called simple order reaction.

Page 11: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Sheet 1Sheet 1Problem 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

t

Slope = k

Page 12: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

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

t

Slope = k

Page 13: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

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

t

Slope = k1/CAo

Page 14: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

a) Plot C vs t

b) Plot 1/C vs t

The order of reaction is 2nd order

Slope = k = 1.415E-06 lit/gmole. Sec

Page 15: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Problem no. 2Problem no. 2Given E= 75000 calK at 650oC and 500oCHence, 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 =

Page 16: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

k1/k2 =

Since E= 73000 cal & R= 1.987 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).

Page 17: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Chemical Third Year, CHE2012Any Question

Page 18: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

For any problem, please contact me:

Eng. Hassan El- ShimiE-mail: [email protected]

Mobile: 011 80 87 86 2

Page 19: CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT Organic Chemical Process Industries TA. Jasmine Amr Eng. Hassan El- Shimi Section

Thank You