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CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT

CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT. Organic Chemical Process Industries. Section 1. TA. Jasmine Amr Eng. Hassan El- Shimi. Wednesday, September 22 nd , 2010. Today’ Agenda. General view on Organic Industries Course contents Marks

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CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT

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  1. CAIRO UNIVERSITYFACULTY OF ENGINEERINGCHEMICAL 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. Lectures Sections • 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) Laboratory • Peroxide value • Saponification value • O & G Estimation • BOD5 & COD • Dying • Mwt. Determination Course Contents 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 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 rAα 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

  6. 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)

  7. 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 Mechanism of reaction

  8. 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= 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

  9. 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’ CAn, 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.

  10. 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

  11. 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

  12. 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

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

  14. 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 =

  15. 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).

  16. Chemical Third Year, CHE2012 Any Question

  17. For any problem, please contact me: Eng. Hassan El- Shimi E-mail: hassanshimi@gmail.com Mobile: 011 80 87 86 2

  18. Thank You

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