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

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

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

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

Chemical Reaction Engineering Chemical Reaction Kinetics Optimization Reactor Design Control

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

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)

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

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

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.

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

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

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

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

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 =

k1/k2 = Since E= 73000 cal & R= 1. 987 cal/gmol 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).

Chemical Third Year, CHE2012 Any Question

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

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