document-1700092808-000_Chemical_reaction_engineering

3 Contents Maple installation 1 Coursework assignments 1 List of symbols 6 Introduction: basic chemistry, mathematics and Maple 9 1 A review of the material for simple reactions in simple reactors 15 1.1 Refresh and extend basic chemical kinetics for one homogeneous reaction 15 1.2 Temperature dependence of rate and equilibrium constants 22 1.3 Simple reactors and their mole balances 23 2 Multiple reactions 35 2.1 Two reactions 35 2.1.1 Parallel reactions (branching reactions). Formulation in terms of component concentrations/amounts 35 2.1.2 Consecutive (series) reactions: formulation in terms of yield 41 2.2 Many reactions in a batch, PF or CST reactor 44 3 Reaction mechanisms and reactive intermediates 52 3.1 Active intermediates 52 3.2 Rate of the overall process 55 3.3 Quick equilibria 57 3.4 Figuring out the mechanism from experimental data 58 3.5 Chain branching & reactor safety 61 4 Energy balance. Adiabatic reactors 68 4.1 Brief introduction 68 4.2 Irreversible reaction in an adiabatic reactor 72 4.3 Reversible reaction in an adiabatic reactor 77 5 Heat exchange and external work 86 5.1 Batch reactor 86 5.2 Plug flow reactor with thermal fluid jacket 87 5.3 Continuous stirred-tank reactor 95 5.4 Radiator coil heat exchanger 95 6 Residence time distribution 103 6.1 Brief introduction to probability 103 6.2 The distribution of the residence time and how to measure it 106 6.3 Ideal reactors 109 6.4 Using the residence time distribution for diagnostics 113 7 Mixing models for nonideal reactors 118

6 List of symbols A area, m 2 A m specific surface area, m 2 /kg a activity, mol/m 3 (concentration-based) or dimensionless (molar fraction-based) C concentration, mol/m 3 C p total heat capacity at constant pressure c p molar heat capacity at constant pressure E activation energy, J/mol F total flow rate, mol/s (total molar flow rate) or kg/s (total mass flow rate) j molar flux, mol/m 2 s H enthalpy, J/mol (reaction enthalpy) or J (heat) K equilibrium constant K I association constant of an inhibitor-enzyme complex K M Michaelis constant (for enzymes), or Monod ’s constant (for cell growth) k rate constant, various units m mass, kg n amount of substance, mol p pressure, Pa p i partial pressure, p i = x i p R gas constant, 8.314 J/molK S AB overall selectivity of A with respect to B, S AB =  BA n A / n B =  B n A /  A n B =  B C A /  A C B s AB instantaneous selectivity of A with respect to B, s AB = |  BA d n A /d n B | = |  BA r A / r B | r rate of a reaction, V − 1 d  /d t , mol/m 3 s, or rate of production of a component, V − 1 d n /d t T temperature, K t time, s V volume V m i molar volume of a component, V m i = d V /d n i v volume flow rate, m 3 /s w i weight fraction, w i = m i / m X conversion, X =  /  ∞ x i mole fraction, x i = n i / n Y overall yield fraction, Y = |  l | n u /  u ( n l0 − n l ), u ≡ useful, l ≡ limiting y instantaneous yield fraction, y =  l d n u /  u d n l =  l r u /  u r l , u ≡ useful, l ≡ limiting  length of the radical chain  ji stoichiometric coefficient of the i th component in the j th reaction  reaction yield, mol  V reaction yield per unit volume, mol/m 3  density  space-time, residence time  volume fraction of voids, porosity  A fraction of useful area of the catalyst Indices and brackets: X 0 initial value of X X 1/2 value at half-reaction (half-life) X ∞ value of X after a complete irreversible reaction, until the key component is exhausted X e equilibrium value of X

9 Introduction: basic chemistry, mathematics and Maple Your job is to: fundamentals … revise basic chemistry (units, moles, concentration, ideal gas, pH); skills … refresh your Maple skills (syntax, analysis of a function). This section is for your own use. “ Problems i) basic chemistry ” should cause no difficulty whatsoever and are there for you to just check whether you are on the page you need to be. Some of these are solved in the Maple examples file. The second set, “ Problems ii) mathematics and Maple ” , summarize the mathematical apparatus we will use in the course. There should be nothing new to you in this section in terms of mathematics. It, however, also teaches you how to use Maple on a basic level, so it is recommended to try and solve the problems with Maple. Read more Check the introduction sections in Examples.mw on QMplus. In Maple, read Help menu > Maple Help > Manuals > 1…11. Problems 1. Which are the SI units for: a) activation energy E ; b) entropy S ; c) molar fraction-based activity a x ; d) molality based activity, a m ; e) volume V ; d) mass m ; e) molar mass M ; f) mass fraction w ; g) molar concentration C ; h) Avogadro’s 1 number; i) gas constant R ; j) heat capacity C p of an object at constant pressure; k) temperature T ; l) rate of a reaction r ; m) enthalpy of a reaction. u 2. Switch to SI units: a) M = 18 g/mol; b)  r H = 13 kcal; c) C = 0.3 mol/L; d) p = 2.4 atm; e)  = 24 g/L; f) v = 3 L/min; g) T = – 18.2 °C. u 3. Harmonizing units: a) What mass of water is contained in 1 gallon of 460 mM NaCl solution of density 1.03 g/cm 3 ? b) A heater delivers 343 kcal of heat to 300 lbs of iron of heat capacity 25 J/molK. What is the resulting increase in temperature? e c) Component A in a liquid mixture degrades, due to a reaction of constant rate of 0.015  mol·pint -1 ·min -1 , in a vessel of volume 1 barrel. The vessel initially contained 0.010 lb-mol of A; how much of A will be left after 2 h 55 min 12 s? n 4. Molar mass computations a) What is the weight of 0.001 mmol Pb? b) What is the weight of 30 atoms of Cl? c) What is the weight of an oxygen molecule? d) How many atoms are there in 6 kg of iron? e) How many moles contain 10 g of ethanol? e 1 /av ɔˈ g ɑː dr ɔ /

12 12. Maple Analysis of a general function For each of the following functions, answer the questions: - What values of x produce a real value of y ? - What is the value of y ( x = 0), and where y ( x ) = 0? Plot the function and its zeros. - Find the extrema and the inflections of the functions. Plot them as points together with the function. a) y = sin( x )/ x (solved example); b) 2 2 ln( 1) y x x = − − (solved in part); c) e x x − (| x | is written as abs( x ) in Maple); d) 3/4 x x − + ; e) 1 1 1 x x x + + + . e 13. Analysis of functions that chemical engineers actually care about a) The rate of an enzymatic reaction changes with temperature following the equation / / e 1 e a b E RT E RT aT r b − − = + . Here, r [mM/h] is rate of the reaction, T [K] is temperature, R [J∙K − 1 ∙mol − 1 ] is the gas constant, E a [J/mol] is the activation energy of the enzymatic reaction, E b is the activation energy for denaturation of the enzyme, a [mM∙h − 1 ∙K − 1 ] and b [1] are kinetic parameters 2 . At low temperatures, this formula corresponds to the Arrhenius equation. At high temperatures, however, the enzyme partially denatures which leads to a fall in rate. There is an optimal temperature at which the rate has an extremum, and this is the temperature at which we want to operate our bioreactors. Plot r as a function of T , in the range 0…100 °C (careful with the units!). Mark on the plot the position of the extremum and the two inflection points. Parameters: a = 1.8×10 15 mM∙h − 1 ∙K − 1 ; b = 2.2×10 20 ; E a = 91 kJ/mol; E b = 120 kJ/mol; R = 8.314 J∙mol − 1 ∙K − 1 . Careful with the units – use SI! b) The Arrhenius equation for the rate constant reads ( ) 0 a exp / k k E RT = − , T = 0…∞ K. It predicts an inflection point of k ( T ) at a certain temperature. Find this point analytically. Use the values k 0 = 1 s − 1 , E a = 38 kJ/mol to plot k ( T ). Mark the position of the inflection point on the picture. c) If you pick a random molecule in a gas, the probability it to have velocity v is proportional to the so-called probability density  v of the Maxwell distribution , 2 2 /2 e Mv RT v v  − = , v = 0…∞ m/s. Here, M is the molar mass of the gas, T is the temperature of the gas (25 °C), R is the gas constant. - Plot  v vs. v for oxygen and nitrogen at room temperature, on the same plot. - The dependence  v ( v ) has a maximum corresponding to the most probable velocity v m of the molecules. Find its value for oxygen and nitrogen. Mark the position of the maximum with a point. Careful, the SI units for M are kg/mol, for T – kelvin! e 2 Units [1] mean that b is dimensionless.

15 1 A review of the material for simple reactions in simple reactors Your job is to: fundamentals… revise and extend basic chemical kinetics, revise basic reactor engineering (batch, PFR, CSTR); skills … design a one-reaction reactor. You have already been introduced to chemical reactors, mostly with examples where only one chemical reaction is taking place. In part one of CRE-II, we will focus on simultaneously occurring interdependent reactions. However, we will first introduce several new concepts on a couple of familiar examples. 1.1 Refresh and extend basic chemical kinetics for one homogeneous reaction Yield; stoichiometric 3 coefficients. Let us mix n A0 moles of a chemical species A and n B0 moles of B. Assume that the following reaction between A and B is possible: 2A + B → ½C. (1-1) We say that   moles of the reaction (1-1) took place when 2  moles of the reagent A and  moles of B reacted to form ½  of the product C. This can be written as: n A = n A0 − 2  , n B = n B0 −  , n C = ½  . (1-2) The quantity  is called the yield or extent of the reaction (1-1), and is a function of time: right after A and B were mixed,  ( t = 0) = 0 (zero moles of reaction at the beginning of it), and then the reaction proceeds to completion (  ∞ ) or to equilibrium (  e ). The relations (1-2) can be written concisely as: n i = n i 0 +  i  , (1-3) or per unit volume, C i = C i 0 +  i  V , (1-4) where  V [mol/m 3 ] are the moles of reaction per m 3 of mixture. Here, i = A,B,C; n i 0 is the initial amount of component i (where n C0 = 0);  i is the stoichiometric coefficient of i . The stoichiometric coefficients of the reagents are, by definition, negative – for the reaction (1-1),  A = − 2,  B = − 1,  C = + ½. (1-5) The stoichiometric coefficients of any inert species present in the mixture (a solvent, an inert carrier gas, or an admixture) is null. One standard way to write a chemical reaction (the second most common after the one you know) is  i  i ⇄ 0, (1-6) where  i is the i th component. For example, eq. (1-1) can be written as − 2A − B + ½C ⇄ 0. (1-7) This is the default way to input chemical equations in many chemical engineering software packages. 3 / ˌstɔɪkɪəʊˈmetrɪk / 4 better /ksai/, although /sai/ and /zai/ are more common;  is  capital

18 r ni =  i r n , and after dividing by V , r i =  i r . (1-22) For example, for the reaction 2A + B → ½C, this corresponds to r A = − 2 r , r B = − r , r C = ½ r . (1-23) Very often, people speak of rate of a reaction with respect to component A, B or C . This corresponds to the rates of the equivalent reactions A + ½B → ¼C, 2A + B → ½C, and 4A + 2B → C, respectively, i.e. rate with respect to component i refers to a formulation where |  i | = 1. The three formulations refer to three definitions of yield:  A ,  B , and  C . The three yields are related as 4  A = 2  B =  C . Rate law and mass action law. The rate of every reaction is a well-defined function of the conditions in the mixture – the composition, the temperature etc. The relationship between rate and composition, r = f( C A , C B, C C …), (1-24) is called the rate law that the reaction follows. Nearly all elementary (single-stage) reactions follow a simple power law dependence on the concentrations of the components, called the mass action law . For the irreversible elementary reaction  A A +  B B → C, (1-25) the mass action law states that: A B ν ν A B r kC C = . (1-26) Here, k is the rate constant of the reaction. The quantity |  A +  B | is the order of the reaction; just |  A | is the order with respect to A. Reactions that follow the mass action law are called simple . The elementary reactions are almost always simple, but the simple reactions are often multi-staged, and their rate constant k is usually a combination of several elementary rate constants. The elementary rate constants can often be predicted theoretically via micromechanical modelling of chemical reactions, e.g., via the transition state theory of Eyring 6 . Reactions that do not follow a simple rate law may still proceed according to a power law dependence. For example, if the reaction (1-25) is complex, it can follow a rate law of the sort A B C D r kC C C C     = , (1-27) where  and  are the orders with respect to the reagents A and B,  is the order with respect to the product C, and  is the order with respect to a catalyst or inhibitor D. Unlike the stoichiometric coefficients, the reaction orders  ,  ,  and  can have values that are positive or negative, integer or fraction. In addition, the rate law might be more complicated than a power law – polynomial fractions are especially common (see sec. 3 below). An important limitation of power laws for the rate of chemical reactions is that they are inaccurate for strongly non-ideal mixtures and solutions, especially where the dependence of the activation energy on the concentration is significant; reactions involving formation of a new phase are often such. Consider now a reversible elementary reaction, of equilibrium constant K :  A A +  B B ⇄  C C +  D D, C D A B ν ν Ce De ν ν Ae Be C C K C C = . (1-28) This condition for equilibrium is only an approximation – ideal mixture is assumed. Generally, activities a i have to be used instead of concentrations C i , and then the equilibrium condition alters to ν e i i a K =  . (1-29) 6 / ˈ airing/

21 Fig. 1. Trajectories of the amounts of substance and the reaction yield of the gas-phase reaction A → 2B under constant pressure and temperature, in nondimensionalized coordinates. Example 1-2. Consider the elementary reversible reaction A ⇄ B in liquid phase; neglect any change of volume and assume isothermal conditions. Analyse the evolution of the amount of A and B. Find the time for half-reaction. Solution. As stated, the problem implies again a batch reactor system. Since the process is isochoric, we will work with  V [mol/L] instead of  [mol]. The relationships between C i and  V are: C A = C A0 –  V , C B =  V . (1-43) Let us first find the composition after equilibrium has been reached. The equilibrium condition for the reaction A ⇄ B reads: Be Ae C K C = , or e A0 e V V K C   = − . (1-44) The equilibrium yield (the full reaction , the moles per litre of reaction until equilibrium is reached) is the solution to eq. (1-44), namely: e A0 1 V K C K  = + , so e e 1 V K X K    = = + , (1-45) where  ∞ = C A0 . The rate law for A ⇄ B reads ( ) ( ) f A B f A0 f e d 1 1 / d V V V V K K r k C C K k C k t K K     + +   = = − = − = −     . (1-46) Here, we first substituted C A and C B with eqs. (1-43) and simplified, and then used (1-45) to get rid of C A0 . This is solved by separation of variables similarly to the previous problem: ( ) e f f r e 1 ln V V V K k t k k t K    − + = − = − + ; (1-47) ( ) f r e 1 e k k t V V   − +   = −   , or after division by  ∞ = C A0 , ( ) f r e 1 e k k t X X − +   = −   . (1-48) The time for half-reaction (where  =  V e /2) is easiest to obtain from eq. (1-47): 1/2 f r ln 2 t k k = + . (1-49) n i / n A0 t / t 1/2 X =  /  ∞ t / t 1/2 half- reaction half- reaction

22 Fig. 2. Trajectories of the concentrations and the reaction yield of the liquid-phase reaction A ⇄ B, in nondimensionalized coordinates; K = 3. 1.2 Temperature dependence of rate and equilibrium constants In the simplest case, the rate constant follows the Arrhenius 7 equation ( ) f f 0 f exp / k k E RT = − , (1-50) where E f [J/mol] is the activation energy of the forward reaction; the preexponential factor k f0 is the collision frequency of molecules in the correct orientation. Similarly, for the reverse reaction, ( ) r r0 r exp / k k E RT = − . (1-51) Chemical engineers often use these equations in their equivalent forms f f f 1 1 exp E k k R T T     = − −         ; r r r 1 1 exp E k k R T T     = − −         , (1-52) where k ○ is the value of k at one standard temperature T ○ (usually 298 K). E f and E r are positive, and therefore, the rate constants rapidly increase with T . Some radical reactions are of zero activation energy (and are very fast); for them, k = k 0 may drop with the increase in T , due to a weak power law dependence of k 0 on the temperature ( k 0 ~ T n , where n = − 2…+3) . For the equilibrium constant K = k f / k r , an equation of similar form holds: Δ 1 1 exp H K K R T T     = − −         , (1-53) where K ○ = k f ○ / k r ○ is the equilibrium constant at room temperature, and  H = E f – E r is the heat of the reaction. This is van ’t Hoff’s 8 equation. The relation between heat of reaction and activation energies is often illustrated with the dependence of the energy U of a system on the reaction coordinate – a certain molecular geometric characteristic, such as distance between two atoms or an angle, that changes continuously upon a reaction event. For example, for the 7 / ɑ:ˈrenius / 8 / vɑ:n ə t ˈhɔf /, note the spelling of van ’t X =  V /  V ∞ t / t 1/2 half- reaction C i / C A0 t / t 1/2 C Be C Ae

23 decomposition AB → A + B of a diatomic molecule AB, this would be the distance d AB between the two atoms A and B, see Fig. 3. As the distance increases from the equilibrium bond length (state AB) to infinity (state A + B), the system is forced to pass through a maximum energy state, corresponding to the so-called transition complex [A---B] ‡ (elongated unstable configuration of the two atoms). The energy of [A---B] ‡ is by E f higher than that of the stable molecule AB, and by E r higher than that of two free atoms A + B. The difference between the energies of the initial state AB and the final A + B is equal to the heat of the reaction (approximately, as  U and  H differ by a small term RT due to the expansion of the system). Fig. 3. Relationship between activation energies and heat of a reaction (Polanyi-Wigner diagram). Derive from Maxwell distribution 1.3 Simple reactors and their mole balances Most of our module is dedicated to the question how the reactor type and the reaction nature determine the performance of the reactor, or more specifically, what is the performance equation of the reactor: output = f( input ; design ). Here, input means feed characteristics (composition, temperature, feed flow rate); design means reactor type, size, heat exchange system etc.; output means outflow composition, temperature, flow rate. The performance equation depends on two broad features of the process: the chemistry (i.e. reactions, reaction rates, mixture composition) and the mixing pattern (i.e. reactor design, flow parameters, heat transfer). Let us now briefly summarize what you already know about the basic types of reactors. The main thing the reactor type determines is the way the amount of a given component is delivered or removed from the system, i.e. the reactor type determines the mole balance in the system. The most general mole balance of the i th component in a reactor reads: amount amount change of amount flowing in flowing out produce from in the reactor with feed with product reactions 0 d d d i i i i V n F F r V t = − +  , (1-54) where n i is the amount of component i in a volume V of the reactor, F i 0 is the inflow of i (mol/s), and F i is the outflow; r i is the production rate. For one reaction of rate r taking place in the mixture, we know that r i =  i r . Eq. (1-54) is the integral (total) form of the mole balance. One useful differential (local) form reads 1 i i i i C F C r t t V      − = − +    ; (1-55) A─B ⇄ [A---B] ‡ ⇄ A + B d AB [Å] AB bond transition complex free A and B U [kJ/mol] E r E f  H

24 it assumes simple dependence of F on V (as in a plug flow reactor). The general mole balance (1-54) is valid for any reactor. Different types of reactors allow for various simplifications of eqs. (1-54) and (1-55). Eq. (1-54) is actually a very general count balance and can be applied to anything. An example: the population n of the UK changes for 4 reasons, - immigration ( F 0 = 600,000 y -1 ); - emigration ( F = 350,000 y -1 ); - the formation reaction, 2 humans → 3 humans, of rate  r birth d A = 680,000 y -1 ; - the consumption reaction, 1 human being → free soul, of rate  r death d A = − 350,000 y -1 . The integrals are over the area of the UK. This example is actually with two reactions in the reactor. The total rate of population change is 1 0 birth death d d d 580000 y d A A n F F r A r A t − = − + + =   , (1-56) A under the integral denotes that the integration is over the whole available area (not an integration limit). The data are rough averages for 2005-2015. Based on the equation, you can estimate the population and the structure of the UK society based on sociological models (e.g., Brexit will supposedly restrict the immigration from the EU, so F 0 will decrease by ~ 5-10%). Batch reactor , gas or liquid. A batch reactor is filled with the reactant mixture and then closed. There are no flows in or out of the batch reactor, therefore, F = F 0 = 0. In addition, let us assume that the reactor is well-mixed and therefore, homogeneous ( r i under the integral in eq. (1-54) is independent of the location). Under these assumptions, the mole balance eq. (1-54) simplifies to d d i i n Vr t = . (1-57) This is a trivial result – the mass balance coincides with definition of reaction rate. The balance can be simplified further in case that the volume of the reactor is constant (liquid phase reaction, or a gas phase reaction in a bomb) – in such case, it holds true that d n i /d t = V d C i /d t , so d d i i C r t = . (1-58) Using here eqs. (1-4)&(1-22), we obtain d d V r t  = . (1-59) For isochoric-isothermal reaction, where the rate law specifies a dependence of r on  , this equation is integrated to: 0 1 d ( ) V V V t r    =  . (1-60) For example, for a second order reaction A + B → C, the mass action law gives r = k ( C A0 −  V )( C B0 −  V ). Substituting this in eq. (1-60) and integrating gives immediately the time at which a reaction should be quenched in order to obtain a specified yield – see question 6c). For a single non-isochoric isothermal reaction, eq. (1-57) can be re-written using eqs. (1-3) as d  /d t = Vr . If V and r here depend on  only, we can again integrate: 0 1 d ( ) ( ) t V r     =  . (1-61)

25 Compared to the isochoric case, eq. (1-60), there are three complications. The first is that  is in [mol] and should refer to a specified initial amount of the mixture; if this initial amount is not specified, you should specify it yourself – i.e. consider 1 mole in total of the initial mixture, or 1 kg, or 1 mol of the key (most expensive) reagent. The second is that V is changing with  , and the dependence has to be specified by an additional condition. For example, for isobaric reaction in an ideal gas, V = nRT/p = RT  n i / p = RT  n i 0 / p + RT  i  / p = V 0 + ( RT  / p )×  , (1-62) where  =  i is the change in the number of moles in the reaction. For liquid phase reactions, V is again linear function of  : V =  n i v i =  n i 0 v i +  i  v i = V 0 +  v×  , (1-63) where 9  v =  i v i is the change in volume for mole of the reaction, and v i is the partial molar volume of the i th component (roughly a constant equal to the volume of the i th molecule times N A ). The final complication is that the mass action law is sensitive to the change in density. For example, for the reaction A + B → C, the mass action law has to be used in the form r = kC A C B = kn A n B / V 2 = k ( n A0 −  )( n B0 −  )/ V (  ) 2 . The batch reactor is the most widely used experimental reactor: nearly every new industrial process is born in small laboratory batch reactors. The reason is that we cannot really make a small plug flow or continuous stirred-tank reactor that consumes less than a barrel of feed per experiment. Batch reactors are fairly common for small-scale pharmaceutic and speciality chemical productions, as they allow for better quality control – each batch can be analysed and confirmed to be suitable for use. Natural/household examples for batch reactors: cooking soup in a pot. An engine cylinder can be thought as a batch reactor, but a very untypical one: cycle times are very small; the volume is fixed but time dependent; reactions do not produce useful products, only useful free energy. Semi-batch reactor is one in which some components are continuously added in the mixture, mostly to avoid excess heat production or to increase selectivity when multiple reactions are possible. Therefore, F i 0 in the general balance (1-54) is not zero for one or more components, and the mole balance reads 0 d d i i i n Vr F t = + . (1-64) Fig. 4. A typical batch or semi-batch reactor ( stolen from essentialchemicalindustry.org ). Semi-batch reactors are fairly common for small-volume chemical production and for bioreactions. Natural/household examples for semi-batch reactors: the stomach. Once food is delivered to the stomach, acid and enzymes are secreted continuously. After residence time of about 30 min, the mixture is transported to another reactor joint in series – the intestines. 9 Be careful –  (the Greek letter nu) is stoichiometric coefficient; v (the English vee) is molar volume, say IUPAC.

26 Plug flow tubular reactor (PFR). These reactors are normally operated under steady state conditions, so d n i /d t = 0 in the general balance (1-55). The differential mole balance reads d F i = r i d V , or ( ) d d d d i i i vC F r V V = = ; (1-65) here, v is the volume flow rate, and F i = vC i . If the density of the mixture does not change along the length of the reactor, then v = v 0 (the volume flow rate at the entrance) and the mole balance simplifies to: 0 d d i i C v r V = , or d d i i C r  = . (1-66) Here,  = V / v 0 is the space-time of the reactor . Note that eqs. (1-58) for a batch reactor and eq. (1-66) for PFR differ only by the name of the variable ( t in the former case,  in the latter). This means that C i ( t ) for isochoric-isothermal batch reactor and C i (  ) for isothermal PFR with incompressible flow are the same functions. Therefore, we can write for PFR: d ( ) d V V r    = , 0 d ( ) V V V r     =  , (1-67) analogously to eqs. (1-59)&(1-60). The plug flow reactor mole balance for a mixture that changes density is easiest to deal with if a yield  of units mol/s is introduced. If there were no reaction, the mole flow rate of component i through each cross-section of the reaction would be constant, F i = F i 0 . Due to the reaction, it changes by stoichiometric coefficient times the yield  t of the reaction, 0 ( ) ν ( ) i i i t F V F V   + , (1-68) where  t is in mol/s, i.e. it is a production rate . This is how many moles of reaction take place in a part of the reactor of volume V for one second. The general mole balance d F i = r i d V is valid; substituting F i in it with eq. (1-68) and rearranging gives α d ν ν ν d i t i i i i i r r k C V  = = =  . Note that d  t /d V turns out to be equal to the rate of the reaction r . Now we need to express the concentrations through the new quantity  t : ( ) 0 / ν / i i i i t C F v F v  = = + , and then also the volumetric flow rate v through  t : 0 0 Δ Δ i i i i t t v v F v F v v v   = = + = +   , where v 0 is the volumetric flow rate of the feed, v i is the partial molar volume of the i th component, and  v =  i v i is the change of the partial molar volume due to the reaction. For liquid phase reaction, this is usually a constant; for an ideal gas, v i = RT / p , and it could be the case that pressure p or the temperature T change along the length of the PFR. Combining the last three equations leads to a single equation for  t ( V ): ( ) ( ) α α 0 0 α 0 0 ν d ν d Δ Δ i i i i i t t i i t t t F F r k k V v v v v      +   + = = =    + +     . (1-69) If  v is constant, this is easily integrated to obtain the design equation:

27 ( ) ( ) α 0 α 0 0 0 Δ d d ( ) ν i t t i t t t t i i t v v V r k F         + = = +    . It tells us what should the reactor volume be in order to achieve production rate of  t . However,  v is a constant only for most liquid-phase reactions and for gas-phase isobaric-isothermal reactions. In gas-phase fixed-bed reactors, the pressure drop is important and  v =  RT / p ( V ) has to be used, with p ( V ) specified by a flow model (or sometimes assumed linear). When this is the case, the separation of variables above does not work. For non-isothermal gas-phase reactions, T (and therefore k ) become functions of V as well. In such situations, your non- explicit design equation is really the differential equation (1-69), which has to be solved numerically. PFRs are the most commonly encountered reactors in the chemical industry, because of a long list of advantages. They allow for continuous, fast, reliable, large-scale production; PFRs work well with gaseous, liquid and heterogeneous mixtures; residence times can be very short; the heat exchange system is very easy to design; high pressures can be achieved etc. Membrane plug flow reactor . The normal plug flow reactor has impermeable walls. The membrane plug flow reactor is one in which a component from the reaction mixture is added or removed selectively through the walls. In such case, the total molar flow rate F i in the general eq. (1-54) has two components: one for the cross-section flow F cs and a second for the flow through the walls: cs, d d d i i i F j A r V + = ; (1-70) here, A is the area of the membrane, and j i is the molar flow per unit area of component i through the membrane, in direction out of the reactor. Eq. (1-70) per unit volume of the reactor reads d d d d i i i F A j r V V + = . (1-71) If the reactor is a cylindrical tube of radius R r , d A /d V is simply equal to 2/ R r . Fig. 5. Typical membrane reactor designs. In the first one, the product C is removed from the mixture through the membrane (used with reversible reactions to increase conversion). In the second design, one of the reactants is delivered through the membrane (used to increase selectivity when the rate of a side reaction is minimized at low concentration of A). Stolen from Fogler [2]. An example for a natural membrane tubular flow reactor (although laminar flow, not plug flow) are our intestines. Continuous stirred-tank reactor (CSTR) , or mixed flow, or vat, or backmix reactor. In stationary state, d n /d t = 0, and the mixture is homogeneous, which simplifies the general mole balance (1-54) to 0 0 i i i F F rV = − + . (1-72) If flow rate is v , so F i 0 = vC i 0 , F i = vC i and V = v  , this equation simplifies to the familiar form 0 0 i i i C C r  = − + , i.e. 0 i i i C C r  − = . (1-73) Using eqs. (1-4)&(1-22), we can write this as an equation for  V : / V r   = . (1-74)

28 (stolen from Wikipedia) Natural examples: the surface of the Ocean is a sort of CST bioreactor – light, CO 2 and minerals are constantly flowing in, and O 2 and organic material is constantly flowing out. The human liver is an inflow-outflow tank type reactor, but not a well-stirred one. Example 1-3. Consider the elementary reversible reaction A ⇄ B in liquid phase, CSTR. Analyse the composition of the reactor as a function of its space-time. Find the space-time for half-reaction. Find the production rate of B, F B = vC B , as function of the flow rate. Solution. The relationships between C i and  V and the equilibrium composition are unchanged compared to the previous example 1-2: C A = C A0 –  V , C B =  V ; (1-75) e A0 1 V K C K  = + . (1-76) The rate law for A ⇄ B still reads ( ) f e 1 V V K r k K   + = − . (1-77) The mole balance (1-74) reads simply ( ) f e 1 V V V K k K     + = − , which is a linear equation for  V . To obtain the solution, we solve that for  V and simplify using that K = k f / k r : ( ) f e e 1 1 f r f 1 1 1 1 V V V K k K K k k k K      − − + = = + + + + . This is the design equation for the reaction A ⇄ B in CSTR – it allows computing the conversion  V at any flow rate and volume of the reactor (i.e. any  = V / v ). Combining reactors. Reactors are often used in series or in parallel. Well-designed assemblies of reactors are able to combine the advantages of two or three types of reactors. Assemblies are especially useful when multiple reactions take place.

29 Fig. 6. Nature- inspired reactors in series in artwork: “Cloaca” by Wim Delvoye . A room-sized assembly in Hobart, Tasmania, that replicates the journey of food through our gastrointestinal tract. This art-piece shares one thing in common with industrial reactors: as a rule, they are smelly! Judging from the photo, what type of reactors have been immortalized? Picture stolen from ABC News. The main parameters optimized when a reactor is designed: safety, selectivity, yield, temperature control and cost. Minimum volume. Typical advantages, disadvantages and features of the main reactors (based on table 3-1 of Schmidt) BR CSTR PFR reactor size for given conversion (typically) + - + simplicity and cost + + - continuous? - + + easy to clean? + + - on-line analysis - + + product certification + - - typical residence times hours from 10 min to 10 h s to tens of min

30 Read more Levenspiel’s ch. 1 -6 [1]. Fogler’s ch. 1 -7 [2]. Supafastquiz A reaction follows the rate law r = k [A] 2 /( K + [B]). What are three ways to accelerate it? A reversible reaction 2A ⇄ B is slow and exothermic. What temperature would be optimal for its rate: the highest achievable, the lowest achievable, or somewhere in the middle? Which of the following processes will be approximately isothermal, and which – adiabatic? a) synthesis of an organic molecule in dilute aqueous solution; b) an explosion of H 2 + O 2 ; c) a liquid phase reaction of enthalpy  r H = 10 J/mol; d) a slow exothermic reaction taking place for 24 h in a well-cooled batch reactor; e) compression of air in the engine cylinder for 10 ms during the intake stroke. Three reactors that allow components to be added to the mixture after the start of the reaction? A CSTR is of volume 5 L and flow rate is 10 L/h. What is the residence time? A batch reactor is of volume 5 L and residence time 10 h. What is the flow rate? The following three reactions are completely equivalent: A → ½B, r 1 = k 1 A 3/2 ; 2A → B, r 2 = k 2 A 3/2 ; 3/2 A → ¾ B, r 1.5 = k 1.5 A 3/2 . However, k 2 = k 1 /2, and therefore, the rate of the second reaction is slower than the rate of the first reaction. Now how is that possible? If the rate is not invariant to the way we choose to write the reaction, then what is invariant? What is the realationship between k 1.5 and k 2 ? PS are  H 1 ,  H 2 ,  H 1.5 . Problems 1. Easy questions a) What are the stoichiometric coefficients of each component in the following reactions: 3C 2 H 2 → C 6 H 6 ; − H 2 − Cl 2 + 2HCl → 0; Ca 2+ + SO 4 2- → CaSO 4 (S) and Ca 2+ + SO 4 2- + 2H 2 O → CaSO 4 ·2H 2 O (S) . b) Are the following reactions different? (i) A → ½C and (ii) 2A → C. Is their yield different? How are  i and  ii related? c) Are the following reaction different?

31 (i) 2A ⇄ 3C and (ii) 3C ⇄ 2A. How are  i and  ii related? 2. Derivations a) Derive eq. (1-13) from eq. (1-3). b) The relation (1-3) between amount of substance and yield, n i = n i 0 +  i  , can be thought of as a particular case of the mole balance eq. (1-54). Derive (1-3) from (1-54) for batch reactor. Derive the same for CSTR. Show that for any homogeneous reactor (semi-batch, non-stationary CSTR), instead of n i = n i 0 +  i  , it is true that ( ) 0 0 0 ν d t i i i i i n n F F t  = + + −  . c*) Space time is V / v 0 ; if the volumetric flow rate v changes along the length of a PFR, then the space time is not equal to the mean residence time  of the mixture in the reactor. Can you relate the function v ( V ) to  ? Can you relate  to v (   )? Ans. 1 0 d V v V  − =  ; ( ) 1 1 0 0 Δ d t t t v v r     − − = +  . 3. Basic irreversible reactions a) Consider the irreversible reaction 4A + 3B → 2C in liquid phase. The initial composition of the mixture is [A] 0 = 0.3 M, [B] 0 = 0.25 M, [C] 0 = 0.001 M. Find the composition of the mixture after completion of the reaction. b) In an engine cylinder, 10 mg isooctane react completely with 0.195 g of air (excess!). What will be the composition of the final mixture (give mole fractions of each component)? Air is 21:79 O 2 :N 2 by mole. c) The electrochemical reaction Zn + Cu 2+ → Zn 2+ + Cu consists of two conjugated half- reactions, Zn → Zn 2+ (aq) + 2e − (zinc anode), Cu 2+ (aq) + 2e − → Cu (copper cathode). The two electrons are transported from the anode to the cathode through an external circuit. Let the mass of the anode be 10 g, and the electrolyte solutions be of very large amount. What is the maximum yield of the reaction? What amount of Cu will be deposited on the cathode? What amount of electricity will pass through the external circuit? One mole of electrons corresponds to F coulombs, where F is Faraday’s constant. 4. Basic equilibria a) The following equilibrium has been established in aqueous solution: 2Fe 3+ + 2I – ⇄ 2Fe 2+ + I 2 . The concentration of iodine is determined via titration: [I 2 ] e = 0.0032 M. We also know the initial concentrations of Fe 3+ and I – : [Fe 3+ ] 0 = 0.0150 M, [I – ] 0 = 0.0110 M. Determine the equilibrium yield of the reaction, the conversion and the equilibrium constant of the process. n b) Gaseous HCl and O 2 are mixed in ratio 1:1 at 870 K, 1 bar. The following reaction takes place under constant pressure: 4HCl (G) + O 2 (G) ⇄ 2Cl 2 (G) + 2H 2 O (G) . The molar part of the Cl 2 is 0.238 when equilibrium is reached. Find the molar fraction-based equilibrium constant K x , the pressure based equilibrium constant K p , and the concentration- based equilibrium constant K C .

32 c) Consider the heterogeneous reaction A (G) + 2B (S) ⇄ 2C (G) (B is solid, A and C are gases). It takes place under constant pressure of 100 kPa; its equilibrium constant is K p = 1.3 (st. state 101325 Pa). The initial partial pressures of A and C are p A0 = 9.5 kPa and p C0 = 12.5 kPa (the rest is inert gas). Find the equilibrium composition of the mixture (the molar fractions and the partial pressures). d) Dimethyl ether, CH 3 OCH 3 or Me 2 O, is used as fuel, refrigerant, aerosol propellant, precursor for dimethyl sulfate, Me 2 SO 4 , and as solvent for some low-temperature pharma reactions. It is produced from methanol (one of the many useful products that stems from syngas): 2CH 3 OH ⇄ CH 3 OCH 3 + H 2 O. Your boss told you to investigate the possibility to produce Me 2 O in a gas phase reaction at increased temperature (above the boiling point of methanol, 65 °C), and your first job is to find the yield of the process as function of T . The equilibrium constant as function of temperature is given by K p = 0.0000196× T 1.18 exp(3226/ T ), where T is in units K; standard conditions do not matter. Plot the equilibrium conversion, X e =  e /  ∞ , as function of T up to 800 °C. Plot also the equilibrium partial pressures and mole fractions of all components as function of T , if the total pressure is 1 atm. What will be the conversion of methanol at 150 °C? Hint: two solutions, only one is physical! 5. Post-Arrhenius kinetics a) k = k 0 T n exp( − E / RT ) 6. An irreversible reaction in batch or PF reactor a) The liquid-phase reaction A + 2B → 2C follows the rate law r = k [A]. If the initial concentrations are [A] 0 = [B] 0 , find the time of completion of the reaction in a batch reactor (when [B] = 0). What is the time for completion of the reaction if [B] 0 = 3[A] 0 ? b) Maple The reaction ½ A + ½ B → C is simple. Let it take place in PFR; the density of the fluid is constant. Unless [A] 0 = [B] 0 , this reaction is completed for a finite residence time  fin . Find  fin as function of the ratio [A] 0 /[B] 0 . What happens when [A] 0 = [B] 0 ? Find the evolution of the concentrations in this case. c) The gas phase reaction A + B → 3C, which is simple, takes place in a batch reactor under constant volume (bomb). Find t (  ) using eq. (1-60). Plot the composition of the mixture and the pressure in the reactor against time using parametric plot. Parameters: n A0 = 1.5 mol, n B0 = 4 mol, V = 100 L, k = 0.01 m 3 ·mol -1 ·h -1 , T = 300 K. 45 d) The liquid-phase reaction A → 2B follows the rate law r = k [A] 2 . Find the composition of the mixture along the length of a PFR reactor (use the space-time as a variable), and plot it/sketch it. Find the time for half-reaction. Parameters: [A] 0 = 0.1 M, k = 0.01 M -1 ·h -1 . e) Maple The simple reaction 3A → 2B takes place in gaseous phase under constant pressure and temperature; no inerts. Find the evolution of the composition and the volume as function of time and the time for half-reaction – first analytically, and then plot/compute them. What should be the quenching time needed to obtain 90% conversion? Parameters: p = 1 atm, k = 1×10 -4 m 6 ·mol -2 ·h -1 . Hint: use eq. (1-61). f) The photochemical reaction C 2 Cl 4 + Cl 2 → C 2 Cl 6 follows the rate law r = k [Cl 2 ] 3/2 . If the initial concentrations are [C 2 Cl 4 ] 0 = ½[Cl 2 ] 0 , and the reaction takes place under isochoric- isothermal conditions, find the time of completion of the reaction (when [C 2 Cl 4 ] = 0). 7. A reversible reaction in batch or PF reactors

33 a) The reaction A ⇄ B is second-order ( not first! not simple! ) and reversible. It takes place in PFR under constant T without change in density. Find the equilibrium yield. Find the relationship between the equilibrium constant K and the rate constants k f and k r ; use it to get rid of k r . Find the dependence of  on  using eq. (1-67), and the time for half-reaction. Plot [A] and [B] vs.  . Find the flow rate needed to produce 60% conversion. Parameters: K = 2, [A] 0 = 1 M, k f = 0.003 M -1 ·s -1 , V = 300 L. 45 8. A reaction in CSTR a) The simple reaction 2A → B takes place in CSTR. Find [A], [B],  , X , and the specific production rate F B V = [B]/  as a function of  . Plot for [A] 0 = 1 M. What initial concentration of A is needed to produce 85% conversion? Parameters: V = 100 L, v = 5 L/min, k = 1 M -1 ·min -1 . b) The same for the simple reversible reaction 2A ⇄ B. 9. Autocatalytic reaction a) Consider the simple reaction A + B → 2B in a PFR. If the density of the system is constant, express the concentrations C A and C B through the yield  V . Write the rate law, and the explicit differential equation for  V (  ). Solve it. Plot the results for  V , C A , C B and for the rate r as functions of the space-time for C A0 = 1 M, C B0 = 10 -6 M, k = 0.1 M -1 min -1 . The rate r has a maximum at a specific space-time, called the initiation time  in of the autocatalytic reaction. Find it. Ans. for conversion: ( ) ( ) 0 0 0 0 [A] [B] [A] [B] 0 0 1 e [A] 1 e [B] − + − + − = + kt kt X . b) Consider the same reaction in a CSTR. Plot C A and C B as function of  . c) Consider 2A + B → 2B in a PFR. 10. Reactions used as clocks: dating rocks When a volcano erupts, it throws out a mixture of molten rocks. Due to the high temperature, the inorganic ions mix well in this mixture. However, as it cools down, various minerals phase- separate from it. Two common volcanic minerals that phase-separate from the lava are the plagioclase (a mixture of NaAlSi 3 O 8 and CaAl 2 Si 2 O 8 ) and the quintessence (FeTiO 3 ). Both minerals contain significant amount of impurities. In particular, small amounts of ions of Sr and Rb from the lava are included in the two minerals as admixture defects. Rb has a much higher affinity to the quintessence. Millions of years ago, the two ions, Sr and Rb, were well-mixed in the lava; in result, the lava had a well defined isotopic ratio 87 Sr/ 86 Sr. However, Rb started to mess up the isotopic ratio once the phase separation occurred, because 87 Rb is a little bit radioactive and produces 87 Sr, according to the nuclear reaction 87 87 37 38 Rb Sr k e − ⎯⎯→ + , a 1 st -order process with half-life of 87 Rb of t 1/2 = 48  10 9 y. As Rb is accumulated preferentially in the quintessence, there the isotopic ratio 87 Sr/ 86 Sr increases more quickly than in the plagioclase. A geologist or an astronaut dug out probes from the two rocks (when rocks are dated, two phases of the same age are called isochron ) and brought them to you. Your laboratory analysed the isochron and gave you the following concentration ratios: 87 Rb/ 86 Sr 87 Sr/ 86 Sr

34 Plagioclase 0.004 0.699 Quintessence 0.180 0.707 When did the lava solidify? Hint : use the results from Example 1-1.

35 2 Multiple reactions Your job is to: fundamentals … predict the yields in multi-reaction mixture; skills … formulate and solve a system of algebraic/ordinary differential equations, optimize selectivity/yield. Figuring out which are the important reactions taking place in the reactor and what are their rates is a significant chunk of the job of the reactor engineer. In particular, process optimization is often all about increasing conversion and selectivity, or avoiding certain undesirable or dangerous reactions by purifying the feed or altering the temperature and the pressure. Moreover, no major industrial chemical process is single-step; instead, you would usually have complicated mechanism with many consecutive stages and side reactions. This section is about the alphabet of multiple reactions; the next one will be about reaction mechanisms and the rates of non-simple reactions. 2.1 Two reactions There are only a handful of ways in which two reactions can be interconnected 10 : consecutive: A → B → C; parallel: A ← B → C; independent: A → B ← C; and A → B; C → D. (2-1) Under isochoric-isothermal conditions, the independent reactions are treated as simple 1-stage reactions; the first reaction does not affect the second in any way and the two just sum up. However, th e “independent” processes suddenly start to interact if we have gas-phase reactions where the volume changes (due to dilution effects) or under non-isothermal conditions (where T is altered by the heats of both reactions simultaneously). We will not consider this sort of interactions here, which leaves us with just two types. 2.1.1 Parallel reactions (branching reactions). Formulation in terms of component concentrations/amounts Parallel reactions are extremely common in organic chemistry – as a rule, in any synthesis, a number of isomers are produced together with the main product. An example is substitution reactions with arenes and alkanes. Whenever organic molecules and oxygen are co-existing at high temperature, oxidation takes place in parallel to the main process. If there is no O 2 , then another nasty side reaction might take place – coking, the production of carbonaceous material. Quantifying selectivity. The main problem that has to be solved by the engineer is to minimize the formation of the side products and make the process more selective . One everyday example: when we cook meat, we would like to minimize the oxidation reactions leading to aromatization and carbonization (producing foul-smelling carcinogenic chemicals) simultaneously with the desired partial hydrolysis, protein denaturation and esterification. There are several characteristics of the reaction that are commonly used as parameters to optimize. All of these are measuring how much desirable product D is formed in comparison 10 A reaction scheme is a type of graph , having vertices (the components) and edges (the reactions). The analysis of complicated reaction schemes utilizes the mathematical graph theory .

36 with various undesirable ones. Especially “undesirable” are those side products that can poison the catalyst, and products that are hard to separate from D. The first set of characteristics in use is conversions of the main (most expensive) reagent A to desirable product D and undesirable U: D A DA A0 Δ n X n = , where  D n A =  AD n D ; U A UA A0 Δ n X n = , where  U n A =  AU n U . (2-2) Here,  D n A is the amount of A consumed to produce D, and  AD is the stoichiometric ratio for A and D (how many moles of A are needed to produce 1 mole of D if no side reactions take place), and similarly for  D n A and  AU . For uncomplicated reaction schemes,  AD = |  A |/  D . If one desirable and several undesirable products are formed, the sum X DA + X U 1 A + X U 2 A + … will be equal to the total conversion of the reagent A, X A =  n A / n A0 = ( n A0 − n A )/ n A0 . Table 1. Standard characteristics of selectivity of multiple reactions. The stoichiometric coefficients are skipped for clarity. conversion X A (amount of A converted)/(initial amount of A) ~(A 0 − A)/A 0 ~ (D+U)/A 0 desirable conversion X DA (amount of A converted to D)/(initial amount of A) ~D/A 0 overall yield fraction Y D (amount of A converted to D)/(amount of A converted) ~D/(A 0 − A) ~ D/(D+U) overall selectivity S DU (amount of A converted to D)/(amount of A converted to U) ~D/U instantaneous yield fraction y D (rate of conversion of A to D)/(rate of conversion of A) ~ r D /| r A | ~ r D /( r D + r U ) instantaneous selectivity s DU (rate of conversion of A to D)/(rate of conversion of A to U) ~ r D / r U Two other common characteristics are the instantaneous and the overall selectivities (Table 1). The instantaneous selectivity s DU is defined as D D DU UD UD U U d ν ν d n r s n r = = , (2-3) where r D ≡ V -1 d n D /d t and r U ≡ V -1 d n U /d t are the rates of production of the two components D (desirable) and U (undesirable), and  UD =  UA /  DA (  DA is how much D is produced by a mole of A if no side products are formed; same for  UA ). For simple reaction schemes,  UD =  U /  D . The other selectivity in use is called the overall selectivity , defined as DU UD D U ν / S n n = . (2-4) For CSTR, S DU = s DU (where does that follow from?) . If multiple undesirable products are formed, the definition of selectivity has to be modified accordingly, e.g., 1 2 1 2 U U D DU D U U ... ν ν ν n n n S   = + +       . (2-5) The higher this selectivity, the lower the cost of the reagents per unit mass of useful product (as no feed is lost to produce side products), and the lower the cost of the separation procedures needed to remove U 1 , U 2 etc.

37 The final two common characteristics are the instantaneous yield fraction : D D D AD AD A A d ν ν d n r y n r = = − , (2-6) and the respective overall yield fraction : D D D AD AD A A0 A ν ν Δ n n Y n n n = = − . (2-7) Here,  n A is the total amount of A consumed (spent on both desirable and undesirable products). The last equation can be written as 1 2 1 2 U U D D D D D U U ... ν ν ν ν n n n n Y   = + + +       . (2-8) Unfortunately, there is no firmly established terminology for these fractions. Your colleagues might call the yield Y D selectivity and overall conversion , and use the letter S for it; our desirable conversion X DA is often called a yield . You might also see the stoichiometric factors dropped out in the definition. Therefore, always doublecheck. Example 2-1. Analysis of parallel reactions in a batch reactor. Consider the following two reactions occurring in the same liquid mixture: (2-9) Analyse the evolution of the mixture composition and characterize the selectivity. Solution. These reactions consume A simultaneously. Let the initial concentration of A be C A0 . After  V 1 moles per litre of the 1 st and  V 2 of the 2 nd reaction have taken place, the mixture content will be C A = C A0 – 2  V 1 – 2  V 2 ; C B =  V 1 ; C C = 3  V 2 . (2-10) Differentiation of these three equalities with respect to time t gives us the relation between the rates of production of A, B and C and the rates r 1 = d  V 1 /d t and r 2 = d  V 2 /d t of the two reactions: r A = – 2 r 1 – 2 r 2 ; r B = r 1 ; r C = 3 r 2 . (2-11) For the two reactions, two similarly looking rate laws hold true (just different k ’s) : 2 1 1 A r k C = and 2 2 2 A r k C = . (2-12) The substitution of eq. (2-12) in eq. (2-11) leads to 3 differential equations for the 3 unknown concentrations: ( ) 2 A 1 2 A d 2 d C k k C t = − + ; 2 B 1 A d d C k C t = ; 2 C 2 A d 3 d C k C t = . (2-13) The initial conditions for these ordinary differential equations are C A (0) = C A0 ; C B (0) = C C (0) = 0. This is the first common formulation of a chemical kinetic problem: in terms of unknown concentrations. The first of eqs. (2-13) can be integrated immediately by separating the variables: ( ) ( ) 0 A 1 2 1 2 2 A A A0 d 1 1 2 d 2 t t C k k t k k t C C C =  = − +  − + = − + ; 2A B 3C 1 2

38 ( ) A0 A 1 2 A0 1 2 C C k k C t = + + . This result is substituted in the second of the eqs. (2-13) for B: ( ) 2 A0 B 1 1 2 A0 d d 1 2 C C k t k k C t   =   + +   , and the integration from t = 0 to t gives ( ) ( ) ( ) ( ) 2 1 2 A0 A0 1 A0 B 1 2 1 2 A0 1 2 0 0 1 2 A0 d 1 2 d 1 2 2 1 2 t t k k C t C k C C k t k k C t k k k k C t + +       = = =   + + + + +         ( ) ( ) ( ) 2 1 A0 1 A0 1 2 1 2 A0 1 2 A0 0 1 2 1 2 1 2 t k C k C t k k k k C t k k C t   = − =   + + + + +   . (2-14) Likewise, ( ) 2 2 A0 C 1 2 A0 3 1 2 k C t C k k C t = + + . (2-15) Fig. 7. Evolution of the concentrations of A, B and C for two parallel reactions, liquid phase, batch reactor ( k 1 = 0.4 M -1 ·h -1 , k 2 = 0.15 M -1 ·h -1 , C A0 = 1 M). Let B be the desired product. The ratio S BC = 3 C B / C C is the overall selectivity of B with respect to the undesired product C, eq. (2-4); the coefficient  CB = 3 is used to account that 3 moles of C are formed from the amount of A that is needed to produce a mole of B. The selectivity for the considered reaction scheme turns out to be constant independent of the time of the reaction (but dependent on temperature): B 1 BC C 2 3 C k S C k = = . (2-16) The ratio Y B = 2 C B /( C A0 − C A ) is the overall yield fraction of B; the stoichiometric ratio  AB = 2 stands for 2 moles of A required to produce a mole of B. Using the results for C B and C A : B 1 B A0 A 1 2 2 ... C k Y C C k k = = = − + . The maximum yield of B (the maximum possible moles or concentration of B in the product) is achieved at infinite time where 1 in the denominator of eq. (2-14) becomes negligible: A0 1 B 1 2 2 C k C k k  = + . (2-17). C i / C A0 t / t 1/2 C B C C 2A B 3C 1 2