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Chapter 1: Chemical reactions and equilibrium, and stoichiometry. Chemical Engineering Kinetics and Reactor Design. 1-. Chemical Reactions and Kinetics. หมายถึง การที่สารตั้งต้นเปลี่ยนไปเป็นสารผลิตภัณฑ์ เมื่อเวลาผ่านไป ปริมาณของสารตั้งต้นจะลดลงขณะที่ปริมาณสารใหม่จะเพิ่มขึ้นจนในที่สุด.

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## Chemical Engineering Kinetics and Reactor Design

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**Chapter 1: Chemical reactions and equilibrium, and**stoichiometry Chemical Engineering Kinetics and Reactor Design 1-**Chemical Reactions and Kinetics**หมายถึง การที่สารตั้งต้นเปลี่ยนไปเป็นสารผลิตภัณฑ์ เมื่อเวลาผ่านไป ปริมาณของสารตั้งต้นจะลดลงขณะที่ปริมาณสารใหม่จะเพิ่มขึ้นจนในที่สุด 1. ปริมาณสารตั้งต้นหมดไป หรือเหลือสารใดสารหนึ่งและมีสารใหม่เกิดขึ้นเรียกว่า ปฏิกิริยาเกิดสมบูรณ์ เช่น A + B C 2. ปริมาณสารตั้งต้นยังเหลืออยู่ (ทุกตัว) เกิดสารใหม่ขึ้นมา เรียกว่า ปฏิกิริยาเกิดไม่สมบูรณ์ ซึ่ง ความเข้มข้นของสารในระบบจะคงที่ เช่น 1-**The Basic Classifications of Chemical Reactions**1. Homogenous Reactions These reactions occur in a single phase, for example the reaction between a β-ketone and HOCl: Other examples include: 1-**2. Heterogeneous Reactions**These reactions occur at an interface between two phases. Examples include photocatalysis and catalysis. 2.1 Mars-van Krevelen reaction pathway Catalytic oxidation of volatile organic compounds can occur via this reaction pathway which follows a two step process. 1-**2.2 Langmuir-Hinshellwood reaction pathway**(Langmuir-Hinshellwood-Hougen_Watson) In contrast to Mars-van Krevelen, a surface reaction may be proceed by a single step process while the reactants are adsorbed onto the surface of the catalyst. 3. Catalyst A catalyst speeds up a chemical reaction but it is neither consumed nor produced in the reaction. 1-**4. Reversible Reactions**Reversible reactions are perfectly reversible as the name implies and may occur when 5. Irreversible Reactions Irreversible reactions do not proceed in the reverse direction at any measurable rate and are often found when 1-**6. Parallel Reactions**Parallel reactions are reactions that occur in parallel. For example, TiO2 degradation of organics: Overall Reaction : Initiation: Parallel reactions: 1-**7. Series Reactions**Series reactions occur in several steps. For example, nitrification occurs by a two step process. Nitrosomonas: Nitrobacter: 1-**8. Acid base reactions**Reactions where an acid or base is produced; involving proton transfer 9. Precipitation reactions Coordination of anions and cations from an amorphous solid 1-**10. Redox reactions**Reactions that have transfer of electrons The reduction half reaction (reactant are gaining e-) The oxidation half reaction (reactant are loosing e-) 1-**Rate of Chemical Reaction**หมายถึง ปริมาณของสารใหม่ที่เกิดขึ้นในหนึ่งหน่วยเวลา หรือปริมาณของสารตั้งต้นที่ลดลงในหนึ่งหน่วยเวลา (1-1) Unit of Reaction Rate Mass or Moles Time For aqueous and solid Volume Time For gas 1-**Rate Determination**(การหาอัตราการเกิดปฏิกิริยา) สามารถหาได้จากสารทุกตัวในปฏิกิริยา แต่มักจะใช้ตัวที่หาได้ง่ายและสะดวกเป็นหลัก ซึ่งจะมีวิธีวัดอัตราการเกิดปฏิกิริยาหลายอย่าง เช่น • วัดจากปริมาณก๊าซที่เกิดขึ้น • วัดจากความเข้มข้นที่เปลี่ยนไป • วัดจากปริมาณสารที่เปลี่ยนไป • วัดจากความเป็นกรด-เบสของสารละลาย • วัดจากความดันที่เปลี่ยนไป • วัดจากตะกอนที่เกิดขึ้น • วัดจากการนำไฟฟ้าที่เปลี่ยนไป 1-**Rate**Law(กฎอัตราเร็วของปฏิกิริยา) aA + bB cC + dD (1-2) where is the rate constant at infinite dilution and is the activity coefficient for the activated complex in the forward direction. 1-**In dilute systems, the activity coefficients go to 1 and the**above rate law reduces to the following expression. (1-3) This rate law is “a” order with respect to the concentration of A and “b” order with respective to B and has a total order of “a+b”. In any reaction, the rate proceeds in accordance with the activities of A and B. For dilute solutions or ideal gases, the activity is equal to the molar concentration or partial pressure of A. 1-**Rate Determination**1. ถ้าปฏิกิริยากำหนดให้เกิดในขั้นตอนเดียว ค่า m, n จะเท่ากับค่าสัมประสิทธิ์ หน้าสาร A, B เช่น 2A + 3B C + 4D (1-4) Rate = d[A]/dt = k[A]2[B]3 2. ถ้าปฏิกิริยาเกิดหลายขั้นตอน ค่า m, n จะเท่ากับสัมประสิทธิ์หน้าสาร A, B ในขั้นตอนที่เกิดช้าที่สุด เช่น A + B C + D slow Step 1 A + A E + D fast Step 2 E + B C + A 1- Rate = d[A]/dt = k[A]2**3.**ถ้าปฏิกิริยาเป็นปฏิกิริยาย้อนกลับได้ (Reversible reaction) (1-5) (1-6) 1-**The relationship between the rates of reaction for a single**reaction step (1-7) 1-**Example 1-1**Assuming that for every 50 mg of algal carbon that is fixed, 5 mg of nitrogen and 1 mg of phosphorus will be used. Accordingly, the empirical stoichiometry is the rate of production of algal carbon or loss of nitrogen or phosphorus due to photosynthesis would be: If rcarbon is 50 mg/L then the following relationships may be expressed. rcarbon = 50 mg Carbon fixed/(L-d) 1-**Example 1-1**rnitrogen = 5 mg Nitrogen utilized/(L-d) rphosphorus = 1 mg Phosphorus utilized/(L-d) The oxygen production as compared to the carbon fixation is given by the stoichiometry of this equation: 1-**For a zero order reaction**A Products [A] – [A]o = -kt (1-8) when [A] = 0.5[A]o, we find the half-life to be (1-9) t1/2 = 0.5[A]o/k 1- Unit of k = mass or moles/(L-time)**For a first order reaction**A Products (1-10) (1-11) 1- Unit of k = 1/time**For a second order reaction**A +A Products (1-12) Unit of k = L/(mass or moles-time) 1-**A +B Products**If reactant B is present in large excess over reactant A, then (1-13) Where k’ = k[B] and k’ is a pseudo first-order reaction rate constant. This equation can be integrated to give (1-14) 1-**Reaction Rate Dependent Terms**• Nature of compounds • Concentrations of compounds • Surface Area • Temperature • Catalyst • Pressure 1-**Table 1-2: Effect of Reaction Rate Dependent Terms on Rate**of Reactions 1-**Temperature Dependent from Arrehenius’ Law**For many reactions, and particularly elementary reactions, the rate expression can be written as a product of a temperature-dependent term and a composition dependent term, or ri = f1(temperature)*f2(composition) For such reactions, the reaction rate constant has been found in practically all cases to be well represented by Arrhenius’ Law: (1-15) 1- Ea = activation energy**At the same concentration, but at two different**temperatures, Arrehenius’ Law indicates that (1-16) Example 1-2 Milk is pasteurized if it is heated to 63 oC for 30 mins, but if it is heated to 74 oC it only needs 15 s for the same result. Find the activation energy of this sterilization process. Solution t1 = 30 min at a T1 = 336 K t2 = 15 sec at a T2 = 347 K 1-**Example 1-2**Now the rate is inversely proportional to the reaction time so, The activation energy is 422,000 J/mol 1-**Elementary Reactions and the Relationship between Rate**Expressions and Mechanisms Very often the reason a pathway taken by a given chemical reaction is very different than what is given by the stoichiometric equation. This is why the rate expression is not necessarily related to stoichiometric equation. Furthermore, this is why it is important to study the reaction pathway so that an appropriate rate law is formulated. If we consider our general equation as shown below. 1-**What really occurs is that several steps are taken to carry**out the overall reaction and these steps can involve several intermediates as shown here. Reactant (s) Intermediate Intermediate + Reactant Product + Intermediate Intermediate Reactant (or stable product) Each step shown above is referred to as an elementary reaction. Elementary reactions are defined as unimolecular, bimolecular and trimolecular as they involve only one, two or at most three species. 1-**A B (Unimolecular)**A+B C (Bimolecular) A+B+C D+E (Trimolecular) The rates of reaction for unimolecular, bimolecular, and trimolecular reactions are given as: ra = -k[A] (1-17) (1-18) ra = -k[A][B] (1-19) ra = -k[A][B][C] 1-**To illustrate the difference between the reaction pathway**and its stoichiometric equation consider this biological reaction. S P The elementary reactions are given by these pathways: In this case, the presence of the enzyme E lowers the energy barrier to produce product P and therefore it speeds up the reaction. 1-**The rate of substrate loss, rs, is equal to this expression**(1-20) rs = -k1[S][E] However, this expression is useless in the rate expression because the concentration of unbound enzyme [E] would be difficult to determine. Consequently, the pseudo-steady state assumption is invoked as shown below. (1-21) rES = 0 = k1[S][E] –k-1[ES]-k2[ES] (1-22) The total amount of enzyme, Et, is a fixed quantity as given in this expression: 1-**(1-23)**Now it is possible to write [E] in terms of the reactant, [S] and Etwhich does not depend on time. (1-24) (1-25) The maximum velocity of the reaction occurs when all the enzymeis saturated and entirely in the ES form: (1-26) vmax = k2Et 1-**(1-27)**vmax = k2Et (1-28) When combining the above two equations, the resulting expression is the Michaelis-Menton equation: (1-29) The Monod equation relates the cellular specific growth rate to the substrate concentration and is similar to the Michaelis-Menton equation: (1-30) 1-**In biological waste treatment, the following expression is**often used. (1-31) Let’s take the case where the stoichiometric equation and elementary equation are identical. For the reversible reaction: The second reaction is very fast and therefore we find that the conversion of CO2 to follow this reaction. 1-**(1-32)**If we assume the following equation is the elementary reaction Then we can write (1-33) At equilibrium ra = 0 or we can use the thermodynamics to develop this equation: (1-34) 1-**For reactions where radicals play a significant role. The**chainreaction is characterized by the following sequence of elementary reactions: A classical example of chain reaction is the formation of hydrogenbromide from hydrogen and bromine, 1-**This reaction may be explained by the following free-radical**chain reaction. Each of these is assumed to be elementary. The procedure to be used to develop the rate expression is described here. The net rate of formation of HBr is: 1-**The rate law should not contain the concentration of free**radical, [H+] and [Br-], both of which are difficult, if not impossible, to measure. The net rates of formation of the free radicals are given here: The desired rate of formation of [HBr] is obtained by substituting and rearranging followed by simplification. (1-35) 1-**Steps to obtain the rate law.**• Propose reaction pathway. • Write out the rate laws for the pathway. • Invoke pseudo-steady state assumption for reactive intermediates.Or perform a mass balance on the catalytic species. Rearrange these expressions to obtain expressions of the reactive intermediates or catalytic species in terms of the reactants or products which are easily measured. • Obtain rate expression and eliminate the hard to measure species from the rate law using the results from step 3. 1-**REACTANTS**Ea ENERGY LEVEL DH°Reaction PRODUCTS REACTION EXTENT Effect of Temperature and Catalysis on Reaction Rate When molecules collide in pairs, they do not always react to form products. Consider A + B P 1- Figure 1-1: Reaction Intermediates and the Energy Required to Form Them**According to collision theory, only colliding pairs which**havesufficient kinetic energy to overcome the activation energy, Ea, will react. If they follow a Bolzman distribution, they follow the distribution in the following figure. In addition, even for unimolecular elementary reactions, only a fraction of the reactant has enough energy to form the reactive intermediate to proceed to form products. NO. WITH ENERGY GREATER THAN Ea Ae-Ea/RT Ea ENERGY OF COLLISION 1- Figure 1-2: Kinetic Energy Distribution of Molecules Colliding During a Reaction**Let’s relate this concept back to the rate of reactions.**If we have -ra = (frequency of collision)*(fraction of those which react energy and orientation) (1-36) Where A is the Arrhenius factor or the pre-exponential factor.Similarly, if we have: (1-37) 1-**Effect of a catalyst on the reaction energy during a**reaction Ea Ea AFTER INTRODUCTION OF CATALYST REACTANTS ENERGY PRODUCTS REACTION EXTENT 1- Figure 1-3: Effect of a Catalyst on the Reaction Energy During a Reaction**Chemical Equilibrium and Thermodynamics**Will this reaction go? How can it proceed? 1-**The Dynamic Nature of Chemical Equilibrium**rb = kb[C]c[D]d rf = kf[A]a[B]b At equilibrium rf = rb (1-38) kf[A]a[B]b = kb[C]c[D]d (1-39) 1-**The Thermodynamics Basis of Chemical Equilibrium**For a closed system at constant pressure and temperature, the thermodynamic expression for free energy can be expressed as: G = H-TS (1-40) Where G = Gibbs free energy , kcal T = absolute temperature, K S = entropy, kcal/K H = enthalpy, kcal 1-**Enthalpy (H, heat content of a substance) :**natural systems tend to go from a state of higher energy to a state of lower energy. For instance, a ball rolls down a hill spontaneously, but not up. The ball looses potential energy as it rolls down hill. At the bottom of the hill it has zero potential energy. This same idea can be applied to chemical potential. 1-

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