Special Topic 4 of the Registered Chemical Engineer Professional Examination: Chemical Reaction Engineering

(7) Catalysts and their effect on reaction rate (multiple-choice question): They participate in chemical reactions while maintaining their mass and chemical properties before and after the reaction; catalysts can alter the rates of the forward and reverse reactions, but they cannot change the heat of reaction or the initial and final states of the reaction system. They also cannot change the equilibrium state, but they can reduce the time required to reach equilibrium. Catalytic action is selective. (8) Some useful conclusions: For reactions of positive order, the reaction rate always decreases as the conversion rate increases; for irreversible reactions and reversible endothermic reactions, the reaction rate increases with rising temperature, while for reversible exothermic reactions, the reaction rate is highest at the optimal temperature; for reversible reactions, the reaction rate is lowest when equilibrium is reached. 1). In a autocatalytic reaction, when the conversion rate is zero and there are no products in the system, what is the reaction rate? A. Maximum B. It reaches an extreme value C. Negative D. Zero. D 2). In a plug-flow reactor, for a certain reaction with variable volume, if &a>0, what is the relationship between space-time T and average residence time t? A. Equal to B. Less than C. Greater than D. Close to C. 3). Catalysts cannot ( ) A. Change the reaction products B. Change the reaction equilibrium C. Change the reaction rate D. Suppress side reactions. B 4) For a series of reactions A----B----D, with reaction rate constants K1 and K2 respectively, if Ea2 > Ea1, then to favor the formation of B, in principle one should ( ) A. Increase the temperature B. Decrease the temperature C. Remove product D promptly D. Reduce the reaction time. B 5) If the enthalpy change for a certain reaction is -150 kJ/mol, then what is the activation energy for the forward reaction? A. Cannot be determined B. Greater than 150 C. Less than 150 D. Equal to 150. A 6) When CH3CHO vapor undergoes thermal decomposition at 791 K and a pressure of 48.3 kPa, the half-life is 410 seconds; at a pressure of 22.5 kPa, the half-life is 880 seconds. What is the order of this reaction? A. 0 B. 1 C. 2 D. 3. C 7) If the activation energy for a certain reaction is 80 kJ/mol, and the temperature increases from 100°C to 110°C, then the reaction rate becomes approximately ( ) times the original rate. A. 1 time B. 2 times C. 3 times D. 4 times. B 8) The reaction 2A-----P is second-order. The time taken to consume 1/3 of the reactant A differs by 9 seconds from the time taken to consume 2/3 of it. What is the half-life of this reaction? A. 6 seconds B. 8 seconds C. 12 seconds D. 16 seconds. A 9) *In a first-order liquid-phase reaction, at temperature T, the initial reaction rate is r0 = 1*10^-5 mol.dm^-3.s^-1. After 1 hour, the reaction rate is r1 = 3.26*10^-6 mol.dm^-3.s^-1. What is the reaction rate constant? A. 3.11*10^-4 s^-1 B. 2.11*10^-4 s^-1 C. 4*10^-4 s^-1 D. Cannot be determined. A 10) *It is known that at 25°C, the decomposition rate constant for NaOCl is k = 0.0093 s^-1, and at 30°C, it is k = 0.014 s^-1. Then, at 40°C, how many seconds are needed to decompose 99% of the substance? A 60 B 139 C 270 D 360 B

2. Case knowledge 1). A homogeneous reaction takes place in a batch stirred-tank reactor: A + B ----- P. The reaction kinetics are given by -rA = k*CA*CB, with kmol.m^-3.h^-1; k = 10^4 m^3.kmol^-1.h^-1. When the initial concentrations of both reactants are 4 mol.m^-3 and the conversion rate of A is 0.8, this reactor can process 41.04 mol of reactant A per hour on average. The reaction will now be carried out in a plug-flow reactor with an inner diameter of 125 mm, while all other operating conditions remain unchanged; determine the length of the reactor. Since CA0 = CB0, it follows that -rA = k*CA^2. The reaction time tr is given by tr = CA0*f dXA/(-rA) = XA/KCA0*(1-XA) = 0.1 hours. This reaction time corresponds to the residence time in a plug-flow reactor. Thus, r0 = FA0/CA0 = 41.04/4 = 10.26 m³/h. V = r0*t = 10.26*0.1 = 1.026 m³. The length of the pipe, L, is given by V/0.785*d² = 0.1026/0.785/0.125/0.125 = 84 m. 2) For a certain second-order irreversible liquid-phase reaction, it takes 285 seconds to reach a certain concentration when the initial concentration is 5 kmol/m³; it takes 283 seconds to reach the same concentration when the initial concentration is 1 kmol/m³. So, how much time is required to go from an initial concentration of 5 kmol/m³ to 1 kmol/m³? . Thus, T=285-283=2s; it is a reversible reaction in which the volume remains constant before and after the reaction, and the portion that has already reacted has no effect on the reaction. Any moment during the reaction process can be used as the initial and final moment. 3). How much does the radioactivity decrease when a radioactive fluid with a half-life of 20 hours flows at a rate of 0.1 m3/h through two series-connected plug-flow reactors of 40 m3 each? Radioactivity follows first-order reaction kinetics. Thus, k = Ln2/k = 0.693/20 = 0.03465 h^-1. For a fully mixed flow reactor, T = CA0 – CAf / –rA, for the first reactor. 40/0.1 = (CA0 – CAf) / kCA0(1 – XA); XA = 0.9327. In the second reactor, the conversion rate remains 0.9327. Therefore, the overall conversion rate is = (CA0 – CA0(1 – XA)(1 – XB)) / CA0 = 1 – (1 – 0.9327)(1 – 0.9327) = 0.9955. 4). Complex liquid-phase reaction: A + B → P(1) ； A+P------S(2): The initial concentrations are CA0=2.0 mol/l, CB0=4.0 mol/l, while CP0 and CS0 are both 0. The reaction takes place in a batch reactor, and the measured concentrations are CA=0.3 mol/L and CB=2.4 mol/L. What are the concentrations of P and S? Based on B as the benchmark, what is the selectivity for reactant B and the yield of the product? 1 The amount of A consumed = (CB0 – CB) = 4 – 2.4 = 1.6 ； Reaction 2 consumes A = (2 – 0.3) – 1.6 = 0.1; therefore, Cs = 0.1. The concentration of product P is equal to the amount of P produced minus the amount consumed in reaction 2, so Cp = 1.6 – 0.1 = 1.5. Based on B, the selectivity of the reaction is given by B = Cp/(CB0 – CB) = 1.5/4 – 2.4 = 0.9375. The yield of product P is Cp/CB0 = 1.5/4 = 0.375. 5) A certain gas-phase first-order decomposition reaction: A → 3P ； The reaction was carried out in an isothermal tubular reactor; the feed contained 50% reactant and 50% inert material, with a residence time of 10 minutes. The volumetric flow rate at the system outlet became 1.5 times the original value. Determine the conversion rate of A and the reaction rate constant under these conditions. The expansion factor SA = 3 – 1/1 = 2; the expansion rate EA = SA * yA0 = 2 * 0.5 = 1. V = V0(1 + EA * XA), so V/V0 = 1.5 and XA = 0.5. T = CA0 * f * dXA/(-rA) = f * dXA/(k * (1 – XA)/(1 + XA)); with T = 10 and XA = 0.5, it is determined that k = 0.08863 L/min

This post was last edited by zhanghp30 on 2020-1-16 at 10:40. 6)*. In a PFR, the oxidation dehydrogenation of butene takes place under isothermal conditions at 650°C to produce butadiene: C4H8(A) → C4H6 + H2. The reaction rate is given by -rA = kPA / kmol/m3·h. The feed gas is a mixture of butene and water vapor in a molar ratio of 1, with an operating pressure of 0.10133 MPa. At 650°C, the value of k is 106.48 kmol/m3·h·MPa. The conversion rate of butene is 0.9; what should the space velocity be? T = CA0 * f * dXA / kPA; PA = CA * RT, and SA = 1 + 1 – 1/1 = 1. Therefore, CA = CA0 * (1 – XA) / (1 + 0.5XA). Thus, T = f * (1 + 0.5XA) * dXA / (kRT * (1 – XA)) = 13.2 s. For the liquid-phase autocatalytic reaction A + P → P + P, -rA = k * CACP. In an isothermal batch reactor, the reaction rate was determined to be CA0 = 0.95 mol/L and CP0 = 0.05 mol/L. The reaction rate reached its maximum value after 1 hour; the rate constant at this temperature is to be determined. -rA = kCA*(1-CA); d(-rA)/dCA = 0. Therefore, CA = 0.5 MOL/L. T = -f dCA/k*CA(1-CA). After integration, and by substituting T = 1, it is found that k = 2.944 L/MOL·H. Questions 6 and 7 require a solid foundation in advanced mathematics; although these questions are described simply and their solution processes are straightforward, they are common types of exam questions, so they deserve attention! 8)*. In a CSTR reactor, the following reaction takes place: 4NH3(A) + 6HCHO(B) → (CH2)6N4(P) + 6H2O(S). The volume of the reactor is 490 cm3, and the stirring speed is 1800 rpm. The reaction kinetics are given by –rA = kCACB², where K = 1.42*10^3*E^-3090/T. The concentration of A is 4.06 mol/l, that of B is 6.32 mol/l; both have a flow rate of 1.5 cm3/s. The reaction temperature is 36°C. Determine the concentrations at the reactor outlet, Caf and Cbf. The inlet concentrations are CA0 = 4.06/2 = 2.03, CB0 = 6.32/2 = 3.16. Using the formulas CA = CA0(1–XA) and CB = CB0 – 3/2*CA0, together with the relationship r*(CA0–CA)/–rA = VR (where r = 3 cm3/s), the values of XA, CA, and CB can be determined. The results are XA = 0.82, CA = 0.365 mol/L, and CB = 0.663 mol/L.

9)*. In a fully mixed reactor, a reversible reaction occurs: A + B ↔ R + S. The rates constant for this reaction are k1 = 7 m3/kmol·min and k-1 = 3 m3/kmol·min. The volumetric flow rates of A and B are 0.004 m3/min, with concentrations of 2.8 kmol/m3 and 1.6 kmol/m3 respectively. The volume flow rate of the reactants combined is VR = 0.12 m3/min. What is the conversion rate of B? r=0.008 m3/min, CA0=2.8/2=1.4, CB0=1.6/2=0.8. Let the conversion of B be CB=CB0(1-XB)=0.8(1-XB); then CA=CA0-CB0XB=1.4-0.8XB, and CS=0.8XB=CR - rB=K1CACB-K-1*CR. Thus CS=2.56XB^2-12.32XB+7.84. T=VR/r=0.12/0.008=15 min. Solving CB0XB/-rB gives XB=0.75. 10) For the sequential reaction A------P-------S, with K1=0.15 MIN^-1 and K2=0.05 MIN^-1, an inlet flow rate of 0.5 m3/min, CA0≠0, while CP0=CS0=0; determine the yield of P in a CSTR with a volume of 1 m3. T=1/0.5=2 MIN, YP=CP/CA0=K1T/（1+K1T）（1+K2T）=0.21

This is my weak area; I’m going to print it out and study it carefully~~

It seems the original poster hasn’t provided any summary regarding complex reactions (parallel, sequential).

It’s in the review tutorial; I have a tight work schedule, so I can only focus on the types of questions like these – and these are likely to appear in the exams! If you don’t know, go and look it up in a book.

I lost points in this area in the last exam; I need to make it up this time