Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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- The reversible gas-phase elementary reaction CH₂ CH₂ (a) Write out each step of the algorithm. + H₂ (ethylbenzene styrene + H₂) is carried out in an isothermal CSTR with no pressure drop. The feed enters a volumetric flow rate of vo = 5000 dm³/hour. The feed consists of half ethyl benzene (species A) and half interts on a molar basis and is well mixed before it enters the reactor (I). The pressure in the reactor is 6 atm (so PA0 = 3 atm and P10 = 3 atm, making the initial concentration of A = 0.04 mol/dm^3). The molar flow rate of A is FAO = 200 mol/hr. At the reaction temperature of 640 C, the rate constant isk₁= 5.92 mol/ (dm³ hr * Atm). The equilibrium constant is Kf = 9 atm and the equilibrium conversion is X₂ = * 0.84.arrow_forward13) The following gas-phase complex reactions follow elementary rate laws A+ 2B-C 2A+ C D RB=WBa=k'C(CB-CBS)α = KCB CB and take place isothermally in an isothermal membrane reactor. An equal molar feed of A and B enter the reactor, and species C is able to diffuse through the membrane. Pressure drop is significant in the reactor. Set up the equations that you would need to plot and analyze the molar flow rates of each species and conversion of species A as a function of volume. Include all equations that you would need to enter into Polymath except for explicit equations defining constants (i.e., vo = 1 L/min, etc.) or initial and final conditions. A species: A,B,C,D MB: =rA dFA dFB=B dFc CFD =rc-RC V dv dv CH =rD Rate laws: -VA=KACACB -VA = KACACC N Net Rate laws: rj = Σ Relative Rates: VA=VB = C=D Fj olchiometry: C = CTo FP7 = C=C TO + T=FA+FB+FC+FD FT PO FJ =Cj=CTO! FT isothermal essure Drop: d (P/Po) - dy -- a T FT Ap dw dw Hold this for each specie 2y 24 TO FTO X=> RC = 7arrow_forward17arrow_forward
- 1. The kinetics of an enzyme-catalyzed reaction is described as r = (1+ aI)(K1 + S)* The following data were obtained in rate studies conducted in a well-mixed steady state continuous flow reactor using the same enzyme loading. S (mol/L) I (mol/L) r (mol/{L.min}) | 4.3 0.6 0.3 0.5 0.05 3.8 2.1 The experimental data should provide enough relations for the unknown kinetic parameters, rm, K1, and a (no need to use linear regression). Solve the relations to obtain the parameter values.arrow_forwardProblem IX (10 p): Consider the isomerization of compound A from its cis to its trans isomer at 298 K, which is first order in both directions: ki cis-A = trans-A k2 The rate constant for the forward reaction is k1 = 4.2 × 10* s', and the rate constant for the reverse reaction is k2 = 2.1 × 10-ª s-1. %3D a) Calculate the equilibrium concentrations [cis-A]eq and [trans-A]eq, assuming that at time t= 0 all of A was in the cis-form, i.e., [cis-A]o = [A]o. (Hint: Use the fact that at any time, [cis-A] + [trans-A] = [A]o.) %3D b) Which reaction is spontaneous, the forward reaction or the reverse reaction? Provide your reasoning. Total /100 parrow_forwardLittle Johnny’s parents leave him alone with the babysitter while they go out for an amazing dinner at Ruan Thai restaurant in Wheaton, Maryland. While the babysitter listens to music by Between the Buried and Me and texts her friend, Johnny goes down to the basement and performs the thermite reaction that he read about—it was used to weld railroad ties! How cool is that?! 2 Al(s) + Fe2O3(s) à 2 Fe(s) + Al2O3(s) If he reacts 11.000 g aluminum with 19.000 g iron(III) oxide, will the iron he makes melt? (In other words, will the heat produced be sufficient to raise the temperature of the iron he makes to its melting point of 1537oC?) Assume that the iron absorbs only 60.0% of the heat produced. Assume the iron’s initial temperature is 17.22oC. The specific heatof iron is 0.450 J/goC. Do you think Johnny will get in trouble? [Hint: Find the final temperature of the iron and compare to the melting point. Hint: The masses of both reactants are given. What…arrow_forward
- 14) The gas-phase catalytic reaction A + B + C is believed to have the following rate law. -r= k (KB PAPB - KCPC) Ks 1+ KBPB+ KCPC Propose a mechanism and rate limiting step that are consistent with the rate law above. Note: for full credit, you must prove that the mechanism and rate limiting step match the rate law. where did the negative come from? Molecular adsorption Mechanism: A*A* B+* B* Desorption this is written as an A*+B*→C+* YAD=KAM (PACK-CA*) VAD,B=KAB (PBC*- rs= Ks CA*CB* CB KB sts. adsorptionarrow_forwardSuppose that the reaction 2A 2 3B k- is second-order in the forward direction and third-order in reverse. Derive an expression for the equilibrium constant in terms of the concentrations of A and B by equating the forward and reverse rates.arrow_forwardot pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt Z 40 what is the value of the rate constant? The decomposition of N₂O5 in CCl4 is a first-order reaction. If 1.47 mg of N₂O5 is present initially and 1.45 mg is present after 2.92 minutes at 55 °C, min-1 1 pt D Rate constant = Submit Answer W S Jull 3 X Alt E Try Another Version Q Search 14 с P % F 5 V FO 3 item attempts remaining G [References] SUS ZenBook as H N J S M Alt fuent Ctrl Homearrow_forward
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