QUESTIONS No. 60) – 64): Determine Xe for a PFR with no pressure drop when the reaction is reversible: 2A e B. The system is gas phase isothermal with an initial feed rate of A = 3 moles/min, initial concentration of A = 0.1 mol/dm?, k = 2 dm/mal-s, Ks = 100 dm/mol. 60)The concentration of A at equilibrium is represented by: a) Cae = Cao (1 – X,) b) Cae = CAo(1-X,) c) Cae CAo²(1=2X,) d) (1+ex,) (ex) Cae VAo(1-X,) eCAo(1+X,) 61)The concentration of B at equilibrium is represented by: CAo(1-X,) CAo (1-2x,) a) CBe = 2CA0 (1 – eXe) b) Cge c) Cse (1+ex.) (1+ex,) d) Све (1+ex,) 62)The derived equation of finding the equilibrium constant is represented by: b) Ke = Xe(1+X) Xe(1-X,) а) к, c) K. d) 2CAO(1-X) Cao(1-ex,) Ke = 2(1-X) Cao(1+Xe)? (1+Xe)? 63)The value of e is: a) 1 b) 1.5 64)The conversion of the reaction at equilibrium is: b) 0.82 c) 0.5 d) 2.0 a) 0.89 c) 0.84 d) 1.15

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:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
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Please answer number 62 and 64. Thank you in advance!

QUESTIONS No. 60) – 64): Determine Xe for a PFR with no pressure drop when the
reaction is reversible: 2A e B. The system is gas phase isothermal with an initial feed
rate of A = 3 moles/min, initial concentration of A = 0.1 mol/dm?, k = 2 dm/mol-s, Ks
= 100 dm/mol.
60)The concentration of A at equilibrium is represented by:
a) Cae = Cao (1 – X,)
b) Cae
CAo(1-X)
c) Cae
CAo (1-2x,)
d)
(1+ex,)
(ex)
VAo(1-X,)
Cae
61)The concentration of B at equilibrium is represented by:
а) Све — 2Сло (1 — еX,) b) Сзе
eCao (1+X,)
Cao(1-X,)
CAo (1-2X,)
c) Cpe =
(1+ex.)
(1+ex,)
Cao (x.)
d) Cse =
(1+ex,)
62)The derived equation of finding the equilibrium constant is represented by:
а) к, —
Xe(1+Xe)
b) Ke =
c) Ke =
d)
2CA0(1-X,)"
(1+X,)
CAo(1-ex,)
K =
2(1-X)
63)The value of e is:
a) 1
b) 1.5
64)The conversion of the reaction at equilibrium is:
b) 0.82
c) 0.5
d) 2.0
a) 0.89
c) 0.84
d) 1.15
Transcribed Image Text:QUESTIONS No. 60) – 64): Determine Xe for a PFR with no pressure drop when the reaction is reversible: 2A e B. The system is gas phase isothermal with an initial feed rate of A = 3 moles/min, initial concentration of A = 0.1 mol/dm?, k = 2 dm/mol-s, Ks = 100 dm/mol. 60)The concentration of A at equilibrium is represented by: a) Cae = Cao (1 – X,) b) Cae CAo(1-X) c) Cae CAo (1-2x,) d) (1+ex,) (ex) VAo(1-X,) Cae 61)The concentration of B at equilibrium is represented by: а) Све — 2Сло (1 — еX,) b) Сзе eCao (1+X,) Cao(1-X,) CAo (1-2X,) c) Cpe = (1+ex.) (1+ex,) Cao (x.) d) Cse = (1+ex,) 62)The derived equation of finding the equilibrium constant is represented by: а) к, — Xe(1+Xe) b) Ke = c) Ke = d) 2CA0(1-X,)" (1+X,) CAo(1-ex,) K = 2(1-X) 63)The value of e is: a) 1 b) 1.5 64)The conversion of the reaction at equilibrium is: b) 0.82 c) 0.5 d) 2.0 a) 0.89 c) 0.84 d) 1.15
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