(a) Half-life of a compound decomposing in second order process only depends on the reaction rate (b) Half-life of a compound decomposing in 0-th order process only depends on the reaction rate (c) If a reaction enthalphy <0 and reaction entropy > 0, the reaction is only spontancous if temperature is larger than certain temperature T
(a) Half-life of a compound decomposing in second order process only depends on the reaction rate (b) Half-life of a compound decomposing in 0-th order process only depends on the reaction rate (c) If a reaction enthalphy <0 and reaction entropy > 0, the reaction is only spontancous if temperature is larger than certain temperature T
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
Problem 1.1P
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mark each of True or False and all parts pls.
![(a) Half-life of a compound decomposing in second order process only depends on the
reaction rate
(b) Half-life of a compound decomposing in 0-th order process only depends on the reaction
rate
(c) If a reaction enthalphy < 0 and reaction centropy > 0, the reaction is only spontancous if
temperature is larger than certain temperature T*
(d) If reaction enthalpy is > 0 and reaction entropy > 0, the reaction is always spontancous
(e) If reaction enthalpy is > 0 and reaction entropy > 0, the reaction is never spontancous
(f) We have a system that can exist in two isomers, A and B, with reaction rate A -> B
equal to ki, and reaction rate of B -> A cqual to k2. To be able to determine he ratio of
equilibrium concentrations [A]/[B], we also need to specify the initial concentrations of
A and B.
(g) Perpetum mobile of the second kind satisfies first law of thermodynamics, but breaks
the second law.
(h) One of the equivalent formulations of the third law of thermodynamics states it is
impossible to reach T=OK.
(1) We are studying two different reactions A <-> B, and C <>D. The Gibbs free energy
difference between the starting state and end state of the process are the same, but in one
case (C -> D) the reaction includes much larger barrier during the conversion of C to D.
Hence, the equilibrium ratio D/C is going to be smaller than B/A.
6) I have a reaction of compounds in solution, following this reaction: A + B<->C, which
reaches its equilibrium concentrations A_eq, B_eq, and C_eq. Then, I will add some
more C to the equilibrated mixture, and let it equilibrate into a new values of
concentrations A*_eq, B*_eq, and C*_eq. This is enough information for us to be able
to say that A°_eq will be larger than A_eq.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F221b47ae-c881-4152-8ff2-d1d3bc3262d6%2Fd1850659-f253-4c70-a39e-283ab2a94bd6%2Fcu5z3js_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(a) Half-life of a compound decomposing in second order process only depends on the
reaction rate
(b) Half-life of a compound decomposing in 0-th order process only depends on the reaction
rate
(c) If a reaction enthalphy < 0 and reaction centropy > 0, the reaction is only spontancous if
temperature is larger than certain temperature T*
(d) If reaction enthalpy is > 0 and reaction entropy > 0, the reaction is always spontancous
(e) If reaction enthalpy is > 0 and reaction entropy > 0, the reaction is never spontancous
(f) We have a system that can exist in two isomers, A and B, with reaction rate A -> B
equal to ki, and reaction rate of B -> A cqual to k2. To be able to determine he ratio of
equilibrium concentrations [A]/[B], we also need to specify the initial concentrations of
A and B.
(g) Perpetum mobile of the second kind satisfies first law of thermodynamics, but breaks
the second law.
(h) One of the equivalent formulations of the third law of thermodynamics states it is
impossible to reach T=OK.
(1) We are studying two different reactions A <-> B, and C <>D. The Gibbs free energy
difference between the starting state and end state of the process are the same, but in one
case (C -> D) the reaction includes much larger barrier during the conversion of C to D.
Hence, the equilibrium ratio D/C is going to be smaller than B/A.
6) I have a reaction of compounds in solution, following this reaction: A + B<->C, which
reaches its equilibrium concentrations A_eq, B_eq, and C_eq. Then, I will add some
more C to the equilibrated mixture, and let it equilibrate into a new values of
concentrations A*_eq, B*_eq, and C*_eq. This is enough information for us to be able
to say that A°_eq will be larger than A_eq.
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