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3. Calculate the standard free energy change of the reaction: 1 < Ni > +;(02) =< Ni0 > At 327 C (600K) from the following data AH298, S298, = 7.12 cal/deg/mol S298,(0,) = 49.02 cal/deg/mol S298, Cp Cp.(0,) = 7.26 + 7.0 X 10-37 – 0.4 X105T-2 cal/deg/mol Cp. = -57500 cal/mol %3D = 9.10 cal/deg/mol = 6.03 + 10.44 X 10-672 – 2.5 X10-3T cal/deg/mol %3D = 12.91 cal/deg/mol %3D

3. Calculate the standard free energy change of the reaction: 1 < Ni > +;(02) =< Ni0 > At 327 C (600K) from the following data AH298, S298, = 7.12 cal/deg/mol S298,(0,) = 49.02 cal/deg/mol S298, Cp Cp.(0,) = 7.26 + 7.0 X 10-37 – 0.4 X105T-2 cal/deg/mol Cp. = -57500 cal/mol %3D = 9.10 cal/deg/mol = 6.03 + 10.44 X 10-672 – 2.5 X10-3T cal/deg/mol %3D = 12.91 cal/deg/mol %3D

Introduction to Chemical Engineering Thermodynamics
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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3. Calculate the standard free energy change of the reaction: < Ni > +÷(02) =< Ni0 > At 327 C (600K) from the following data AH298,<Nio> = -57500 cal/mol S298,<Ni> = 7.12 cal/deg/mol S298,(0,) S298,<Nio> = 9.10 cal/deg/mol Cp <Ni> = 7.26 + 7.0 X 10-3T – 0.4 X105T-2 cal/deg/mol = 49.02 cal/deg/mol = 6.03 + 10.44 X 10-6T2 – 2.5 X10-3T cal/deg/mol CP.(0,) Cp <Nio> %3D = 12.91 cal/deg/mol
Transcribed Image Text:3. Calculate the standard free energy change of the reaction: < Ni > +÷(02) =< Ni0 > At 327 C (600K) from the following data AH298,<Nio> = -57500 cal/mol S298,<Ni> = 7.12 cal/deg/mol S298,(0,) S298,<Nio> = 9.10 cal/deg/mol Cp <Ni> = 7.26 + 7.0 X 10-3T – 0.4 X105T-2 cal/deg/mol = 49.02 cal/deg/mol = 6.03 + 10.44 X 10-6T2 – 2.5 X10-3T cal/deg/mol CP.(0,) Cp <Nio> %3D = 12.91 cal/deg/mol
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