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Calculate the standard free energy change of the reaction: {Pb} + ½ (02) = At 527 °C (800 K) from the following data: AH298, S298, = 16.20 cal/deg/mol S298, = 15.50 cal/deg/mol S298,(0,) = 49.02 cal/deg/mol Cp = 10.60 + 4.0 X 10-3T cal/deg/mol Cp = 5.63 + 2.33 X 10-³T cal/deg/mol Cp.{Pb} = 7.75 – 0.74 X 10-3T cal/deg/mol Cp.(0,) = 7.26 + 7.0 X 10-3T – 0.4 X105T-2 cal/deg/mol -52400 cal/mol

Calculate the standard free energy change of the reaction: {Pb} + ½ (02) = At 527 °C (800 K) from the following data: AH298, S298, = 16.20 cal/deg/mol S298, = 15.50 cal/deg/mol S298,(0,) = 49.02 cal/deg/mol Cp = 10.60 + 4.0 X 10-3T cal/deg/mol Cp = 5.63 + 2.33 X 10-³T cal/deg/mol Cp.{Pb} = 7.75 – 0.74 X 10-3T cal/deg/mol Cp.(0,) = 7.26 + 7.0 X 10-3T – 0.4 X105T-2 cal/deg/mol -52400 cal/mol

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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Calculate the standard free energy change of the reaction: {Pb} + ½ (02) = <PbO> At 527 °C (800 K) from the following data: AH298,<Pb0> S298 <Pbo> = 16.20 cal/deg/mol S298 <Pb> = 15.50 cal/deg/mol S298,(0,) = 49.02 cal/deg/mol Cp<Pbo> = 10.60 + 4.0 X 10-3T cal/deg/mol Cp<Pb> = 5.63 + 2.33 X 10-3T cal/deg/mol Cp{Pb} = 7.75 – 0.74 X 10-3T cal/deg/mol Cp.(o,) = 7.26 + 7.0 X 10-3T – 0.4 X105T-2 cal/deg/mol = -52400 cal/mol
Transcribed Image Text:Calculate the standard free energy change of the reaction: {Pb} + ½ (02) = <PbO> At 527 °C (800 K) from the following data: AH298,<Pb0> S298 <Pbo> = 16.20 cal/deg/mol S298 <Pb> = 15.50 cal/deg/mol S298,(0,) = 49.02 cal/deg/mol Cp<Pbo> = 10.60 + 4.0 X 10-3T cal/deg/mol Cp<Pb> = 5.63 + 2.33 X 10-3T cal/deg/mol Cp{Pb} = 7.75 – 0.74 X 10-3T cal/deg/mol Cp.(o,) = 7.26 + 7.0 X 10-3T – 0.4 X105T-2 cal/deg/mol = -52400 cal/mol
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