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Is the reaction Fe₂O3(s) + 3H2(g) = 2Fe(s) + 3H₂O(g) spontaneous at 112°C and 1 atm pressure? (Give your calculation results also) Given: HᵒT, H2O(g) = -253971+ 38.5×T J/mol HOT, H2(g) =-8855 +29.7×T J/mol == HOT, Fe(s) = 10023 +33.6×T J/mol HᵒT, Fe2O3(s) = - 869984 +145.3×T J/mol SOT, H2O(g) = 176.9 +0.055×T J/mol.K SOT, H2(g) = 122.9+ 0.043×T J/mol.K SOT, Fe(s) 14.1 +0.052×T J/mol.K SOT, Fe203(s) = 40.68 + 0.208×T J/mol.K

Is the reaction Fe₂O3(s) + 3H2(g) = 2Fe(s) + 3H₂O(g) spontaneous at 112°C and 1 atm pressure? (Give your calculation results also) Given: HᵒT, H2O(g) = -253971+ 38.5×T J/mol HOT, H2(g) =-8855 +29.7×T J/mol == HOT, Fe(s) = 10023 +33.6×T J/mol HᵒT, Fe2O3(s) = - 869984 +145.3×T J/mol SOT, H2O(g) = 176.9 +0.055×T J/mol.K SOT, H2(g) = 122.9+ 0.043×T J/mol.K SOT, Fe(s) 14.1 +0.052×T J/mol.K SOT, Fe203(s) = 40.68 + 0.208×T J/mol.K

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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This reaction is spontaneous at given condition. true or false

Is the reaction Fe₂O3(s) + 3H2(g) = 2Fe(s) + 3H₂O(g) spontaneous at 112°C and 1 atm pressure? (Give your calculation results also) Given: HᵒT, H2O(g) = -253971+ 38.5×T J/mol HOT, H2(g) - 8855 +29.7×T J/mol HOT, Fe(s) HOT, Fe203(s) = - 869984+ 145.3×T J/mol SOT, H2O(g) = 176.9 +0.055×T J/mol.K SOT, H2(g) = 122.9 +0.043×T J/mol.K SºT, Fe(s) = = 14.1 +0.052×T J/mol.K SOT, Fe203(s) = 40.68 +0.208×T J/mol.K - 10023+33.6×T J/mol ==
Transcribed Image Text:Is the reaction Fe₂O3(s) + 3H2(g) = 2Fe(s) + 3H₂O(g) spontaneous at 112°C and 1 atm pressure? (Give your calculation results also) Given: HᵒT, H2O(g) = -253971+ 38.5×T J/mol HOT, H2(g) - 8855 +29.7×T J/mol HOT, Fe(s) HOT, Fe203(s) = - 869984+ 145.3×T J/mol SOT, H2O(g) = 176.9 +0.055×T J/mol.K SOT, H2(g) = 122.9 +0.043×T J/mol.K SºT, Fe(s) = = 14.1 +0.052×T J/mol.K SOT, Fe203(s) = 40.68 +0.208×T J/mol.K - 10023+33.6×T J/mol ==
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