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QUESTION 1 Calculate the standard "enthalpy" heat of reaction for the following reaction at 25°C (298 K) and 1 atm pressure. Al and Fe are their pure states. 3 + 2 = + 3 Given: AH°298 = (-264.84 kJ/mol). AH°298 < A1203>= (-1673.6 kJ/mol). Atomic weight of Fe = 56 I ◄ per mole of Al2O3 formed per mole of Fe formed per mole of FeO reacted per mole of Al reacted per gram of Fe formed A. -293 kJ B. -879 kJ/mol C. -293 kJ D. 1408.8 kJ E. -5.2 kJ F. 293 kJ G. 879 kJ/mol H. 469.6 kJ 1. -439.54 kJ

QUESTION 1 Calculate the standard "enthalpy" heat of reaction for the following reaction at 25°C (298 K) and 1 atm pressure. Al and Fe are their pure states. 3 + 2 = + 3 Given: AH°298 = (-264.84 kJ/mol). AH°298 < A1203>= (-1673.6 kJ/mol). Atomic weight of Fe = 56 I ◄ per mole of Al2O3 formed per mole of Fe formed per mole of FeO reacted per mole of Al reacted per gram of Fe formed A. -293 kJ B. -879 kJ/mol C. -293 kJ D. 1408.8 kJ E. -5.2 kJ F. 293 kJ G. 879 kJ/mol H. 469.6 kJ 1. -439.54 kJ

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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QUESTION 1 Calculate the standard "enthalpy" heat of reaction for the following reaction at 25°C (298 K) and 1 atm pressure. Al and Fe are their pure states. 3<FeO> + 2<AI> = <Al2O3> + 3<Fe> Given: AH°298 <FeO> = (-264.84 kJ/mol). AH°298 <A1203> = (-1673.6 kJ/mol). Atomic weight of Fe = 56 per mole of Al2O3 formed per mole of Fe formed per mole of FeO reacted per mole of Al reacted per gram of Fe formed A. -293 kJ B. -879 kJ/mol C. -293 kJ D. 1408.8 kJ E. -5.2 kJ F. 293 kJ G. 879 kJ/mol H. 469.6 kJ I. -439.54 kJ Save Answer
Transcribed Image Text:QUESTION 1 Calculate the standard "enthalpy" heat of reaction for the following reaction at 25°C (298 K) and 1 atm pressure. Al and Fe are their pure states. 3<FeO> + 2<AI> = <Al2O3> + 3<Fe> Given: AH°298 <FeO> = (-264.84 kJ/mol). AH°298 <A1203> = (-1673.6 kJ/mol). Atomic weight of Fe = 56 per mole of Al2O3 formed per mole of Fe formed per mole of FeO reacted per mole of Al reacted per gram of Fe formed A. -293 kJ B. -879 kJ/mol C. -293 kJ D. 1408.8 kJ E. -5.2 kJ F. 293 kJ G. 879 kJ/mol H. 469.6 kJ I. -439.54 kJ Save Answer
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