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Regarding the following reaction at 298.15 K: 4 HCI (g) + O2(g) → 2Cl2 (g) + 2H₂O (1) The following table lists the stand enthalpy of formation, the standard entropy and the standard formation Gibbs energies

Regarding the following reaction at 298.15 K: 4 HCI (g) + O2(g) → 2Cl2 (g) + 2H₂O (1) The following table lists the stand enthalpy of formation, the standard entropy and the standard formation Gibbs energies

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 2 Regarding the following reaction at 298.15 K: 4 HCI (g) + O2 (g) →→ 2Cl₂ (g) + 2H₂O (1) The following table lists the stand enthalpy of formation, the standard entropy and the standard formation Gibbs energies Cl2 (g) O2 (g) 0 H₂O (1) -285.83 0 HCI (g) -92.31 186.91 -95.3 69.91 4+Ho(kJmol-1) sº (JK-1mol-1) 4Gº (kJmol-1) 205.14 0 223.07 0 ? Please calculate the standard reaction entropy 4,S°(JK-1mol-1), the standard reaction enthalpy Hº(kJmol-1), the standard reaction Gibbs energy 4,Gº(kJmol-1), and the maximum nonexpansion work that can be gained from the oxidation of HCI (g) at 298.15K. Instruction: Please enter all answers with two decimal places, for example: -344.101 is written as -344.10, please pay attention to the units, particularly As in J/(K.mol) while others is kJ/mol (1) 4,5°(JK-1 mol-1) = (2) 4,H°(kJmol-1) = Please using your results from (1) and (2) to calculate the the standard reaction Gibbs energy:
Transcribed Image Text:QUESTION 2 Regarding the following reaction at 298.15 K: 4 HCI (g) + O2 (g) →→ 2Cl₂ (g) + 2H₂O (1) The following table lists the stand enthalpy of formation, the standard entropy and the standard formation Gibbs energies Cl2 (g) O2 (g) 0 H₂O (1) -285.83 0 HCI (g) -92.31 186.91 -95.3 69.91 4+Ho(kJmol-1) sº (JK-1mol-1) 4Gº (kJmol-1) 205.14 0 223.07 0 ? Please calculate the standard reaction entropy 4,S°(JK-1mol-1), the standard reaction enthalpy Hº(kJmol-1), the standard reaction Gibbs energy 4,Gº(kJmol-1), and the maximum nonexpansion work that can be gained from the oxidation of HCI (g) at 298.15K. Instruction: Please enter all answers with two decimal places, for example: -344.101 is written as -344.10, please pay attention to the units, particularly As in J/(K.mol) while others is kJ/mol (1) 4,5°(JK-1 mol-1) = (2) 4,H°(kJmol-1) = Please using your results from (1) and (2) to calculate the the standard reaction Gibbs energy:
(1) 4,5°(JK-¹1mol-1) = (2) 4,H°(kJmol-1) = Please using your results from (1) and (2) to calculate the the standard reaction Gibbs energy: (3) 4,Go(kJmol-1) = Please use result from (3) and data below to calculate stand formation Gibbs energy of H₂0 (1): (4) AGO (kJmol-1) = (5) w'max (kJg-1)that can be gained from the oxidation of HCI (g) on a per gram basis = equation contains 4 moles of HCI) (Note the reaction.
Transcribed Image Text:(1) 4,5°(JK-¹1mol-1) = (2) 4,H°(kJmol-1) = Please using your results from (1) and (2) to calculate the the standard reaction Gibbs energy: (3) 4,Go(kJmol-1) = Please use result from (3) and data below to calculate stand formation Gibbs energy of H₂0 (1): (4) AGO (kJmol-1) = (5) w'max (kJg-1)that can be gained from the oxidation of HCI (g) on a per gram basis = equation contains 4 moles of HCI) (Note the reaction.
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