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
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:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4dfc165-1571-419f-99fc-a4587d76c340%2Fafabc16b-84a2-46d2-8dd2-9300496b318d%2Fnc97v4_processed.jpeg&w=3840&q=75)
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4dfc165-1571-419f-99fc-a4587d76c340%2Fafabc16b-84a2-46d2-8dd2-9300496b318d%2Fg4fbmrk_processed.jpeg&w=3840&q=75)
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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