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For the product solution, H₁ = 2.178 kJ/mol r= (44,400 mol H₂O)/(5480 mol HCI) = 8.10 HCl(g. 25°C) +8.10 H₂O(1, 25°C), HCl(aq, 25°C)- Table 8.11 AH, = AĤ,(25°C, r = 8.1) HCl(aq, 40°C) -67.4 kJ/mol HCI The heat capacities of aqueous hydrochloric acid solutions are listed on p. 2-183 of Perry's Chemical Engineers' Handbook (see Footnote 5) as a function of the mole fraction of HCI in the solution, which in our problem is

For the product solution, H₁ = 2.178 kJ/mol r= (44,400 mol H₂O)/(5480 mol HCI) = 8.10 HCl(g. 25°C) +8.10 H₂O(1, 25°C), HCl(aq, 25°C)- Table 8.11 AH, = AĤ,(25°C, r = 8.1) HCl(aq, 40°C) -67.4 kJ/mol HCI The heat capacities of aqueous hydrochloric acid solutions are listed on p. 2-183 of Perry's Chemical Engineers' Handbook (see Footnote 5) as a function of the mole fraction of HCI in the solution, which in our problem is

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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How was -67.4 kJ/mol HCl obtained?  I do not see it on the table.  

||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 56°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw T Read aloud Q Search r(mol H₂O/mol solute) 0.5 1 1.5 2 3 4 5 TABLE B.11 Integral Heats of Solution and Mixing at 25°C² (AĤs) NaOH(s) kJ/mol NaOH 10 20 25 30 40 50 100 200 500 1000 2000 5000 10 000 50 000 100 000 500 000 673 of 695 ■ (AĤs) HCK(g) kJ/mol HCI -26.22 -48.82 -56.85 -61.20 -64.05 -69.49 -71.78 -72.59 -73.00 -73.26 -73.85 -74.20 -74.52 -74.68 -74.82 -74.93 -74.99 -75.08 -75.10 CD -75.14 28.87 -34.43 -37.74 -42.51 -42.84 42.72 -42.59 -42.51 42.34 -42.26 -42.38 -42.47 -42.55 -42.68 -42.72 -42.80 -42.89 (AĤm) H₂SO4 kJ/mol H₂SO4 -15.73 -28.07 - 36.90 -41.92 - 48.99 - 54.06 -58.03 -67.03 -72.30 -73.34 -73.97 -76.73 -78.57 -84.43 -87.07 -93.64 -95.31 -96.19 "Adapted from J. C. Whitwell and R. K. Toner, Conservation of Mass and Energy, pp. 344-346. Copyright © 1969 by McGraw-Hill, Inc. Used with permission of McGraw-Hill. {" J 63 60 D ENG Sign in (0) + O 7:07 AM 5/22/2023
Transcribed Image Text:||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 56°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw T Read aloud Q Search r(mol H₂O/mol solute) 0.5 1 1.5 2 3 4 5 TABLE B.11 Integral Heats of Solution and Mixing at 25°C² (AĤs) NaOH(s) kJ/mol NaOH 10 20 25 30 40 50 100 200 500 1000 2000 5000 10 000 50 000 100 000 500 000 673 of 695 ■ (AĤs) HCK(g) kJ/mol HCI -26.22 -48.82 -56.85 -61.20 -64.05 -69.49 -71.78 -72.59 -73.00 -73.26 -73.85 -74.20 -74.52 -74.68 -74.82 -74.93 -74.99 -75.08 -75.10 CD -75.14 28.87 -34.43 -37.74 -42.51 -42.84 42.72 -42.59 -42.51 42.34 -42.26 -42.38 -42.47 -42.55 -42.68 -42.72 -42.80 -42.89 (AĤm) H₂SO4 kJ/mol H₂SO4 -15.73 -28.07 - 36.90 -41.92 - 48.99 - 54.06 -58.03 -67.03 -72.30 -73.34 -73.97 -76.73 -78.57 -84.43 -87.07 -93.64 -95.31 -96.19 "Adapted from J. C. Whitwell and R. K. Toner, Conservation of Mass and Energy, pp. 344-346. Copyright © 1969 by McGraw-Hill, Inc. Used with permission of McGraw-Hill. {" J 63 60 D ENG Sign in (0) + O 7:07 AM 5/22/2023
||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 56°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw (T) Read aloud + References: HC1, H₂0 at 25°C and 1 atm Substance A in H₁kJ/mol HCI 0 HCl(g) H₂O(1) HCl(aq) Calculate H₁ and ₂: For the product solution, Q Search nin 5480 mol/HCL 44,400 mol H₂O L Energy Balance: 466 of 695 0.73 kcal kg. °C HCI(g, 2C) → HCl(g, 100°C) 100°C Ĥ₁ = AĤ = C₂1 H₁ = 2.178 kJ/mol 5480 mol HCl/h (5480+ 44,400) mol/h AHb o r = (44,400 mol H₂O)/(5480 mol HCI) = 8.10 AĤa AHb HCl(g, 25°C) + 8.10 H₂O(1, 25°C). HCl(aq, 25°C)→→→ HCl(aq, 40°C) ΔΗ = ΔΗ,(25°C, r = 8.1) -67.4 kJ/mol HCI Table B.11. The heat capacities of aqueous hydrochloric acid solutions are listed on p. 2-183 of Perry's Chemical Engineers' Handbook (see Footnote 5) as a function of the mole fraction of HCI in the solution, which in our problem is 1000 kg solution 5480 mol HCI 40°C Q = AH = = Σñ‚á - Σñ‚¡ Codt J25°C for HCl(g) from Table B.2 J25°C OD nout 5480 mol HCI H₂kJ/mol HCI) ty H d 4.184 kJ kcal = 0.110 mol HCl/mol Ĥ out 0.557 Cp dT = 8.36(kJ/mol HCI clau kJ mol HCI-C Ĥ₂ = AĤ₁ + AĤb = (-67.4+8.36) kJ/mol HCl = -59.0 kJ/mol HCl {" Ơ Ⓡ J 63 50 D ENG Sign in (0) + O 7:27 AM 5/22/2023
Transcribed Image Text:||! PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X + 56°F Sunny File | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdf Draw (T) Read aloud + References: HC1, H₂0 at 25°C and 1 atm Substance A in H₁kJ/mol HCI 0 HCl(g) H₂O(1) HCl(aq) Calculate H₁ and ₂: For the product solution, Q Search nin 5480 mol/HCL 44,400 mol H₂O L Energy Balance: 466 of 695 0.73 kcal kg. °C HCI(g, 2C) → HCl(g, 100°C) 100°C Ĥ₁ = AĤ = C₂1 H₁ = 2.178 kJ/mol 5480 mol HCl/h (5480+ 44,400) mol/h AHb o r = (44,400 mol H₂O)/(5480 mol HCI) = 8.10 AĤa AHb HCl(g, 25°C) + 8.10 H₂O(1, 25°C). HCl(aq, 25°C)→→→ HCl(aq, 40°C) ΔΗ = ΔΗ,(25°C, r = 8.1) -67.4 kJ/mol HCI Table B.11. The heat capacities of aqueous hydrochloric acid solutions are listed on p. 2-183 of Perry's Chemical Engineers' Handbook (see Footnote 5) as a function of the mole fraction of HCI in the solution, which in our problem is 1000 kg solution 5480 mol HCI 40°C Q = AH = = Σñ‚á - Σñ‚¡ Codt J25°C for HCl(g) from Table B.2 J25°C OD nout 5480 mol HCI H₂kJ/mol HCI) ty H d 4.184 kJ kcal = 0.110 mol HCl/mol Ĥ out 0.557 Cp dT = 8.36(kJ/mol HCI clau kJ mol HCI-C Ĥ₂ = AĤ₁ + AĤb = (-67.4+8.36) kJ/mol HCl = -59.0 kJ/mol HCl {" Ơ Ⓡ J 63 50 D ENG Sign in (0) + O 7:27 AM 5/22/2023
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