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The reaction of calcination is represented by the equation below: CaC03(s) → Cao, + CO2 K = Pco2 The equilibrium constant of the reaction is equal to the partial pressure of the carbon dioxide product The standard free energy for the reaction is related to the equilibrium constant as follows AG° = -RT In Pcoz The standard enthalpy and entropy changes of the reaction can be calculated using the following equations as derived from the standard free energy: a(AG°/T) a(1/T) ΔΗ' a(AG°) AS° = a(T) Given the results below, Calculate the standard free energy, entbalpy, and entropy Time (min) T (°C) Pco2(psi) 2 700 15.69595 4 720 15.69595 6 740 15.69595 8 760 15.69595 10 780 33.12302 12 800 43.41041 14 820 53.13908 16 840 62.31456 18 860 87.30181 20 880 120.1292 22 900 145.9958 24 920 199.7872 26 940 271.6728 28 960 355.9691 30 980 462.2001 32 1000 583.0125 34 1020 572.8791 36 1040 613.7714

The reaction of calcination is represented by the equation below: CaC03(s) → Cao, + CO2 K = Pco2 The equilibrium constant of the reaction is equal to the partial pressure of the carbon dioxide product The standard free energy for the reaction is related to the equilibrium constant as follows AG° = -RT In Pcoz The standard enthalpy and entropy changes of the reaction can be calculated using the following equations as derived from the standard free energy: a(AG°/T) a(1/T) ΔΗ' a(AG°) AS° = a(T) Given the results below, Calculate the standard free energy, entbalpy, and entropy Time (min) T (°C) Pco2(psi) 2 700 15.69595 4 720 15.69595 6 740 15.69595 8 760 15.69595 10 780 33.12302 12 800 43.41041 14 820 53.13908 16 840 62.31456 18 860 87.30181 20 880 120.1292 22 900 145.9958 24 920 199.7872 26 940 271.6728 28 960 355.9691 30 980 462.2001 32 1000 583.0125 34 1020 572.8791 36 1040 613.7714

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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The reaction of calcination is represented by the equation below: CaCO3() → Cao, + CO2 K = Pco2 The equilibrium constant of the reaction is equal to the partial pressure of the carbon dioxide product. The standard free energy for the reaction is related to the equilibrium constant as follows AG° = -RT In pco, The standard enthalpy and entropy changes of the reaction can be calculated using the following equations as derived from the standard free energy: a(AG°/T) ΔΗ a(1/T) a(AG°) AS° = a(T) Given the results below, • Calculate the standard free energy, entbalpy and entropy Time (min) T (°C) Pcoz(psi) 700 15.69595 4 720 15.69595 740 15.69595 760 15.69595 10 780 33.12302 12 800 43.41041 14 820 53.13908 16 840 62.31456 18 860 87.30181 20 880 120.1292 22 900 145.9958 24 920 199.7872 26 940 271.6728 28 960 355.9691 30 980 462.2001 32 1000 583.0125 34 1020 572.8791 36 1040 613.7714
Transcribed Image Text:The reaction of calcination is represented by the equation below: CaCO3() → Cao, + CO2 K = Pco2 The equilibrium constant of the reaction is equal to the partial pressure of the carbon dioxide product. The standard free energy for the reaction is related to the equilibrium constant as follows AG° = -RT In pco, The standard enthalpy and entropy changes of the reaction can be calculated using the following equations as derived from the standard free energy: a(AG°/T) ΔΗ a(1/T) a(AG°) AS° = a(T) Given the results below, • Calculate the standard free energy, entbalpy and entropy Time (min) T (°C) Pcoz(psi) 700 15.69595 4 720 15.69595 740 15.69595 760 15.69595 10 780 33.12302 12 800 43.41041 14 820 53.13908 16 840 62.31456 18 860 87.30181 20 880 120.1292 22 900 145.9958 24 920 199.7872 26 940 271.6728 28 960 355.9691 30 980 462.2001 32 1000 583.0125 34 1020 572.8791 36 1040 613.7714
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