Homework Help is Here – Start Your Trial Now!
Engineering
Chemical Engineering
Question 5. The basic chemical reaction involved in a typical wet process for SO₂ removal (emission of SO₂ to atmosphere is a main cause for formation of acid rain) can be explained in the following: AĤ,= -47.3 kJ/mol, SO₂ (g) + H₂O (1) + (NH4)2SO3(aq) 2 NH4HSO3(aq) = where AĤ, is the standard heat of this reaction. A gaseous mixture of N₂ and SO₂ (0.60 mol %) fed to a reactor at a rate of 25000 mol/h at 50°C is treated with the above process. An aqueous stream which contains 15.0 mol % (NH4)2SO3 is sent into the reactor at 25°C. Through this process, 93% of the SO₂ can be converted to NH4HSO3, and the molar ratio of NH4HSO3 to (NH4)2SO3 is 0.65 in Stream c. (i) Calculate flow rates for Streams a, b, and c in mol/h. (ii) What will be the required rate of heat removal (or addition) in kJ/h or kW for this process? You can assume that all aqueous streams have a specific heat capacity of 4.12 J/g.°C, and evaporation of water may be neglected in your calculation. N₂, SO₂, 50°C (NH4)2SO3 (aq), 25°C Gas SO₂ SO₂ N₂ To save your time in calculation, some specific enthalpy data of gaseous compounds over the related temperature ranges have been pre-calculated and tabulated in the following table: N₂ 3 a Reactor b C Specific enthalpy change (T in °C) 50 ▲Â = §C„dT =1.01kJ/mol 25 35 AĤ= fcdT=0.40 kJ/mol 25 50 AĤ=fc₂dT= jc,dr 25 35 AĤ = §CÂdT = 0.292 kJ/mol 25 N₂, SO2, 35°C NH4HSO3(aq) & (NH4)2SO3 (aq), 35°C = 0.73 kJ/mol

Question 5. The basic chemical reaction involved in a typical wet process for SO₂ removal (emission of SO₂ to atmosphere is a main cause for formation of acid rain) can be explained in the following: AĤ,= -47.3 kJ/mol, SO₂ (g) + H₂O (1) + (NH4)2SO3(aq) 2 NH4HSO3(aq) = where AĤ, is the standard heat of this reaction. A gaseous mixture of N₂ and SO₂ (0.60 mol %) fed to a reactor at a rate of 25000 mol/h at 50°C is treated with the above process. An aqueous stream which contains 15.0 mol % (NH4)2SO3 is sent into the reactor at 25°C. Through this process, 93% of the SO₂ can be converted to NH4HSO3, and the molar ratio of NH4HSO3 to (NH4)2SO3 is 0.65 in Stream c. (i) Calculate flow rates for Streams a, b, and c in mol/h. (ii) What will be the required rate of heat removal (or addition) in kJ/h or kW for this process? You can assume that all aqueous streams have a specific heat capacity of 4.12 J/g.°C, and evaporation of water may be neglected in your calculation. N₂, SO₂, 50°C (NH4)2SO3 (aq), 25°C Gas SO₂ SO₂ N₂ To save your time in calculation, some specific enthalpy data of gaseous compounds over the related temperature ranges have been pre-calculated and tabulated in the following table: N₂ 3 a Reactor b C Specific enthalpy change (T in °C) 50 ▲Â = §C„dT =1.01kJ/mol 25 35 AĤ= fcdT=0.40 kJ/mol 25 50 AĤ=fc₂dT= jc,dr 25 35 AĤ = §CÂdT = 0.292 kJ/mol 25 N₂, SO2, 35°C NH4HSO3(aq) & (NH4)2SO3 (aq), 35°C = 0.73 kJ/mol

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
See similar textbooks
icon
Related questions
Question
Question 5. The basic chemical reaction involved in a typical wet process for SO₂ removal (emission of SO₂ to atmosphere is a main cause for formation of acid rain) can be explained in the following: AĤ=-47.3 kJ/mol, SO₂ (g) + H₂O (1) + (NH4)2SO3(aq) 2 NH4HSO3(aq) where AF, is the standard heat of this reaction. A gaseous mixture of N₂ and SO₂ (0.60 mol %) fed to a reactor at a rate of 25000 mol/h at 50°C is treated with the above process. An aqueous stream which contains 15.0 mol% (NH4)2SO3 is sent into the reactor at 25°C. Through this process, 93% of the SO₂ can be converted to NH4HSO3, and the molar ratio of NH4HSO3 to (NH4)2SO3 is 0.65 in Stream c. (i) Calculate flow rates for Streams a, b, and e in mol/h. (ii) What will be the required rate of heat removal (or addition) in kJ/h or kW for this process? You can assume that all aqueous streams have a specific heat capacity of 4.12 J/g °C, and evaporation of water may be neglected in your calculation. N2, SO2, 50°C (NH4)2SO3 (aq), 25°C Gas SO₂ SO₂ N₂ = a N₂ To save your time in calculation, some specific enthalpy data of gaseous compounds over the related temperature ranges have been pre-calculated and tabulated in the following table: Reactor b C Specific enthalpy change (T in °C) 50 =jc,dT= AĤ = [C,dT 25 35 AĤ = C,dT = 0.40 kJ/mol 25 50 AĤ = [C₁dT=0.73 kJ/mol 25 = 1.01 kJ/mol N₂, SO2, 35°C NH4HSO3(aq) & (NH4)2SO3 (aq), 35°C 35 ▲Â = §C₁dT = 0.292 kJ/mol 25
Transcribed Image Text:Question 5. The basic chemical reaction involved in a typical wet process for SO₂ removal (emission of SO₂ to atmosphere is a main cause for formation of acid rain) can be explained in the following: AĤ=-47.3 kJ/mol, SO₂ (g) + H₂O (1) + (NH4)2SO3(aq) 2 NH4HSO3(aq) where AF, is the standard heat of this reaction. A gaseous mixture of N₂ and SO₂ (0.60 mol %) fed to a reactor at a rate of 25000 mol/h at 50°C is treated with the above process. An aqueous stream which contains 15.0 mol% (NH4)2SO3 is sent into the reactor at 25°C. Through this process, 93% of the SO₂ can be converted to NH4HSO3, and the molar ratio of NH4HSO3 to (NH4)2SO3 is 0.65 in Stream c. (i) Calculate flow rates for Streams a, b, and e in mol/h. (ii) What will be the required rate of heat removal (or addition) in kJ/h or kW for this process? You can assume that all aqueous streams have a specific heat capacity of 4.12 J/g °C, and evaporation of water may be neglected in your calculation. N2, SO2, 50°C (NH4)2SO3 (aq), 25°C Gas SO₂ SO₂ N₂ = a N₂ To save your time in calculation, some specific enthalpy data of gaseous compounds over the related temperature ranges have been pre-calculated and tabulated in the following table: Reactor b C Specific enthalpy change (T in °C) 50 =jc,dT= AĤ = [C,dT 25 35 AĤ = C,dT = 0.40 kJ/mol 25 50 AĤ = [C₁dT=0.73 kJ/mol 25 = 1.01 kJ/mol N₂, SO2, 35°C NH4HSO3(aq) & (NH4)2SO3 (aq), 35°C 35 ▲Â = §C₁dT = 0.292 kJ/mol 25
Expert Solution
steps

Step by step

Solved in 4 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami…
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Process Dynamics and Control, 4e
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:
9781119285915
Author:
Seborg
Publisher:
WILEY
Industrial Plastics: Theory and Applications
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:
9781285061238
Author:
Lokensgard, Erik
Publisher:
Delmar Cengage Learning
Unit Operations of Chemical Engineering
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:
9780072848236
Author:
Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:
McGraw-Hill Companies, The
SEE MORE TEXTBOOKS