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1 mol of gas mixture (contains (5Y) vol.% CO and 100- 5Y vol.% N,) is burnt completely with stoichiometric amount of dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,. CO + ½ 0, = CO, AH® CO (g) =-110000 J.mol C,, CO = 34.00 AH CO.(g) = -393500 J.mol" C,, CO,- 45.14 + 9.04x10'T AH° 0, (g) =0 J.mol J.K'.mol" J.K'.mol C, o, = 29.96 + 4.18x10'T- 1.67x10'T J.K'.mol AH° N, (g) =0 J.mol C, N, -27.87 + 4.27x10'T J.K'.mol Calculate the maximum flame temperature (K).

1 mol of gas mixture (contains (5Y) vol.% CO and 100- 5Y vol.% N,) is burnt completely with stoichiometric amount of dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,. CO + ½ 0, = CO, AH® CO (g) =-110000 J.mol C,, CO = 34.00 AH CO.(g) = -393500 J.mol" C,, CO,- 45.14 + 9.04x10'T AH° 0, (g) =0 J.mol J.K'.mol" J.K'.mol C, o, = 29.96 + 4.18x10'T- 1.67x10'T J.K'.mol AH° N, (g) =0 J.mol C, N, -27.87 + 4.27x10'T J.K'.mol Calculate the maximum flame temperature (K).

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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1 mol of gas mixture (contains (5Y) vol.% CO and 100- 5Y vol.% N,) is burnt completely with stoichiometric amount of dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,. CO + ½ 0, = CO, AH° CO (g) = -110000 J.mol' C, CO = 34.00 AH° CO.(g) = -393500 J.mol" C, Co.= 45.14 + 9.04x10'T AH° 0, (g) = 0 J.mol" J.K".mol" 2 J.K'.mol" C,, O, = 29.96 + 4.18x10'T - 1.67x10'T J.K".mol AH° N, (g) =0 J.mol' C,, N, - 27.87 + 4.27x10'T J.K".mol' Calculate the maximum flame temperature (K).
Transcribed Image Text:1 mol of gas mixture (contains (5Y) vol.% CO and 100- 5Y vol.% N,) is burnt completely with stoichiometric amount of dry air at 298 K. Dry air contains 21 vol.% O, and 79 vol.% N,. CO + ½ 0, = CO, AH° CO (g) = -110000 J.mol' C, CO = 34.00 AH° CO.(g) = -393500 J.mol" C, Co.= 45.14 + 9.04x10'T AH° 0, (g) = 0 J.mol" J.K".mol" 2 J.K'.mol" C,, O, = 29.96 + 4.18x10'T - 1.67x10'T J.K".mol AH° N, (g) =0 J.mol' C,, N, - 27.87 + 4.27x10'T J.K".mol' Calculate the maximum flame temperature (K).
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