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Calculate the equilibrium vapour pressure of oxidation reaction (1) at 675, 875 and 1075 K.

Calculate the equilibrium vapour pressure of oxidation reaction (1) at 675, 875 and 1075 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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Calculate the equilibrium vapour pressure of oxidation reaction (1) at 675, 875 and 1075 K.

 

second image is b values

For below given chemical reactions: (1) 2Ni(s) + 02(9) = 2N10(s) AG; = -471200 + 172T (J/mol) (2) C(s) + 02(9) = CO2(9) AG² = -394100 – 0.84T (J/mol) %3D (3) C(s) + /½ 0zc0) = CO9) AG; = -111700 - 87.65T (J/mol)
Transcribed Image Text:For below given chemical reactions: (1) 2Ni(s) + 02(9) = 2N10(s) AG; = -471200 + 172T (J/mol) (2) C(s) + 02(9) = CO2(9) AG² = -394100 – 0.84T (J/mol) %3D (3) C(s) + /½ 0zc0) = CO9) AG; = -111700 - 87.65T (J/mol)
Again relate equation (2) with equation (3) RTINP0;(g) = -471200 + 172T Substitute this expression with values of temperature and R = 8. 314 J. mol-'. K-| At 675 K (8.314 J.тol-1. K-1x675к) In Poss) — — 471200 + 172 x (675) In Po,g) -355100 atm 5611.95 In Po,) = -63. 276 28 Po:(2) = 3. 31 x 10-2 At 875 K (8.314 J. mol-. K-l × 875K) In Po:(2) = -471200 + 172 x (875) In Po,e) -320700 atm 7274.75 In Po,) = -44. 08 Po:® = 7. 18 x 10-20 At 1075 K (8.314 J. mol-. K-l x 1075K) In Po,) = -471200 + 172 × (1075) In Po,«) In Po,e) = -32. 03 -286300 atm 8937.55 Po,e = 1.23 x 10-14 %3D
Transcribed Image Text:Again relate equation (2) with equation (3) RTINP0;(g) = -471200 + 172T Substitute this expression with values of temperature and R = 8. 314 J. mol-'. K-| At 675 K (8.314 J.тol-1. K-1x675к) In Poss) — — 471200 + 172 x (675) In Po,g) -355100 atm 5611.95 In Po,) = -63. 276 28 Po:(2) = 3. 31 x 10-2 At 875 K (8.314 J. mol-. K-l × 875K) In Po:(2) = -471200 + 172 x (875) In Po,e) -320700 atm 7274.75 In Po,) = -44. 08 Po:® = 7. 18 x 10-20 At 1075 K (8.314 J. mol-. K-l x 1075K) In Po,) = -471200 + 172 × (1075) In Po,«) In Po,e) = -32. 03 -286300 atm 8937.55 Po,e = 1.23 x 10-14 %3D
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