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
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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F492e85a0-6387-49c3-95ee-c3c5004c485c%2F9c28dd2d-e517-40e5-b70d-822da78fab43%2Fgtghsy_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F492e85a0-6387-49c3-95ee-c3c5004c485c%2F9c28dd2d-e517-40e5-b70d-822da78fab43%2Fsehoyt9_processed.png&w=3840&q=75)
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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