Calculate the equilibrium constant of oxidation reaction (3) at 675, 875 and 1075 K and from Van’t Hoff equation calculate the heat of reaction by using the graphical method.

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 constant of oxidation reaction (3) at 675, 875 and 1075 K and from
Van’t Hoff equation calculate the heat of reaction by using the graphical method.

AG
AG vs T
4G
675
-355100
Reaction 1
875
-320700
200
400
G00
800
1000
1200
50000
1075
-286300
-100000
-150000
675
-394667
Reaction 2
875
-394835
-200000
1075
-395003
-250000
-300000
675
-170863.75
350000
Reaction 3
875
-188393.75
400000
1075
-205923.75
450000
Transcribed Image Text:AG AG vs T 4G 675 -355100 Reaction 1 875 -320700 200 400 G00 800 1000 1200 50000 1075 -286300 -100000 -150000 675 -394667 Reaction 2 875 -394835 -200000 1075 -395003 -250000 -300000 675 -170863.75 350000 Reaction 3 875 -188393.75 400000 1075 -205923.75 450000
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)
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