Calculate the enthalpy of formation of PCL5 (s), given the heats of the following reactions at 25C: 2P (s) + 3 C12 = 2PCI3 (1) H = - 635.13 KJ/mol PCI3 (1) + 2 C12 = PCI5 (s) H = - 137.28 KJ/mol Select the correct response: -772.41 KJ/mol -497.85 KJ/mol O -386 2K /mol
Calculate the enthalpy of formation of PCL5 (s), given the heats of the following reactions at 25C: 2P (s) + 3 C12 = 2PCI3 (1) H = - 635.13 KJ/mol PCI3 (1) + 2 C12 = PCI5 (s) H = - 137.28 KJ/mol Select the correct response: -772.41 KJ/mol -497.85 KJ/mol O -386 2K /mol
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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Transcribed Image Text:Table 6.3 Standard Enthalpies of Formation of Some
Inorganic Substances at 25°C
Substance
AH; (kJ/mol)
Substance
AH? (kJ/mol)
- 187.6
Ag(s)
AgCl(s)
H,O2(/)
- 127.04
Hg(/)
Al(s)
2(s)
Al,03(s)
- 1669.8
HI(g)
25.94
Br2()
Mg(s)
HBr(g)
C(graphite)
-36.2
MgO(s)
-601.8
M9CO (s)
N2(g)
NH3(g)
NO(g)
NO2(g)
N204(g)
-1112.9
C(diamond)
1.90
CO(g)
Co,(G)
Ca(s)
-110.5
-46.3
-393.5
90.4
33.85
CaO(s)
-635.6
9.66
CaCO3(s)
-1206.9
N20(g)
81.56
O(g)
249.4
HCIG)
O2(g)
O3(g)
-92.3
Cu(s)
142.2
CuO(s)
-155.2
S(rhombic)
S(monoclinic)
SO-(g)
SO;(g)
F2(g)
0.30
HF(g)
H(g)
-268.61
-296.1
218.2
-395.2
H2(g)
H,S(g)
-20.15
H2O(g)
-241.8
ZnO(s)
-347.98
H20(1)
-285.8
Transcribed Image Text:Question 6 of 10
Calculate the enthalpy of formation of PCL5 (s), given the heats of the following reactions at 25C:
2P (s) + 3 C12 = 2PC13 (I) H = - 635.13 KJ/mol
PCI3 (1) + 2 C12 = PCI5 (s) H = - 137.28 KJ/mol
Select the correct response:
-772.41 KJ/mol
-497.85 KJ/mol
-386.2 KJ/mol
-248.92 KJ/mol
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