Using values from Appendix C, calculate the value of ΔH° foreach of the following reactions:(a) CaO(s) + 2 HCl(g) ---->CaCl2(s) + H2O(g)(b) 4 FeO(s) + O2(g) ----> 2 Fe2O3(s)(c) 2 CuO(s) + NO(g) -->Cu2O(s) + NO2(g)(d) 4 NH3(g) + O2(g) ----> 2 N2H4(g) + 2 H2O(l) APPENDIX C Thermodynamic Quantitles for Selected Substances at 298.15 K (25 "C)1091дн;(kj/mol)AG;S*S°днAG;(kj/mol)-245.6(kj/mol)(i/mol-K)SubstanceSubstance(i/mol-K)(kj/mol)VanadiumSOCI (1)H,S(8)H2SO4(aq)H,SO,(1)205.620.1182.2-20.17-33.01514.2453.1V(g)V(s)-909.3-744.528.9156.1-689.9-814.0ZincTitanium130.795.2160.9Zn(g)Ti(g)Ti(s)468180.342241.63Zn(s)ZnCl,(s)ZnO(s)-369.4-415.1111.530.76-763.2-726.8354.9TICL4(8)TICI4(1)TIO2(s)-318.2-348.043.9-804.2-728.1221.9-889.450.29-944.7

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Chemistry for Engineering Students

3rd Edition

ISBN:9781285199023

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter10: Entropy And The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 10.39PAE: How does the second law of thermodynamics explain a spontaneous change in a system that becomes more...

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Using values from Appendix C, calculate the value of ΔH° for
each of the following reactions:
(a) CaO(s) + 2 HCl(g) ---->CaCl₂(s) + H₂O(g)
(b) 4 FeO(s) + O₂(g) ----> 2 Fe₂O₃(s)
(c) 2 CuO(s) + NO(g) -->Cu₂O(s) + NO₂(g)
(d) 4 NH₃(g) + O₂(g) ----> 2 N₂H₄(g) + 2 H₂O(l)

APPENDIX C Thermodynamic Quantitles for Selected Substances at 298.15 K (25 "C)
1091
дн;
(kj/mol)
AG;
S*
S°
дн
AG;
(kj/mol)
-245.6
(kj/mol)
(i/mol-K)
Substance
Substance
(i/mol-K)
(kj/mol)
Vanadium
SOCI (1)
H,S(8)
H2SO4(aq)
H,SO,(1)
205.6
20.1
182.2
-20.17
-33.01
514.2
453.1
V(g)
V(s)
-909.3
-744.5
28.9
156.1
-689.9
-814.0
Zinc
Titanium
130.7
95.2
160.9
Zn(g)
Ti(g)
Ti(s)
468
180.3
422
41.63
Zn(s)
ZnCl,(s)
ZnO(s)
-369.4
-415.1
111.5
30.76
-763.2
-726.8
354.9
TICL4(8)
TICI4(1)
TIO2(s)
-318.2
-348.0
43.9
-804.2
-728.1
221.9
-889.4
50.29
-944.7

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