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KAP 15.
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CITAZIONE (Schrödinger @ 11/12/2007, 15:43)Correggimi se sbaglio...
(Literature: Johnson D. Thermodynamice aspects of anorganic chemistry. M.: the
World, 1985.- 328p.)
ΔНreac= Σ√I ΔHiend -Σ√I ΔHibeg
The standard enthalpy of element formation with
р = 1,01325 ·105Pа (1kgf\см2) и 298 К is equal to null, as devices cannot be
transmuted each other by means of chemical reactions.
The Molar enthalpy for exothermic reactions is yielded with plus (+) for the
exothermic - with a minus (-).
1) СО2 → С + О2
ΔН (С(тв)) = 0, ΔН (О2) = 0, ΔН(СО2)= -393,51 кJ\mol
ΣΔН0
t298(reaction №1)= 0 –(-393,51) = 393,51 кJ\mol
2) СО2(g) + 8Н→ СН4(g) + 2Н2О(g)
ΔН (СО2(g))=- 393,51 kJ\mol, ΔН (Н) = 0,
ΣΔHbeg = -393,51 кJ\mol
ΔН (СН4(g))= - 74,81 кJ\mol, ΔН (2Н2О(g)) = - 482,64
ΣΔHend = -393,51 кJ\mol
ΣΔН0
t298 (reaction №2) = (-557,45) – (-393,51) = -163,94 кJ\mol
3) СН4(g) + О2(g) → С(тв) + 2Н2О(g)
ΣΔHbeg = - 74,81 кJ\mol
ΣΔHend = - 482,64 кJ\mol
ΣΔН0
t298 (reaction №3)= (-482,64) – (-74,81) = -407,83 кJ\mol
ΣΔН0
t298 (reaction №3 и №4) = (-407,83) + (-163,94) = -571,77 кJ\mol
Conclusion:
By results of calculation of heat effects of reactions, it is visible, that energy deposited
at formation and an oxidizing of methane
ΣΔН0
t298 = -571,77 кJ\mol (reaction №2 и №3) is enough for the reaction flow №1
(recovery of dioxide of carbon ΣΔН0
t298 = 393,51 кJ\mol ).
That is, probably, recovery of dioxide of carbon goes through a transient formation of
methane. A necessary requirement for the yielded direction of response - is presence
of atomic hydrogen. Formation of the last occurs in the system under requirements
presented in the basic part (see above).
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