free of charge reaction enthalpy, Water vs Hydrogen peroxide (thermo-40) Es reacting H2(g) with O2(g). Assessments with Help The free reaction enthalpy whether or not Now H2O(g) or H2O2(g) of which: will be. The Reaction temperature of which: 800 Kelvin , please ..

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Es reacting H₂(g) with O₂(g). Assessments with Help The free reaction enthalpy whether or not Now H₂O(g) or H₂O₂(g) of which: will be. The Reaction temperature of which: 800 Kelvin , please ..

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Lösung:

Es to leave The Commission two reactions to formulate:
a) H₂(g) + 1/2·O₂(g) → H₂O(g) and
b) H₂(g) + O₂(g) → H₂O₂(g)
Respect: With this The Results with each other Compared will be can must It's The Commission on The equal to Number of Education and training H₂O Responsibility. H₂O₂ Related!

Table of values, All of them Participants Particles are In the Gaseous State of operation:

	ΔHof(kJ·mol-1)	So(J·K-1·mol-1)
H2 (g)	0	131
O2 (g)	0	205
H2O (g)	-242	189
H2O2 (g)	-136	233

1. reaction a), It 's occurring H₂O

       Definition: ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)

       ΔHoR(H₂O) = (-242) – (0 + 1/2·0) = -242 kJ/mol

       ΔSoR(H₂O) = (189) – (131 + 1/2·205) = -44.5 J/(mol·K)

       Definition: ΔGoR = ΔHoR – T· ΔSoR

       ΔGoR(H₂O) = -242 – 800·-0.0445 = -206.4 kJ/mol

2. reaction b), It 's occurring H₂O₂

       Definition: ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)

       ΔHoR(H₂O₂) = (-136) – (0 + 1/2·0) = -136 kJ/mol

       ΔSoR(H₂O₂) = (233) – (131 + 205) = -103 J/(mol·K)

       Definition: ΔGoR = ΔHoR – T· ΔSoR

       ΔGoR(H₂O₂) = -136 – 800·-0.103 = -53.6 kJ/mol

The calculation shows, that However, both reactions Exergonically ( ΔGoR < 0) and This means that Voluntary running I have to.. reaction a) is But almost to One Factor 4 Stronger Exergonically and will be This means that Preferably running.

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