Gibbs free energy of reaction, water [translation unavailable] Hydrogen peroxide (thermo-40) Es reacting H2(g) with O2(g). Assessments with help the free reaction enthalpy whether or not well, you know, H2O(g) or H2O2(g) of which: is. The reaction temperature is 800 Kelvin.

  Exercise: thermo-40    

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Es reacting H₂(g) with O₂(g). Assessments with help the free reaction enthalpy whether or not well, you know, H₂O(g) or H₂O₂(g) of which: is. The reaction temperature is 800 Kelvin.

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Solution:

Es to leave itself two reactions to formulate:
(a) H₂(g) + 1/2·O₂(g) → H₂O(g) and
b) H₂(g) + O₂(g) → H₂O₂(g)
Note: In order to compare the results with each other, one must look at the same number and H₂O respectively. H₂O₂ related!

Table of values showing that all the particles involved are in a gaseous state:

	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), H₂O is produced

       [translation unavailable]: 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)

       [translation unavailable]: ΔGoR = ΔHoR – T· ΔSoR

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

2. reaction b), H₂O₂ is produced

       [translation unavailable]: ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)

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

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

       [translation unavailable]: ΔGoR = ΔHoR – T· ΔSoR

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

The calculation shows that although both reactions would have to be exergonic ( ΔGoR < 0) and therefore voluntary, reaction a) is almost a factor 4 more exergonic and is therefore preferred.

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