There’s a mistake in that part of TianDa version 1-69
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For the Tianjin University version 1-69, at 298.15 K, the standard Gibbs free energy change for the reaction H2(g) + O2(g) = H2O(l) is ΔGm = -237.13 KJ/mol, and ΔSm = -163.3 KJ/K·mol. Assuming that ΔCp is 0, when the reaction temperature rises to 398 K, the value of ΔGm(398) is ( ) A -237.13 KJ/mol B 237.13 KJ/mol C -220.80 KJ/mol D -253.51 KJ/mol. ΔGm(398) = ΔHm(398) – TΔSm(398) = ΔHm(298.15) + ∫ΔCp dT – TΔSm(298.15) + ∫ΔCp/T dT = -273.13 + (-163.3)*298.15 + 0 – 398*(-163) = 16086.375; there is no correct answer. I don’t know where I made the mistake; I hope someone experienced can give me some advice. @zlm322 @zshiwei2008 @tiyiss @zwp997
The thermodynamics part is the most challenging for me; please have an expert help answer it! @Lin Xuefeng
I checked it several times and compared it with the examples in the textbook; I don’t see any mistakes. The answer just doesn’t match.
dΔH/dT = ΔCp = 0, dΔS/dT = ΔCp/T = 0, and ΔG = ΔH – TΔS; therefore G2 = G1 = -237
This post was last edited by tianyalln on 2015-6-5 at 16:34. Is the answer A? When the temperature rises from 298.15 to 398, △Cp = 0; △S = ∫△Cp/T dT, and △H = ∫△Cp dT. Therefore, △S = 0 and △H = 0, so the change in Gibbs free energy is also 0. This is just my personal opinion; it’s for reference only
I also seek guidance from those who are more experienced. I calculated it several times as well, and the result is always 16086.375; there’s no answer, haha:D
I checked the textbook on physical chemistry (Tianjin University edition); the correct result should be 16086.375
This post was last edited by zwp997 on 2015-6-10 17:12. What an amazing calculation process – did the physical education teacher teach this?:lol
Those who know the subject can understand it at a glance, while those who don’t will find it incomprehensible.
This post was last edited by zwp997 on 2015-6-11 at 15:30. dDelta H/dT=Cp, not dDelta H/dT=Deta Cp; just as in the linear equation y=kx, (y/x)’=k is definitely not Detak. △Cp=0 means that the Cp value remains constant at any moment during the reaction process; this is a different concept from the fact that Cp1=Cp2 at the start and end of the process. Assuming the heating process is reversible: △Gm(398) = △Hm(398) – T△Sm(398) = △Hm(298.15) + ∫Cp dT – T∫Cp/T dT] = △Hm(298.15) = △Gm(298.15) + T△Sm(298.15)