nent submission s assignment, you submit answers by question parts The number of

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nent submission s assignment, you submit answers by question parts The number of submissions remaining for each question part only dangestyou dangete ment Scoring ast submission is used for your score. H16 points ZumChemPe E.033. Calculate the energy required to heat 1.30 kg ethane gas (czHe) from 29.0 c to 74.0 cfrst under conditions of constant volume and then at a below for of relevant data.) Molar Heat Capacites of Various Gases at 298 K K mol K mol He, Ne, Ar 12.47 20.80 8.33 8.32 20.54 8.32 29.03 20.71 8.32 30.38 N2O 3 .27 28.95 8.32 52.92 44.60 C2He constant constant P Need Help? Road L it Answer Save Progress Practice Another Version

Explanation / Answer

(1)

Ethane gas

m=1.30 kg

n=1.30/30=0.0433 kmol

At constant Volume

U=nCvT=0.0433 kmol * 44.60 J/kmol*(74-29)=86.9031 J

Q=U=86.9031 J

W=0

H=nCpT=0.0433*52.92*(74-29)=103.11 J

At constant pressure

H=nCpT=0.0433*52.92*(74-29)=103.11 J

Q=H=103.11 J

U=nCvT=0.0433 kmol * 44.60 J/kmol*(74-29)=86.9031 J

U=Q-W

W=Q-U=103.11-86.9031=16.2069 J

(2)

Fe2O3(s)+3C(graphite)=2Fe(s)+3CO(g)

H=(2*Hfe+3*HCO-3*HC-HFe2O3)

H=(2*0+3*(110.54)-3*0-(824.25))=492.63 kJ/mol

(3)

C2H6(OH)2+H2O2----->C2H4O2+2H2O(l)

Here set of four reaction is there.So adding these reaction we get aove reaction.

So

reaction(1)-reaction(2)(We need backward reaction)+2*(reaction 3)+reaction(4)

H=177.4-(-191.2)+2*(-241.8)-43.8=-158.8 kJ

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