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Question

Determine ΔU at 300 K for the following reaction using the listed enthalpies of reaction.

4CO(g)+8H2(g)3CH4(g)+CO2(g)+2H2O(l)

C(graphite)+12O2(g)CO(g);ΔH1=110.5kJ

CO(g)+12O2(g)CO2(g);ΔH2=282.9kJ

H2(g)+12O2(g)H2O(l);ΔH03=285.8kJ

C(graphite)+2H2(g)CH4(g);ΔH04=74.8kJ


A
653.5 kJ
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B
686.2 kJ
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C
747.4 kJ
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D
None of these
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Solution

The correct option is D None of these
ΔH=ΔU+ΔngRT
Reaction : 4CO(g)+8H2(g)3CH4(g)+CO2(g)+2H2o(l)
ΔrH=3ΔfHCH4+3ΔfHCo2+2ΔfHHo2(l)8ΔfHH2(s)4ΔfHcc
ΔfHCH4=7.8kJ
ΔfHco2=ΔfHco+ΔH2=393.4kJ
ΔfHH2o(l)=ΔH3=285.8kJ
ΔfH(s)=0 ΔfHco=ΔH,=110.5
ΔrH=3(74.8)+(393.4)+2(285.8)4(110.5)
=747.4kJ
now
ΔU=ΔHΔngRT Δng=8
=747.48×8.314×3001000
=767.35kJ

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