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From the enthalpies of reaction H2(g)+F2(g)→2HF(g)ΔH=−537kJ C(s)+2F2(g)→CF4(g)ΔH=−680kJ 2C(s)+2H2(g)→C2H4(g)ΔH=+52.3kJ calculate ΔH for the reaction of ethylene with F2: C2H4(g)+6F2(g)→2CF4(g)+4HF(g)

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Chemistry

From the enthalpies of reaction H2(g)+F2(g)→2HF(g)ΔH=−537kJC(s)+2F2(g)→CF4(g)ΔH=−680kJ 2C(s)+2H2(g)→C2H4(g)ΔH=+52.3kJcalculate ΔH for the reaction of ethylene with F2:C2H4(g)+6F2(g)→2CF4(g)+4HF(g)

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4.0 Ratings (607 Votes)

H2(g)+F2(g)?2HF(g)?H=?537kJ

C(s)+2F2(g)?CF4(g)?H=?680kJ 2

C(s)+2H2(g)?C2H4(g)?H=+52.3kJ

2H2(g)+2F2(g)?4HF(g)    ?H=?1074kJ

2C(s)+4F2(g)?2CF4(g)    ?H=?1360kJ

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2C(s) + 2H2(g) +6F2(g) --------> 2CF4(g) + 4HF(g) ?H = -2434Kj

2C(s)+2H2(g )          ?C2H4(g)                                     ?H=+52.3kJ      substract

(-)        (-)                   (-)                                                        (-)

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C2H4(g)+6F2(g)?2CF4(g)+4HF(g)                           ?H    = -2486
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