In: Chemistry
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)
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