H2 (g) + F2 (g) → 2HF (g) ΔH = –537 kJ
C (s) + 2F2 (g) → CF4 (g) ΔH = –680 kJ
2C (s) + 2H2 (g) → C2H4 (g) ΔH = +52.3 kJ
+36916 First balance the equation
C2H4---> 2C+ 2H2
C2H4 + 6 F2 ----------------------------------> 2 CF4 + 4 HF
Note that we got 4 HF's Given 2 HF's = - 537kj - 537 x 2
we got 2 CF4 Given 1 CF4 = - 680 kj - 680 x 2
___________
deltaH = - 2434kj
But I will assume the intermediate reaction occurs too:
C2H4 -----> 2C + 2H2 we have ONE of these reactions for - 52.3 kj
deltaH becomes - 2434 - 52.3 kj = -2486.3 kj
