Answer to Question #189841 in General Chemistry for ijda

Question

C2H4 + 6F2 2CF4 + 4HF

Using the given equation

H2 + F2 2 HF ΔH= -573 KJ

2C + 2H2 C2H4 ΔH= + 52 KJ

CF4 C+ 2F2 ΔH= +680 KJ

Answer 1

 ∆H = -537 x2 kJ = -1074 kJ. 

 ∆H = +680 x 4 kJ = 2720 kJ.