Question

Given the following equations, calculate the ΔH of reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas. Show your work.

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
I dont know where to even start this is so confusing.

Answer 1

you need to first write (and balance) the equation the question is asking for,

"reaction of ethylene gas (C2H4) with fluorine gas to make carbon tetrafluoride (CF4) gas and hydrogen fluoride gas",

then rearrange all equations given so that you have a "net" equation that looks like the one that is being asked for.

for example, if you want: a -> c

but you have: (1) 2a -> 2b ΔH=x
(2) c -> 2b ΔH=y

divide (1) by 2, and reverse (2):

(1) a -> 2b \(\overbrace{\color{red}{ΔH=\dfrac{x}{2}}}^{note~it~was~divided}\)
(2) 2b -> c \(\underbrace{\color{red}{ΔH=-y}}_{note~the~sign~was~reversed}\)


Add them up and treat the arrow like an equal sign:
a + 2b -> c + 2b

cancel out things you have on both sides:
a + \(\cancel{2b}\) -> c + \(\cancel{2b}\)

net equation: a -> c

Add up their individual enthalpies:

\(ΔH_{reaction}=\dfrac{x}{2}+(-y)\)