Use the thermochemical equations shown below to determine the enthalpy for the reaction:?

Question

(You may or may not need all of the reactions, #1-3.)


Eq. #4 4 CO2 (g) + 6 H2O (g) → 2 C2H6(g) + 7 O2(g)
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Eq. #1 C2H6(g) → C2H2(g) + 2 H2(g)
ΔH = 189 KJ

Eq. #2 C2H2(g) + 5/2 O2(g) => 2 CO2(g) + H2O(g)
ΔH = - 470 KJ

Eq. #3 H2O(g) => H2(g) + ½ O2(g)
ΔH = 142.5 KJ

Im totally lost. I hope you guys can help. Please explain how you arrive at the answer. I'd like to KNOW how to do this, rather than just simply have the answer. THANKS

Answer 1

Eq#1 4H2(g) + 2C2H2(g) ===> 2C2H6(g)
DH= 2(-189kJ) = -378kJ

Eq#2 2H2O(g) + 4CO2(g) ===> 2C2H2(g) + 5O2(g)
DH=2(470) = +540kJ

Eq#3 4H2O(g) ===> 4H2(g) + 2O2
DH=4(142.5) = +570kJ

Add them all up: +732kJ

When writing an equation forward, keep the amount of kJ the same and the sign.

When writing an equation backwards, reverse the sign.

When doubling or tripling an equation, double or triple the amount of kJ.