Thermochemistry Help

Can someone please explain this problem and show their work? I don't understand it and my book has nothing in it so if someone could please show me how to do this, it would be greatly appreciated! Thanks!


Given:

H2(g) + ½ O2(g) ----> H2O(l) ΔH= -286 kJ

CO2(g) -----> C(s) + O2(g) ΔH= +394 kJ

2CO2(g) + H2O(l) -----> C2H2(g) + 5/2 O2(g) ΔH= 1300kJ

Find the ΔH for:

C2H2(g) ------> 2C(s) + H2(g)

Use Hess’s law.

You're going to have to flip equations and multiply them by factors, so as to be able to add the three equations together to get the final answer.

Flip equations:
1) You have to flip the third equation in order to move the C2H2 to the reactant side
2) The second equation is OK because the C is on the product side.
3) The first equation has to be flipped to the H2 will be n the product side.

I'm going to flip them, then discuss the multiplying.

H2O(l) ---> H2(g) + ½ O2(g) ΔH= +286 kJ
CO2(g) ---> C(s) + O2(g) ΔH= +394 kJ
C2H2(g) + 5/2 O2(g) ---> 2CO2(g) + H2O(l) ΔH= -1300kJ

Notice how I changed the signs of the ΔH for the two reactions I flipped.

When, I multiply, I'm looking to create the final equation as well as cancel out all items not in the final equation. The only multiply is to the second equation and it's by 2. Watch what happens to the ΔH.

H2O(l) ---> H2(g) + ½ O2(g) ΔH= +286 kJ
2CO2(g) ---> 2C(s) + 2O2(g) ΔH= +788 kJ
C2H2(g) + 5/2 O2(g) ---> 2CO2(g) + H2O(l) ΔH= -1300kJ

The H2O, O2 and CO2 all cancel when we add the three equations together. The ΔH is:

286 + 788 + (-1300) = -226 kJ

Lots more Hess' law here:

http://www.chemteam.info/Thermochem/Ther…

Just sum up the enthalpy change from each component in the reaction.

ΔH = ΔH(products) - ΔH(reactants)

There is one reactant and two products, so ΔH = 2ΔH(C) + ΔH(H2) - ΔH(C2H2), so:

ΔH(C2H2) = -1300 kJ since it has to be reversed
2ΔH(C) = 788 kJ
ΔH(H2) = 286 kJ

Hence, ΔH = 2ΔH(C) + ΔH(H2) - ΔH(C2H2)
ΔH = (788 kJ) + (286 kJ) - (1300 kJ)
ΔH = -226 kJ is the answer.