Find the enthalpy using:
Target Reaction: 2 CO2(g) + H2O(g)→C2H2 (g) + 5/2 O2 (g)
Step Reactions: C2H2 (g) + 2H2 (g) → C2H6(g) ΔH = -94.5 kJ
H2O (g) → H2(g) + 1/2O2(g) ΔH = 71.2kJ
2C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O(g) ΔH =-566 kJ
Never mind I figured it out
target: 2CO2(g) + H2O(g )→ C2H2 (g) + 5/2 O2 (g)
This gets too complicated to try to do it on the computer. Here are the directions for you to follow.
When I say multiply the equation by 3 you also must multiply the dH value. When I say reverse the equation you do that BUT you ALSO change the sign of dH. Here are the directions.
Reverse eqn 1 and add to the reverse of 3 times equation 2 and add to the reverse of equation 3 divided by 2 then add to equation 2 (yes, two separate equations 2 one is reversed and the other is not) as is. Keep up with the dH values and add them (remembering to change signs appropriately. Check your work and you SHOULD end up with the target equation and the sum of the dH values will be the dH value for the target reaction.