[DevaDevil]

3H2 (g) + CO (g)       -->       CH4 (g) + H2O (g) was given to figure out


2H2 + CO2 --> CH4  @ /\Hf 319.1kJ  the one you calculated

Definitely not the same, right?

And this is aside from the fact that the O2 doesn't cancel out (1O2 consumed in the first, 2 produced in the second)

Also the book says no diatomic gases (or elements) count towards heat formation and are considered zero. Is there any reason for this? and does this mean anything that is 'double' is negated down to 0?

it's the definition since those molecules are in their basic state. And for basic states: the heat of formation for any element in its basic state is arbitrarily set to 0 kJ / mole.

[Jabus]

Oh haha they end up being the same in the end, well that's one way to easily tell if it's wrong or not hah. Okay so:

C (s) + O2 (g)         -->       CO2 (g)                      -393.5 kJ
2CO (g)                     --> 2C (s) + O2 (g)               +221.0 kJ
2H2 (g) + O2 (g)         -->       2 H2O (g)               -483.6 kJ
CO2 (g) + 2 H2O (g)    -->       CH4 (g) + 2O2 (g)   +802.7 kJ
________________________________________
3H2 (g) + CO (g)       -->       CH4 (g) + H2O (g)


C (s) + O2 (g)         -->       CO2 (g)                      -393.5 kJ
CO (g)                     --> C (s) + 1/2O2 (g)               +221.0 kJ * .5
3H2 (g) + 1.5O2 (g)         -->       3 H2O (g)               -483.6 kJ * 1.5
CO2 (g) + 2 H2O (g)    -->       CH4 (g) + 2O2 (g)   +802.7 kJ
________________________________________
3H2 (g) + CO (g)       -->       CH4 (g) + H2O (g)

-->                                                      -393.5 kJ
CO (g)  -->                                           +221.0 kJ * .5
3H2 (g)          -->       H2O (g)               -483.6 kJ * 1.5
    -->       CH4 (g)                                 +802.7 kJ
________________________________________
3H2 (g) + CO (g)       -->       CH4 (g) + H2O (g)

3H2 + CO --> CH4 + H2O    /\Hf = -205.7kJ

I think that attempt made more sense to me. How's that?

[DevaDevil]

looks fine