Data:
2Fe(s) + O2(g) => 2FeO(s) ΔH = -545.6 kJ
4Fe(s) + 3O2(g) => 2Fe2O3(s) ΔH = -1,643.6 kJ
3Fe(s) + 2O2(g) => Fe3O4(s) ΔH = -1,117.7 kJ
Given the data above, determine the heat of reaction, ΔH, for the reaction:
2 Fe3O4(s) => 2 Fe2O3(s) + 2 FeO(s)
I didn't write any of this on a sheet of paper but you should do so. I may have made an error trying to do it in my head.
Reverse eqn 3 and multiply by 2. Change the sign of dH.
Add in equation 1 as is.
Add in equation 2 as is.
The O2 will be on opposite sides and will cancel.
If you reverse a reaction be sure to change the sign of delta H.
If you add in a reaction leave delta H as is.
The sum of dH for eqn 1 and 2 and 3 (with the reversed sign) will be the new dH for the new reaction.