Yes, you have the right formula. Substitute the delta H values you have listed in the problem. The equation is
2C8H18 + 25O2 ==> 16CO2 + 18H2O so delta H for THIS reaction (2 moles that is) is 2*-547 kJ.
please help me? :)
topic: standard enthalpies of formation
"Calculate ΔHf° of octane, C8H18(l), given the entalpy of combustion of octane to CO2(g) and H2O(l) is -5471kJ/mol. The standard enthalpies of formation ofCO2 and H2O are given: CO2(g)ΔHf°=-393.5kJ/mol and H2O(l) ΔHf°=-285.8kJ/mol"
I'm not really sure how to go about this... do I use ΔHrxn°= (sum)nΔHf°prod - (sum)mΔHf°reactant ? Also, can you go through this with me step by step if need be? Thank you for your time!
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