Asked by marie
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!
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!
Answers
Answered by
DrBob222
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.
2C8H18 + 25O2 ==> 16CO2 + 18H2O so delta H for THIS reaction (2 moles that is) is 2*-547 kJ.
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