Asked by Anonymous
For a particular isomer of C8H18, the combustion reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standard conditions.
C8H18(g) + 25/2 O2(g) ⟶ 8CO2(g)+9H2O (g) Δ𝐻rxn=−5113.3 kJ/mol
What is the standard enthalpy of formation of this isomer of C8H18(g)?
C8H18(g) + 25/2 O2(g) ⟶ 8CO2(g)+9H2O (g) Δ𝐻rxn=−5113.3 kJ/mol
What is the standard enthalpy of formation of this isomer of C8H18(g)?
Answers
Answered by
DrBob222
dHo formation for the isomer = (n*dHo products) - (n*dHo reactants)
Look up dHo standard in tables for CO2 and H2O. That of O2 is zero. Plug those numbers into the above equation I gave you and solve for dHo
formation. Post your work if you get stuck.
Look up dHo standard in tables for CO2 and H2O. That of O2 is zero. Plug those numbers into the above equation I gave you and solve for dHo
formation. Post your work if you get stuck.
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