C4H10(g) + (13/2) O2(g) -> 4CO2(g) + 5H2O(g) ΔH°rxn = –2613 kJ/mol
Given the following enthalpies of formation:
ΔH°f[CO2(g)] = -393.5 kJ/mol
ΔH°f[H2O(g)] = -241.8 kJ/mol
Calculate the value of enthalpy of formation of butane from the data given.
dHrxn = [(n*dHfo CO2 + (n*dHfo H2O]-[n*dHfo C4H10]
You know all of the numbers except for dHo C4H10. Substitute and solve for that.
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