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)?
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.
To find the standard enthalpy of formation (ΔHf) of a compound, you need to use the balanced equation for the combustion reaction and the given enthalpy change (ΔHrxn).
The standard enthalpy of formation (ΔHf) is the enthalpy change when one mole of a substance is formed from its elements in their standard states. In this case, we want to find the ΔHf of the isomer of C8H18.
The combustion reaction equation gives you the balanced equation for the formation of CO2 and H2O from C8H18.
C8H18(g) + 25/2 O2(g) ⟶ 8CO2(g) + 9H2O(g) ΔHrxn = -5113.3 kJ/mol
First, let's calculate the enthalpy change for the formation of one mole of C8H18 from its elements. To do this, we need to reverse the combustion reaction equation:
8CO2(g) + 9H2O(g) ⟶ C8H18(g) + 25/2 O2(g)
Since the enthalpy change for the reverse reaction is the negative of the given enthalpy change, we have:
ΔHf(C8H18) = -(-5113.3 kJ/mol) = 5113.3 kJ/mol
Therefore, the standard enthalpy of formation of this isomer of C8H18 is 5113.3 kJ/mol.