Standard enthalpy of formation of C8H18?

Question

For a particular isomer of C8H18 the combustion reaction produces 5093.7 kJ of heat per mole of C8H11(g) consumed, under standard conditions.

C8H18(g)+ 25/2O2(g) —> 8CO2(g) + 9H2O(g)

Delta H= - 5093 * 7kJ / mol

What is the standard enthalpy of formation of this isomer of C8H18(g)?

Answer 1

ΔH = n Σ ΔH^o products - nΣΔH^o reactants

Using this formula you can then calculate the ΔH^o formation of any product or reactant. In this question you are given the ΔH of the reaction, the ΔH^o of formation of the products can be found in your standard table

Δ H^o CO₂(g) = -393.5 kJ/mol ; Δ H^o H₂O(g) = -241.8 kJ/mol ;Δ H^o O₂(g) = 0 kJ/mol

Using the chemical equation C₈H₁₈ + 25/2 O₂(g) → 8 CO₂(g) + 9 H₂O (g) ΔH = -5093.7 kJ and the above general equation the ΔH⁰ of formation can be calculated

ΔH_rxn = n Σ ΔH^o products - nΣΔH^o reactants

-5093.7 kJ = [(8*-393.5 kJ) + (9 * -241.8 kJ)] - [ΔH_f (C₈H₁₈) + (25/2 * 0 kJ )

Rearranging the equation you can now solve for the ΔH_f C₈H₁₈

Answer +230.5 kJ!

Pay attention to the general equation given above!

Hope this helps!