Answer:
the standard enthalpy of formation of this isomer of C₈H₁₈ (g) = -375 kj/mol
Explanation:
The given combustion reaction
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O ............................(1)
Heat of reaction or enthalpy of combustion = -5099.5 kj/mol
from equation (1)
ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation of reactants)
Or, - 5099.5 = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)].................................(2)
Given ΔH⁰f(CO₂) = - 393.5 kj/mol & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0
Using equation (2)
ΔH⁰f(C₈H₁₈) = -621 kj/mol