Answer: The enthalpy of formation of
is -230.68 kJ/mol
Step-by-step explanation:
The chemical equation for the combustion of propane follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_(rxn)=[(8* \Delta H^o_f_((CO_2(g))))+(9* \Delta H^o_f_((H_2O(g))))]-[(1* \Delta H^o_f_{(C_8H_(18)(g))})+((25)/(2)* \Delta H^o_f_((O_2(g))))]](https://img.qammunity.org/2021/formulas/chemistry/high-school/wp0i9o4s25t14q61sh0ndoq3chczgkvpfa.png)
We are given:

Putting values in above equation, we get:
![-5093.7=[(8* (-393.5))+(9* (-241.82))]-[(1* \Delta H^o_f_{(C_8H_(18)(g))})+((25)/(2)* (0))](https://img.qammunity.org/2021/formulas/chemistry/high-school/vhmxon87i24pq3ku6bnblvm7lhw53m9je4.png)

The enthalpy of formation of
is -230.68 kJ/mol