Respuesta :
Answer: The enthalpy of formation of [tex]C_8H_{18}[/tex] is -230.68 kJ/mol
Explanation:
The chemical equation for the combustion of propane follows:
[tex]C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{(CO_2(g))})+(9\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-5093.7kJ[/tex]
Putting values in above equation, we get:
[tex]-5093.7=[(8\times (-393.5))+(9\times (-241.82))]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times (0))[/tex]
[tex]\Delta H^o_f_{(C_8H_{18}(g))}=-230.68kJ[/tex]
The enthalpy of formation of [tex]C_8H_{18}[/tex] is -230.68 kJ/mol
