Respuesta :
Answer: -104.7 kJ
Explanation:
The chemical equation for the combustion of propane follows:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_3H_8(g))})+(5\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-2043.0kJ[/tex]
Putting values in above equation, we get:
[tex]-2043.0=[(3\times (-393.5))+(4\times (-241.8))]-[(1\times \Delta H^o_f_{(C_3H_8(g))})+(5\times (0))]\\\\\Delta H^o_f_{(C_3H_8(g))}=-104.7kJ/mol[/tex]
The enthalpy of formation of [tex]C_3H_8[/tex] is -104.7 kJ/mol
