Question

The standard enthalpy change for the combustion of 1 mole of propane is –2043.0 kJ. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Calculate ΔfH° for propane based on the following standard molar enthalpies of formation. molecule ΔfH° (kJ/mol-rxn) CO2(g) –393.5 H2O(g) –241.8'

Answer 1

Answer: -104.7 kJ

Step-by-step explanation:

The chemical equation for the combustion of propane follows:

`C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(3* \Delta H^o_f_((CO_2(g))))+(4* \Delta H^o_f_((H_2O(g))))]-[(1* \Delta H^o_f_((C_3H_8(g))))+(5* \Delta H^o_f_((O_2(g))))]`

We are given:

`\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`

Putting values in above equation, we get:

`-2043.0=[(3* (-393.5))+(4* (-241.8))]-[(1* \Delta H^o_f_((C_3H_8(g))))+(5* (0))]\\\\\Delta H^o_f_((C_3H_8(g)))=-104.7kJ/mol`

The enthalpy of formation of
`C_3H_8` is -104.7 kJ/mol