Question

For a particular isomer of C 8 H 18 , the combustion reaction produces 5093.7 kJ of heat per mole of C 8 H 18 ( g ) consumed, under standard conditions. C 8 H 18 ( g ) + 25 2 O 2 ( g ) ⟶ 8 CO 2 ( g ) + 9 H 2 O ( g ) Δ H ∘ rxn = − 5093.7 kJ / mol What is the standard enthalpy of formation of this isomer of C 8 H 18 ( g ) ?

Answer 1

Answer: The enthalpy of formation of
`C_8H_(18)` is -230.68 kJ/mol

Step-by-step explanation:

The chemical equation for the combustion of propane follows:

`C_8H_(18)(g)+(25)/(2)O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)`

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))))]`

We are given:

`\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`

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))`

`\Delta H^o_f_{(C_8H_(18)(g))}=-230.68kJ`

The enthalpy of formation of
`C_8H_(18)` is -230.68 kJ/mol