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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 ) ?

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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

User Anjil Panchal
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