Question

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) ΔH°rxn = –1.0940 × 104 kJ/mol What is the standard enthalpy of formation of liquid octane? ΔH°f(CO2(g)) = –393.5 kJ/mol and ΔH°f(H2O(l)) = –285.8 kJ/mol

Answer 1

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Step-by-step explanation:

The given balanced chemical reaction is,

`2C_8H_(18)(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)`

First we have to calculate the enthalpy of reaction
`(\Delta H^o)`.

`\Delta H^o=H_f_(product)-H_f_(reactant)`

`\Delta H^o=[n_(O_2)* \Delta H_f^0_((O_2))+n_(H_2O)* \Delta H_f^0_((H_2O))]-[n_{C_8H_(18)}* \Delta H_f^0_{(C_8H_(18))+n_(O_2)* \Delta H_f^0_((O_2))]`

where,

We are given:

`\Delta H^o_f_((CO_2(g)))=-393.5kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_{(C_8H_(18)(l))}=?kJ/mol\\\Delta H^o_f_((H_2O(l)))=-285.8kJ/mol`

Putting values in above equation, we get:

`-1.0940* 10^4=[(16* -393.5)+(18* -285.8)]-[(25* 0)+(2* \Delat H_f{C_8H_(18)(l)}]`

`\Delta H^o_f_{(C_8H_(18)(l))}=-250.2kJ/mol`

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol