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

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 $(\Delta H^o)$.

$\Delta H^o=H_f_{product}-H_f_{reactant}$

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

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:

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

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

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