Q.The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, $CO_2(g)$ and $H_2O(l)$ are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, $\Delta_f H^\ominus$ of benzene. Standard enthalpies of formation of $CO_2(g)$ and $H_2O(l)$ are –393.5 kJ mol$^{-1}$ and –285.83 kJ mol$^{-1}$ respectively.

$C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l)$

$\Delta_c H^\ominus = -3267.0 \text{ kJ mol}^{-1}$

$\Delta_c H^\ominus = \sum \nu \Delta_f H^\ominus(\text{products}) - \sum \nu \Delta_f H^\ominus(\text{reactants})$

$-3267.0 = [6(-393.5) + 3(-285.83)] - [\Delta_f H^\ominus(C_6H_6) + \frac{15}{2}(0)]$

$\text{C}_6\text{H}_6(l) + \tfrac{15}{2}\,\text{O}_2(g) \rightarrow 6\,\text{CO}_2(g) + 3\,\text{H}_2\text{O}(l),\qquad \Delta_c H^\ominus = -3267.0\ \text{kJ mol}^{-1}.$

$\Delta_c H^\ominus = \big[6\,\Delta_f H^\ominus(\text{CO}_2) + 3\,\Delta_f H^\ominus(\text{H}_2\text{O})\big] - \Delta_f H^\ominus(\text{C}_6\text{H}_6),$

$6\,\Delta_f H^\ominus(\text{CO}_2) = 6 \times (-393.5) = -2361.0\ \text{kJ},$

$3\,\Delta_f H^\ominus(\text{H}_2\text{O}) = 3 \times (-285.83) = -857.49\ \text{kJ},$

$\text{sum over products} = -2361.0 + (-857.49) = -3218.49\ \text{kJ}.$