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.

(N/A) The formation reaction of benzene is given by:
$6 \, C(graphite) + 3 \, H_{2(g)} \rightarrow C_6H_{6(l)}; \Delta_f H^{\ominus} = ? \dots (i)$
The enthalpy of combustion of $1 \, mol$ of benzene is:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)}$ $\rightarrow 6 \, CO_{2(g)} + 3 \, H_2O_{(l)}; \Delta_c H^{\ominus} = -3267.0 \, kJ \, mol^{-1} \dots (ii)$
The enthalpy of formation of $1 \, mol$ of $CO_{2(g)}$ is:
$C(graphite) + O_{2(g)} \rightarrow CO_{2(g)}; \Delta_f H^{\ominus} = -393.5 \, kJ \, mol^{-1} \dots (iii)$
The enthalpy of formation of $1 \, mol$ of $H_2O_{(l)}$ is:
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}; \Delta_f H^{\ominus} = -285.83 \, kJ \, mol^{-1} \dots (iv)$
Using Hess's Law,the enthalpy of combustion is given by:
$\Delta_c H^{\ominus} = [6 \times \Delta_f H^{\ominus}(CO_{2(g)}) + 3 \times \Delta_f H^{\ominus}(H_2O_{(l)})] - [\Delta_f H^{\ominus}(C_6H_{6(l)}) + \frac{15}{2} \Delta_f H^{\ominus}(O_{2(g)})]$
Since $\Delta_f H^{\ominus}(O_{2(g)}) = 0$,we have:
$-3267.0 = [6 \times (-393.5) + 3 \times (-285.83)] - \Delta_f H^{\ominus}(C_6H_{6(l)})$
$-3267.0 = [-2361.0 - 857.49] - \Delta_f H^{\ominus}(C_6H_{6(l)})$
$-3267.0 = -3218.49 - \Delta_f H^{\ominus}(C_6H_{6(l)})$
$\Delta_f H^{\ominus}(C_6H_{6(l)}) = -3218.49 + 3267.0 = 48.51 \, kJ \, mol^{-1}$