• # question_answer 8) The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, $C{{O}_{2}}$(g) and${{H}_{2}}O(l)$ are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation,${{\Delta }_{f}}{{H}^{{}^\circ }}$of benzene. Standard enthalpies of formation of $C{{O}_{2}}$(g) and ${{H}_{2}}O(l)$are $-393.5kJ\,mo{{l}^{-1}}$ and$-285.83kJ\,mo{{l}^{-1}}$ respectively.

Information shadow: (i) ${{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)\to 6C{{O}_{2}}(g)+3{{H}_{2}}O(l);$ $\Delta H=-3267kJ$ (ii) $C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g);\,\,\,\,\Delta H=-393.5kJ$ (iii) ${{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(l)\to {{H}_{2}}O(l);\,\,\,\,\,\Delta H=-285.83kJ$ The required equation is: $6C(s)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(l);\,\,\,\,\,\,\,\,\Delta H=?$ Problem solving strategy: The heat of required equation can be obtained by algebraic method. (ii) $\times$ 6 + (iii) $\times$ 3 + (i) Working it out: $6C(s)+6{{O}_{2}}(g)\to 6C{{O}_{2}}(g);\Delta H=-393.5\times 6kJ$$3{{H}_{2}}(g)+\frac{3}{2}{{O}_{2}}(g)\to 3{{H}_{2}}(l);\,\,\,\,\Delta H=-285.83\times 3kJ$$6C{{O}_{2}}(g)+3{{H}_{2}}O(l)\to {{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g);\,$ $\Delta H=+3267kJ$ On adding,_____________________________ $6C(s)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(l);\,\,\,\,\,\,\Delta H=+48.51kJ\,mo{{l}^{-1}}$Alternatively, $6C(s)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(l)$ ${{\Delta }_{r}}H=\sum$Heat of combustion of reactants $-\sum$ Heat of combustion of products = 6 x (-393.5) + 3 (-285.83) - (- 3267) $=48.51kJ\,mo{{l}^{-1}}$